Chemical equilibrium

Class 11 Chemistry, unit 8, book pages 166–187

Practise 61 MCQs Take the unit quiz Review 32 cards

Many reactions do not go to completion but are reversible, and they settle into a dynamic equilibrium in which the forward and reverse rates are equal and the concentrations stay constant. This unit uses the law of mass action to write and calculate equilibrium constants ( $K_{c}$ , $K_{p}$ , $K_{n}$ , $K_{x}$ ), distinguishes homogeneous and heterogeneous equilibria, and shows how to recognise equilibrium. It then uses Le Chatelier's principle to predict the effect of concentration, pressure, temperature and a catalyst, and applies it to Haber's process and the contact process.

Learning outcomes (14)

Describe what is meant by a reversible reaction and dynamic equilibrium in terms of the rate of forward and reverse reactions being equal and the concentration of reactants and products remaining constant.
Define dynamic equilibrium between two physical states.
State the necessary conditions for equilibrium and the ways that equilibrium can be recognized.
Describe the microscopic events that occur when a chemical system is in equilibrium. Define with examples.
Deduce the equilibrium constant expression $[K_{c}]$ from an equation for a homogeneous reaction.
Determine the relationship between different equilibrium constants ( $K_{c}$ ) for the same reaction at the same temperature.
Write the equilibrium expression for a given chemical reaction in terms of concentration, $K_{c}$ , partial pressure, number of moles and mole fraction.
Differentiate between microscopic and macroscopic events in a chemical reaction.
Propose microscopic events that account for observed macroscopic changes that take place during a shift in equilibrium.
Determine if the equilibrium constant will increase or decrease when temperature is changed, given the equation for the reaction.
State Le Chatelier's principle and be able to apply it to systems in equilibrium with changes in concentration, pressure, temperature, or the addition of a catalyst.
Explain industrial applications of Le Chatelier's principle using Haber's process and the contact process as an example.
Discuss the industrial applications of chemical equilibria and how they can be used to optimize chemical reactions to maximize yields and minimize waste products.
Use the concept of hydrolysis to explain why aqueous solutions of some salts are acidic or basic.

8.1Macroscopic events and microscopic events

Some chemical reactions go on until the limiting reactant is used up completely. Their reactants are completely consumed and converted into products, and such reactions are called irreversible reactions. In many other reactions, however, the net formation of products stops before all the limiting reactant is consumed. These reactions really occur in both directions, forward and backward, and are called reversible reactions.

Such a reaction reaches a stage called chemical equilibrium. At this point the concentrations of reactants and products become constant. The reaction itself does not stop: it continues in both directions without changing the concentrations under the existing conditions. A reversible reaction therefore never stops.

Irreversible and reversible reactions compared
Point	Irreversible reaction	Reversible reaction
Direction	One direction only (forward)	Both directions, forward and backward
Extent	Goes on until the limiting reactant is used up	Net formation of products ends before the limiting reactant is used up
End state	No reactant is left (the limiting one)	Chemical equilibrium: constant concentrations of reactants and products
Does the reaction stop?	Yes, when the limiting reactant is finished	No, it never stops

Macroscopic events

Macroscopic events: Phenomena that can be observed with the naked eye without considering the individual particles or molecules involved in the process.

Examples: a change in colour, the evolution or absorption of heat, the formation of a precipitate, a change in volume or pressure, and a change in the composition of a substance in a chemical reaction.

Microscopic events

Microscopic events: Phenomena that cannot be observed with the naked eye.

Examples: collisions between molecules, breaking and forming of bonds, rearrangement of atoms in molecules, and loss or gain of electrons.

Macroscopic events are the result of many microscopic events happening at the same time. Understanding the microscopic events helps us to explain and predict the macroscopic changes seen in an equilibrium system.

When a change in equilibrium occurs in a chemical reaction, the microscopic events that explain the observed change are the collisions and the formation of new bonds between particles or molecules. These collisions change the rates of the forward and reverse reactions, and the rates are affected by the activation energy and by external factors.

Macroscopic and microscopic events compared
Point	Macroscopic events	Microscopic events
Can be seen?	Yes, with the naked eye	No, they cannot be observed with the naked eye
Concerned with	The whole sample: colour, heat, precipitate, volume, pressure, composition	Individual particles: collisions, bond breaking and forming, rearrangement of atoms, electron transfer
Relation	The result of many microscopic events at once	The cause that explains the macroscopic change

8.2Reversible reactions and dynamic equilibriumOften examined

Reversible reaction: A reaction in which the products, once formed, can react to form the reactants again. It does not go to completion even if stoichiometric amounts of the reactants are taken, and it takes place in both the forward and the backward directions under the existing conditions.

Some examples of reversible reactions are:

N X_{2} (g) + 3 H X_{2} (g) 2 NH X_{3} (g)

2 NO X_{2} (g) N X_{2} O X_{4} (g)

2 NO (g) + Cl X_{2} (g) 2 NOCl (g)

PCl X_{5} (g) PCl X_{3} (g) + Cl X_{2} (g)

Approach to equilibrium: steam and carbon monoxide

Consider the reaction between steam and carbon monoxide under suitable conditions. When they are mixed, macroscopic changes such as changes in concentration are observed. Suppose the reaction is started with the same number of moles of both reactants.

How equilibrium is reached

At the start, the number of collisions per second between $H X_{2} O$ and $CO$ molecules is at its maximum. So the forward reaction has its maximum speed at the beginning, and the concentrations of the reactants decrease.
$H X_{2} O (g) + CO (g) H X_{2} (g) + CO X_{2} (g)$
As $H X_{2} O$ and $CO$ are gradually used up, fewer collisions occur per second and the forward reaction slows down gradually.
As $H X_{2}$ and $CO X_{2}$ molecules accumulate, the reverse reaction also starts. Their concentrations rise, so more and more collisions per second occur between them and the reverse reaction speeds up. It starts at zero speed and increases gradually.
$H X_{2} (g) + CO X_{2} (g) H X_{2} O (g) + CO (g)$
Eventually a time comes when both reactions proceed at the same speed. The reaction is then said to be in chemical equilibrium, and the concentrations of reactants and products become constant.
$H X_{2} O (g) + CO (g) H X_{2} (g) + CO X_{2} (g)$

State of chemical equilibrium: The state of a reversible reaction at which the composition of the reaction mixture does not change. Unless the system is somehow disturbed, no further change in concentration occurs.

Rates of the forward and reverse reactions

0.2

0.4

0.6

0.8

time (s)

rate

equilibrium

forward

reverse

The forward rate falls from its maximum and the reverse rate rises from zero. At equilibrium the two rates are equal and stay constant.

Figure 8.1

t_{e}

0.2

0.4

0.6

0.8

time

conc.

t_{e}

[H X_{2} O], [CO]

[H X_{2}], [CO X_{2}]

Figure 8.1: The plots of concentration of reactants and products versus time. The reactant curve falls and the product curve rises. After the time

t_{e}

both plots are parallel to the time axis (the concentrations no longer change).

What do the plots of Fig. 8.1 show? Why have they become parallel? The concentrations of the reactants ( $H X_{2} O$ and $CO$ ) fall and the concentrations of the products ( $H X_{2}$ and $CO X_{2}$ ) rise until the time $t_{e}$ . After that the plots are parallel to the time axis, because the composition of the mixture no longer changes: the forward and reverse reactions are going at the same speed, so each substance is formed as fast as it is used up.

Why equilibrium is dynamic

When the concentrations of reactants and products become constant, the reaction may appear to have stopped. This is not true. At the microscopic level there is intense activity. Individual reactant molecules continue to combine, and individual product molecules also continue to combine, but the speed of one process is exactly balanced by the speed of the other. Therefore it is called a dynamic equilibrium.

Dynamic equilibrium: The state in which the individual molecules are constantly reacting but the rates of the forward and reverse reactions are the same, so the concentrations stay constant.

Features of a dynamic equilibrium
Feature	At equilibrium
Rates	Rate of forward reaction = rate of reverse reaction (balanced)
Concentrations	Constant, because the system is not disturbed
Macroscopic view	No change seen: the reaction seems to have stopped
Microscopic view	Molecules keep reacting in both directions, so the system is dynamic

8.3The law of mass actionOften examined

The law of mass action was proposed in 1864 by the two chemists C. M. Guldberg and P. Waage as a general description of the equilibrium state.

Law of mass action: The rate at which a substance reacts is proportional to its active mass, and the rate of a chemical reaction is proportional to the product of the active masses of the reacting substances. It can also be stated as: the rate of a chemical reaction is proportional to the molar concentration of each reacting substance raised to a power equal to its stoichiometric coefficient in the balanced chemical equation.
Active mass: The concentration of the reactants and products in $mol d m^{- 3}$ for a dilute solution.

Derivation The equilibrium constant expression $K_{c}$

Aim. Use the law of mass action to obtain the expression for the equilibrium constant of a general reversible reaction.

Assumptions.

The reaction is the general reversible reaction $a A (g) + b B (g) c C (g) + d D (g)$ , where A, B, C and D are chemical species and $a$ , $b$ , $c$ and $d$ are their coefficients in the balanced equation.
The temperature is constant, so the rate constants $k_{f}$ and $k_{r}$ do not change.

Apply the law of mass action to the forward reaction. Its rate is proportional to the concentration of each reactant raised to its coefficient. Introducing the rate constant $k_{f}$ of the forward reaction turns the proportionality into an equation.
$R_{f} \propto [A]^{a} [B]^{b} \Rightarrow R_{f} = k_{f} [A]^{a} [B]^{b} (1)$
Apply it in the same way to the reverse reaction, in which C and D are the reactants, with $k_{r}$ the rate constant for the reverse reaction.
$R_{r} \propto [C]^{c} [D]^{d} \Rightarrow R_{r} = k_{r} [C]^{c} [D]^{d} (2)$
At the equilibrium state the rate of the forward reaction equals the rate of the reverse reaction.
$R_{f} = R_{r} \Rightarrow k_{f} [A]^{a} [B]^{b} = k_{r} [C]^{c} [D]^{d}$
Rearrange so that the two rate constants are on one side and the concentrations on the other.
$\frac{k _{f}}{k _{r}} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}$
The ratio of two constants is itself a constant at a given temperature. It is called the equilibrium constant $K_{c}$ .
$K_{c} = \frac{k _{f}}{k _{r}} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}} (3)$

Result

K_{c} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}

$K_{c}$ is the equilibrium constant and equation (3) is the equilibrium constant expression. The square brackets show the concentrations of the chemical species at equilibrium. The expression for any reaction can be written from its balanced equation: the concentrations of the products go in the numerator and those of the reactants in the denominator.

Conditions for equilibrium

The important features of the equilibrium constant expression are as follows.

$K_{c}$ applies only at the equilibrium state. The subscript c indicates the concentration of reactants and products in $mol d m^{- 3}$ at the equilibrium state.
$K_{c}$ does not depend on the initial concentrations of the reactants and products but depends on temperature. It has only one value at a given temperature. Whether we start the reaction with pure reactants, pure products or any composition in between, the value of $K_{c}$ remains the same.
$K_{c}$ is related to the coefficients of the balanced chemical equation. The concentrations of the products are in the numerator and those of the reactants in the denominator, and each concentration is raised to a power equal to its coefficient in the balanced equation.
The magnitude of $K_{c}$ indicates the position of equilibrium. When $K_{c} < 1$ the denominator is greater than the numerator, so the concentrations of the reactants are greater than those of the products at equilibrium. When $K_{c} > 1$ the numerator is greater, so the concentrations of the products are greater than those of the reactants at equilibrium.

Magnitude of $K_{c}$ and the position of equilibrium
Value of $K_{c}$	What is greater at equilibrium	Position of equilibrium
$K_{c} < 1$	Reactants (denominator greater)	Lies towards the reactants
$K_{c} > 1$	Products (numerator greater)	Lies towards the products

Examples of equilibrium constant expressions

Problem solving strategy

Write the products in the numerator and the reactants in the denominator, within square brackets.
Raise each concentration to the power that corresponds to the coefficient of that species in the balanced chemical equation.

Example 8.1

Write the

K_{c}

expression for the reaction

N X_{2} (g) + 3 H X_{2} (g) 2 NH X_{3} (g)

Put the product $NH X_{3}$ in the numerator. Its coefficient is 2, so its concentration is squared.
$[NH X_{3}]^{2}$
Put the reactants in the denominator. The coefficients are 3 for $H X_{2}$ and 1 for $N X_{2}$ (a power of 1 is not written).
$[H X_{2}]^{3} [N X_{2}]$
Combine the two parts.
$K_{c} = \frac{[ NH X _{3} ] ^{2}}{[ H X _{2} ] ^{3} [ N X _{2} ]}$

K_{c} = \frac{[ NH X _{3} ] ^{2}}{[ H X _{2} ] ^{3} [ N X _{2} ]}

Example 8.2

Write the

K_{c}

expression for the reaction

CO (g) + \frac{1}{2} O X_{2} (g) CO X_{2} (g)

The product is $CO X_{2}$ with coefficient 1, so it goes in the numerator to the power 1.
$[CO X_{2}]$
The reactants are $CO$ (coefficient 1) and $O X_{2}$ (coefficient $\frac{1}{2}$ ). The power of $[O X_{2}]$ is $\frac{1}{2}$ , the coefficient itself.
$[CO] [O X_{2}]^{1/2}$
Combine the two parts.
$K_{c} = \frac{[ CO X _{2} ]}{[ CO ] [ O X _{2} ] ^{1/2}}$

K_{c} = \frac{[ CO X _{2} ]}{[ CO ] [ O X _{2} ] ^{1/2}}

Concept assessment exercise 8.1

1. The following equations represent various industrial reactions at equilibrium. Write the $K_{c}$ expression for each of these reactions. Do not forget to balance the equations.

(i) The equation $SO X_{2} + O X_{2} SO X_{3}$ has 1 S on each side but 4 O on the left ( $2 + 2$ ) and 3 O on the right. Put 2 before $SO X_{2}$ and 2 before $SO X_{3}$ : now there are 2 S and 6 O on each side.
$2 SO X_{2} (g) + O X_{2} (g) 2 SO X_{3} (g)$
Write $K_{c}$ for the balanced equation.
$K_{c} = \frac{[ SO X _{3} ] ^{2}}{[ SO X _{2} ] ^{2} [ O X _{2} ]}$
(ii) For $NH X_{3} + O X_{2} NO + H X_{2} O$ , balance N first (1 each side), then H: 3 H on the left and 2 H on the right, so take 2 $NH X_{3}$ and 3 $H X_{2} O$ (6 H each side). Now 2 N need 2 NO, and O on the right is $2 + 3 = 5$ , so $\frac{5}{2} O X_{2}$ . Multiply by 2 to remove the fraction.
$4 NH X_{3} (g) + 5 O X_{2} (g) 4 NO (g) + 6 H X_{2} O (g)$
Write $K_{c}$ for the balanced equation.
$K_{c} = \frac{[ NO ] ^{4} [ H X _{2} O ] ^{6}}{[ NH X _{3} ] ^{4} [ O X _{2} ] ^{5}}$
(iii) For $CH X_{4} + H X_{2} O CO + H X_{2}$ , C and O are balanced (1 each) but H is 6 on the left and 2 on the right. Put 3 before $H X_{2}$ .
$CH X_{4} (g) + H X_{2} O (g) CO (g) + 3 H X_{2} (g)$
Write $K_{c}$ for the balanced equation.
$K_{c} = \frac{[ CO ] [ H X _{2} ] ^{3}}{[ CH X _{4} ] [ H X _{2} O ]}$

2. Give the balanced equations that correspond to the following equilibrium expressions.

(a) For $K_{c} = \frac{[ CH X _{3} OH ]}{[ CO ] [ H X _{2} ] ^{2}}$ the numerator species is the product (power 1) and the denominator species are the reactants with powers 1 for $CO$ and 2 for $H X_{2}$ . The coefficients are therefore 1, 1 and 2.
$CO (g) + 2 H X_{2} (g) CH X_{3} OH (g)$
(b) For $K_{c} = \frac{[ N X _{2} ] [ H X _{2} O ] ^{2}}{[ NO ] ^{2} [ H X _{2} ] ^{2}}$ the products are $N X_{2}$ (coefficient 1) and $H X_{2} O$ (coefficient 2), the reactants are $NO$ (2) and $H X_{2}$ (2). Check the atoms: N is 2 and O is 2 on each side, and H is 4 on each side.
$2 NO (g) + 2 H X_{2} (g) N X_{2} (g) + 2 H X_{2} O (g)$

The value of $K_{c}$ at a given temperature can be calculated if we know the equilibrium concentrations of the reaction components.

Example 8.3

The following equilibrium concentrations were observed for the reaction

A X_{2} (g) + B X_{2} (g) 2 AB (g)

50 0^{\circ} C

[A X_{2}] = 0.399 M

[B X_{2}] = 1.197 M

[AB] = 0.203 M

. Calculate

K_{c}

Given

$[A X_{2}] = 0.399 mol d m^{- 3}$
$[B X_{2}] = 1.197 mol d m^{- 3}$
$[AB] = 0.203 mol d m^{- 3}$

To find

$K_{c}$

Write the equilibrium constant expression. The product $AB$ has coefficient 2, so its concentration is squared.
$K_{c} = \frac{[ AB ] ^{2}}{[ A X _{2} ] [ B X _{2} ]}$
Substitute the equilibrium concentrations.
$K_{c} = \frac{( 0.203 mol d m ^{- 3} ) ^{2}}{( 0.399 mol d m ^{- 3} ) ( 1.197 mol d m ^{- 3} )}$
Evaluate. The units cancel because the numerator and denominator both contain $(mol d m^{- 3})^{2}$ .
$K_{c} = \frac{0.0412}{0.4776} = 0.086$

K_{c} = 0.086

Worked example: calculating Kc and its unit

At a certain temperature the equilibrium mixture for

PCl X_{5} (g) PCl X_{3} (g) + Cl X_{2} (g)

in a closed vessel contains

0.150 mol d m^{- 3}

PCl X_{5}

0.0750 mol d m^{- 3}

PCl X_{3}

and

0.0750 mol d m^{- 3}

Cl X_{2}

. Calculate

K_{c}

, give its unit and say where the equilibrium lies.

Write the $K_{c}$ expression: products on top, reactant below, each concentration raised to its coefficient (all are $1$ here).
$K_{c} = \frac{[ PCl X _{3} ] [ Cl X _{2} ]}{[ PCl X _{5} ]}$
Substitute the equilibrium concentrations. They must be equilibrium values, not the starting values.
$K_{c} = \frac{( 0.0750 ) ( 0.0750 )}{0.150} = \frac{5.625 \times 1 0 ^{- 3}}{0.150} = 0.0375$
Find the unit by treating the concentration units like numbers. Two concentrations on top and one below leave one.
$\frac{( mol d m ^{- 3} ) ( mol d m ^{- 3} )}{mol d m ^{- 3}} = mol d m^{- 3}$
Interpret the size. $K_{c} < 1$ , so the denominator is larger than the numerator: at this temperature the mixture contains mostly $PCl X_{5}$ and the equilibrium lies towards the reactant.
The same $K_{c}$ would be found if the vessel had been started with only $PCl X_{3}$ and $Cl X_{2}$ , because $K_{c}$ depends only on temperature and not on how equilibrium was reached.

K_{c} = 0.0375 mol d m^{- 3}

8.3.1Equilibrium expressions involving partial pressure, number of moles and mole fractionOften examined

Consider the general gaseous reversible reaction:

\ce{aA(g) + bB(g) <=> cC(g) + dD(g)}

For gases the equilibrium constant expression is often written in terms of the partial pressure of each gas. The book states this through Henry's law: at constant temperature, the partial pressure of a gas is directly proportional to its molar concentration.

Equilibrium constant K_p

The equilibrium constant K_p in terms of partial pressures is:

K_p=\dfrac{P_C^{\,c}\times P_D^{\,d}}{P_A^{\,a}\times P_B^{\,b}}

Here P_A, P_B, P_C and P_D are the partial pressures of the gases A, B, C and D. K_p is related to K_c by the equation:

K_p=K_c\,(RT)^{\Delta n}

where \Delta n is the difference between the total number of moles of the products and of the reactants in the balanced equation.

\Delta n=(c+d)-(a+b)

Equilibrium constant K_n

When the equilibrium amounts of reactants and products are expressed in terms of their moles, the equilibrium constant is written K_n:

K_n=\dfrac{n_C^{\,c}\times n_D^{\,d}}{n_A^{\,a}\times n_B^{\,b}}

Here n_A, n_B, n_C and n_D are the moles of A, B, C and D at the equilibrium state. K_p is also related to K_n:

K_p=K_n\left(\dfrac{P}{n}\right)^{\Delta n}

where P is the pressure of the reaction mixture at equilibrium and n is the total number of moles of reactants and products as shown by the balanced equation.

Equilibrium constant K_x

When the equilibrium amounts are expressed by their mole fractions, the equilibrium constant is written K_x:

K_x=\dfrac{x_C^{\,c}\cdot x_D^{\,d}}{x_A^{\,a}\cdot x_B^{\,b}}

Here x_A, x_B, x_C and x_D are the mole fractions of A, B, C and D. K_p is related to K_x by:

K_p=K_x\,(P)^{\Delta n}

where P is the pressure of the equilibrium mixture.

Derivation How K_p is related to K_c, K_n and K_x

Aim. Show where the three relations K_p=K_c(RT)^{\Delta n}, K_p=K_n(P/n)^{\Delta n} and K_p=K_x(P)^{\Delta n} come from.

Assumptions.

The gases behave ideally, so PV=nRT for each gas.
The reaction is \ce{aA(g) + bB(g) <=> cC(g) + dD(g)} at constant temperature.

For one gas i of the mixture, the ideal gas equation gives its partial pressure in terms of its molar concentration C_i=n_i/V.
P_i=\dfrac{n_i}{V}RT=C_iRT
Put this in K_p. Each concentration brings a factor RT.
K_p=\dfrac{(C_CRT)^c(C_DRT)^d}{(C_ART)^a(C_BRT)^b}
Collect the factors of RT: the power is (c+d)-(a+b)=\Delta n.
K_p=\dfrac{C_C^{\,c}C_D^{\,d}}{C_A^{\,a}C_B^{\,b}}\,(RT)^{(c+d)-(a+b)}=K_c\,(RT)^{\Delta n}
For the other two relations, write the partial pressure of a gas as its mole fraction times the total pressure: P_i=x_iP, where x_i=n_i/n and n is the total number of moles. Substitute in K_p.
K_p=\dfrac{(x_CP)^c(x_DP)^d}{(x_AP)^a(x_BP)^b}=K_x\,P^{\Delta n}
Since x_i=n_i/n, the mole fractions give K_x=K_n\,n^{-\Delta n}. Substituting in the last result gives the relation with K_n.
K_p=K_n\left(\dfrac{P}{n}\right)^{\Delta n}

ResultK_p=K_c(RT)^{\Delta n}=K_n\left(\dfrac{P}{n}\right)^{\Delta n}=K_x\,P^{\Delta n}

When \Delta n=0 (equal numbers of moles of gas on both sides), the factor becomes 1 and K_p=K_c=K_x=K_n.

The four equilibrium constants
Constant	Expressed in terms of	Relation with K_p
K_c	Molar concentrations (\mathrm{mol\,dm^{-3}})	K_p=K_c(RT)^{\Delta n}
K_p	Partial pressures	(the reference)
K_n	Number of moles at equilibrium	K_p=K_n\left(\dfrac{P}{n}\right)^{\Delta n}
K_x	Mole fractions	K_p=K_x(P)^{\Delta n}

Example 8.4

The following reaction was studied at 25^\circ\mathrm{C}: \ce{2NO(g) + Cl2(g) <=> 2NOCl(g)}. The partial pressures at equilibrium were P_{\ce{NOCl}}=1.2\,\mathrm{atm}, P_{\ce{NO}}=5.0\times10^{-2}\,\mathrm{atm} and P_{\ce{Cl2}}=3.0\times10^{-1}\,\mathrm{atm}. Calculate K_p and K_c.

Given

P_{\ce{NOCl}} = 1.2\,\mathrm{atm}
P_{\ce{NO}} = 5.0\times10^{-2}\,\mathrm{atm}
P_{\ce{Cl2}} = 3.0\times10^{-1}\,\mathrm{atm}
T = 25+273=298\,\mathrm{K}
R = 0.0821\,\mathrm{dm^3\,atm\,K^{-1}\,mol^{-1}}

To find

Step 1. Write the K_p expression (products over reactants, each pressure raised to its coefficient).
K_p=\dfrac{(P_{\ce{NOCl}})^2}{(P_{\ce{NO}})^2(P_{\ce{Cl2}})}
Step 2. Substitute the partial pressures of each species.
K_p=\dfrac{(1.2)^2}{(5.0\times10^{-2})^2(3.0\times10^{-1})}
Step 3. Simplify to get K_p. The denominator is 2.5\times10^{-3}\times3.0\times10^{-1}=7.5\times10^{-4}.
K_p=\dfrac{1.44}{7.5\times10^{-4}}=1.9\times10^{3}\ \mathrm{atm^{-1}}
Step 4. Calculate \Delta n: 2 moles of gaseous product and 2+1=3 moles of gaseous reactants.
\Delta n=2-(2+1)=-1
Step 5. Write the expression that relates K_p and K_c.
K_p=K_c(RT)^{\Delta n}
Step 6. Substitute the known values and solve for K_c.
1.9\times10^3=K_c\,(0.0821\times298)^{-1}=\dfrac{K_c}{0.0821\times298}
Multiply both sides by 0.0821\times298=24.47.
K_c=1.9\times10^{3}\times0.0821\times298=4.65\times10^{4}

K_p=1.9\times10^{3}\ \mathrm{atm^{-1}},\quad K_c=4.65\times10^{4}\ \mathrm{dm^3\,mol^{-1}}

Worked example: Kp from Kc and a partial pressure

For \ce{N2O4(g) <=> 2NO2(g)} the value of K_c is 5.8\times10^{-3}\,\mathrm{mol\,dm^{-3}} at 298\,\mathrm{K}. (a) Calculate K_p in atm. (b) At equilibrium the partial pressure of \ce{N2O4} is 0.50\,\mathrm{atm}. Find the partial pressure of \ce{NO2}.

Given

K_c = 5.8\times10^{-3}\,\mathrm{mol\,dm^{-3}}
T = 298\,\mathrm{K}
R = 0.0821\,\mathrm{dm^{3}\,atm\,K^{-1}\,mol^{-1}}
P_{\ce{N2O4}} = 0.50\,\mathrm{atm}

To find

K_p
P_{\ce{NO2}}

Find \Delta n, the moles of gaseous products minus the moles of gaseous reactants in the balanced equation.
\Delta n=2-1=+1
Use K_p=K_c(RT)^{\Delta n} with R=0.0821\,\mathrm{dm^{3}\,atm\,K^{-1}\,mol^{-1}} so that the pressure comes out in atm.
K_p=5.8\times10^{-3}\times(0.0821\times298)^{1}=5.8\times10^{-3}\times24.47=0.142\,\mathrm{atm}
(b) Write the K_p expression for this equation. The power 2 on P_{\ce{NO2}} is the coefficient.
K_p=\dfrac{P_{\ce{NO2}}^{\,2}}{P_{\ce{N2O4}}}
Rearrange and substitute.
P_{\ce{NO2}}^{\,2}=K_p\,P_{\ce{N2O4}}=0.142\times0.50=0.0710\,\mathrm{atm^{2}}
Take the square root.
P_{\ce{NO2}}=\sqrt{0.0710}=0.27\,\mathrm{atm}

K_p=0.142\,\mathrm{atm},\quad P_{\ce{NO2}}=0.27\,\mathrm{atm}

Concept assessment exercise 8.2

The contact process prepares the purest sulphuric acid commercially. The following reaction takes place in the contact chamber in the presence of \ce{V2O5} at 450^\circ\mathrm{C}: \ce{2SO2(g) + O2(g) <=> 2SO3(g)}. Calculate K_p if the following concentrations are found at equilibrium: [\ce{SO2}]=0.59\,\mathrm{M}, [\ce{O2}]=0.05\,\mathrm{M} and [\ce{SO3}]=0.259\,\mathrm{M}.

Given

[\ce{SO2}] = 0.59\,\mathrm{M}
[\ce{O2}] = 0.05\,\mathrm{M}
[\ce{SO3}] = 0.259\,\mathrm{M}
T = 450+273=723\,\mathrm{K}

To find

The data are concentrations, so first find K_c.
K_c=\dfrac{[\ce{SO3}]^2}{[\ce{SO2}]^2[\ce{O2}]}=\dfrac{(0.259)^2}{(0.59)^2(0.05)}=\dfrac{0.0671}{0.01741}=3.85\ \mathrm{dm^3\,mol^{-1}}
Find \Delta n: 2 moles of product and 2+1=3 moles of reactants.
\Delta n=2-3=-1
Convert K_c to K_p with R=0.0821\,\mathrm{dm^3\,atm\,K^{-1}\,mol^{-1}} and T=723\,\mathrm{K}.
K_p=K_c(RT)^{-1}=\dfrac{3.85}{0.0821\times723}=\dfrac{3.85}{59.36}
Evaluate.
K_p=0.0649\ \mathrm{atm^{-1}}

K_p=0.0649\ \mathrm{atm^{-1}}

8.4Types of equilibrium

With respect to the physical states of the reactants and products, there are two types of chemical equilibrium: homogeneous and heterogeneous.

i. Homogeneous equilibria

Homogeneous equilibrium: An equilibrium system in which all of the reactants and products are in the same phase.

Examples:

\ce{N2(g) + 3H2(g) <=> 2NH3(g)}

\ce{2SO2(g) + O2(g) <=> 2SO3(g)}

\ce{2NO(g) + Cl2(g) <=> 2NOCl(g)}

\ce{CH3COOH(l) + C2H5OH(l) <=> CH3COOC2H5(l) + H2O(l)}

ii. Heterogeneous equilibria

Heterogeneous equilibrium: An equilibrium that involves more than one phase.

Examples:

\ce{CaCO3(s) <=> CaO(s) + CO2(g)}

\ce{C(s) + H2O(g) <=> CO(g) + H2(g)}

\ce{3Fe(s) + 4H2O(g) <=> Fe3O4(s) + 4H2(g)}

If pure solids or pure liquids are involved in an equilibrium system, their concentrations are not included in the equilibrium constant expression. This is because a change in the concentration of a pure solid or liquid has no effect on the equilibrium system (its concentration stays constant).

Examples of heterogeneous equilibrium expressions:

Example (i): decomposition of liquid water

Write K_c and K_p for \ce{2H2O(l) <=> 2H2(g) + O2(g)}.

Water is a pure liquid, so it is left out of both expressions. Only the gases appear.
K_c=[\ce{H2}]^2[\ce{O2}]
In terms of partial pressures, the gases \ce{H2} and \ce{O2} appear, each to the power of its coefficient.
K_p=P_{\ce{H2}}^{\,2}\times P_{\ce{O2}}

Example (ii): iron and steam

Write K_c and K_p for \ce{3Fe(s) + 4H2O(g) <=> Fe3O4(s) + 4H2(g)}.

\ce{Fe} and \ce{Fe3O4} are pure solids, so they are left out. The steam and hydrogen are gases.
K_c=\dfrac{[\ce{H2}]^4}{[\ce{H2O}]^4}
In terms of partial pressures:
K_p=\dfrac{P_{\ce{H2}}^{\,4}}{P_{\ce{H2O}}^{\,4}}

Concept assessment exercise 8.3

Write K_c and K_p expressions for each of the following reactions.

(i) \ce{FeO(s) + CO(g) <=> Fe(s) + CO2(g)}. The solids \ce{FeO} and \ce{Fe} are left out; only the gases \ce{CO} and \ce{CO2} appear.
K_c=\dfrac{[\ce{CO2}]}{[\ce{CO}]},\qquad K_p=\dfrac{P_{\ce{CO2}}}{P_{\ce{CO}}}
(ii) \ce{P4(s) + 5O2(g) <=> P4O10(s)}. Both \ce{P4} and \ce{P4O10} are solids, so only \ce{O2} remains, in the denominator with power 5.
K_c=\dfrac{1}{[\ce{O2}]^5},\qquad K_p=\dfrac{1}{P_{\ce{O2}}^{\,5}}
(iii) \ce{CH4(g) + 4Cl2(g) <=> CCl4(l) + 4HCl(g)}. \ce{CCl4} is a pure liquid, so it is left out.
K_c=\dfrac{[\ce{HCl}]^4}{[\ce{CH4}][\ce{Cl2}]^4},\qquad K_p=\dfrac{P_{\ce{HCl}}^{\,4}}{P_{\ce{CH4}}\,P_{\ce{Cl2}}^{\,4}}

8.5Ways to recognize equilibrium

The equilibrium constant expression can be determined by physical as well as chemical methods. Both methods follow how a property of the mixture (or the amount of one substance) changes with time. When it stops changing, equilibrium has been reached.

a) Physical method (spectrometric method)

This method is based on measuring a physical property of the reaction mixture. The property is measured during the reaction without removing the sample from the reaction mixture. In the spectrometric method, the reactant or product absorbs ultraviolet, visible or infrared radiation, and the concentration can be determined by measuring the amount of radiation absorbed.

\ce{N2O4(g) <=> 2NO2(g)}

N2O4 is colourless; NO2 is reddish brown

\ce{N2O4} is a colourless gas whereas \ce{NO2} is a reddish brown gas. The progress of the reaction can be studied by measuring the absorbance at regular intervals. The absorbance is proportional to the concentration of \ce{NO2}. At equilibrium the spectrometer shows a constant value of absorbance.

Absorbance during the reaction (schematic)

Schematic plot (it is drawn to explain the book's text and is not a book figure): the absorbance, which is proportional to [\ce{NO2}], rises and then stays constant when equilibrium is reached.

b) Chemical method

In this method the amount of a reactant or product is determined by a suitable chemical reaction. Consider the reaction between acetic acid and ethanol to form ethyl acetate and water, an example of a reversible reaction in solution.

\ce{CH3COOH(l) + C2H5OH(l) <=> CH3COOC2H5(l) + H2O(l)}

Procedure

Start the reaction by taking a moles of acetic acid and b moles of ethanol in a stoppered flask at room temperature.
Add a small amount of mineral acid to the mixture to catalyse the reaction.
Study the progress of the reaction by finding the concentration of acetic acid after regular intervals. For this, withdraw a small portion of the mixture each time.
Determine the concentration of acetic acid in the portion by titrating it against a standard solution of \ce{NaOH}, using phenolphthalein as the indicator.
The concentration of acetic acid decreases until equilibrium is attained. At equilibrium it becomes constant, which shows that equilibrium has been reached.

Physical and chemical methods compared
Point	Physical method	Chemical method
Principle	Measure a physical property (absorbance of radiation)	Measure the amount of a substance by a chemical reaction (titration)
Sample	Not removed from the reaction mixture	A small portion is withdrawn each time
Example in the book	\ce{N2O4 <=> 2NO2} by a spectrometer	Acetic acid and ethanol, titrating acetic acid with \ce{NaOH}
Sign of equilibrium	Constant absorbance	Constant concentration of acetic acid

8.6Factors affecting equilibrium (the Le Chatelier's principle)Often examined

It is necessary to understand the factors that control the position of a chemical equilibrium. Knowing them helps industrial chemists to choose conditions that favour the desired product as much as possible. We can predict the effect of factors such as concentration, pressure and temperature on a system at equilibrium by using Le Chatelier's principle.

Le Chatelier's principle: If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction which tends to reduce that change.

At equilibrium the forward and reverse reactions are driven by molecular collisions. When we change the temperature, pressure or concentration, the frequency and energy of collisions between the reacting molecules increase or decrease. This changes the reaction rates and causes a shift in the equilibrium position.

Summary of the factors (each is explained in the sections that follow)
Change imposed	Equilibrium shifts	Does K_c change?
Concentration of a substance increased	Away from that substance (the substance is used up)	No
Concentration of a substance decreased	Towards that substance (the substance is formed)	No
Pressure increased	Towards the side with fewer gas molecules	No
Pressure decreased	Towards the side with more gas molecules	No
Temperature increased	In the endothermic direction	Yes
Temperature decreased	In the exothermic direction	Yes
Catalyst added	No shift; equilibrium is reached faster	No

8.6.1The effect of a change in concentrationOften examined

When the concentration of one or more of the reactants or products present in the equilibrium mixture is disturbed, the system does not remain in the equilibrium state. According to Le Chatelier's principle the equilibrium shifts to accommodate the substance added or removed and so restore equilibrium again.

Consider the following gas phase equilibrium:

\ce{CO2(g) + H2(g) <=> CO(g) + H2O(g)}

Adding a reactant

Add \ce{CO2} to the system. It is no longer in equilibrium, because the concentration of \ce{CO2} is higher than before.
The higher concentration of \ce{CO2} increases the rate of the forward reaction relative to the reverse reaction. More \ce{CO2} and \ce{H2} combine, and more \ce{CO} and \ce{H2O} are formed.
As time passes, the concentrations of \ce{CO2} and \ce{H2} decrease, which lowers the rate of the forward reaction. At the same time the increased concentrations of the products speed up the reverse reaction.
Finally the two rates become equal again and equilibrium is re-established. The concentrations of \ce{CO} and \ce{H2O} are higher than before the \ce{CO2} was added. Equilibrium is said to have shifted to the products side.

In all chemical systems, when the concentration of a product is increased the equilibrium shifts towards the reactants. A shift towards the reactants lowers the concentration of the added product.

The opposite happens when we decrease the concentration of a reactant or a product. If the concentration of a reactant is decreased, the system shifts towards the reactants. The removal of a product shifts the equilibrium towards the products.

Effect of changing the concentration
Disturbance	Shift of equilibrium	Reason
Reactant added	To the right (products)	Used up to reduce the added reactant
Product added	To the left (reactants)	Used up to reduce the added product
Reactant removed	To the left (reactants)	Reactant is formed to replace it
Product removed	To the right (products)	Product is formed to replace it

The microscopic events

The rate of a chemical reaction depends on the number of effective collisions between the reacting molecules. At equilibrium the numbers of effective collisions for the forward and the reverse reactions are equal.

An increase in the concentration of a reactant increases the effective collisions for the forward reaction. Thus the equilibrium shifts to the right, with the formation of more molecules of products.
The number of effective collisions for the reverse process also increases, because the products accumulate.
As time passes, the effective collisions of the reactant molecules decrease, lowering the rate of the forward reaction.
Finally the numbers of effective collisions for both processes are equal again, and equilibrium is re-established.

Example 8.5

K_c for the reaction \ce{2ICl(g) <=> Cl2(g) + I2(g)} is 1.0\times10^{-3} at 230^\circ\mathrm{C}. 1.6 moles of \ce{ICl}, 0.05 mole of \ce{I2} and 0.05 mole of \ce{Cl2} are present in the equilibrium mixture in a 2\,\mathrm{dm^3} container at 230^\circ\mathrm{C}. Determine the equilibrium concentrations of \ce{I2}, \ce{Cl2} and \ce{ICl} when the equilibrium is restored after the addition of another mole of \ce{ICl}.

Given

K_c = 1.0\times10^{-3}
V = 2\,\mathrm{dm^3}
n_{\ce{ICl}}\text{ (old equilibrium)} = 1.6\,\mathrm{mol}
n_{\ce{I2}}=n_{\ce{Cl2}} = 0.05\,\mathrm{mol}
\text{added }\ce{ICl} = 1\,\mathrm{mol}

To find

[\ce{ICl}]
[\ce{I2}]
[\ce{Cl2}]\ \text{at the new equilibrium}

Adding \ce{ICl} disturbs the equilibrium and shifts it in the forward direction (Le Chatelier). So the amount of \ce{ICl} decreases by 2x, and the amounts of \ce{I2} and \ce{Cl2} each increase by x (moles).
Make a table of the amounts in moles. The amount of \ce{ICl} just after the addition is 1.6+1=2.6 mol.
\begin{array}{lccc} & 2\ce{ICl} & \ce{Cl2} & \ce{I2}\\[16pt] \begin{aligned}&\text{Initial}\\ &\text{(after adding)}\end{aligned} & 2.6 & 0.05 & 0.05\\[16pt] \begin{aligned}&\text{New}\\ &\text{equilibrium}\end{aligned} & 2.6-2x & 0.05+x & 0.05+x\end{array}
Divide each amount by the volume 2\,\mathrm{dm^3} to get the concentrations.
[\ce{ICl}]=\dfrac{2.6-2x}{2},\quad[\ce{Cl2}]=[\ce{I2}]=\dfrac{0.05+x}{2}
Write the K_c expression and substitute.
K_c=\dfrac{[\ce{Cl2}][\ce{I2}]}{[\ce{ICl}]^2}\ \Rightarrow\ 1.0\times10^{-3}=\dfrac{\left(\dfrac{0.05+x}{2}\right)\left(\dfrac{0.05+x}{2}\right)}{\left(\dfrac{2.6-2x}{2}\right)^2}
Take the square root of both sides. The factor of 2 (the volume) cancels everywhere.
\sqrt{1.0\times10^{-3}}=\dfrac{0.05+x}{2.6-2x}\ \Rightarrow\ 3.16\times10^{-2}=\dfrac{0.05+x}{2.6-2x}
Cross-multiply and solve for x.
0.0316\,(2.6-2x)=0.05+x\ \Rightarrow\ 0.0822-0.0632x=0.05+x\ \Rightarrow\ x=\dfrac{0.0322}{1.0632}=0.030\ \mathrm{mol}
Find the amounts at the new equilibrium and divide by 2\,\mathrm{dm^3}.
n_{\ce{ICl}}=2.6-2(0.030)=2.54\ \mathrm{mol}\ \Rightarrow\ [\ce{ICl}]=\dfrac{2.54}{2}=1.27\ \mathrm{mol\,dm^{-3}}
The two products have the same amount.
n_{\ce{I2}}=n_{\ce{Cl2}}=0.05+0.030=0.080\ \mathrm{mol}\ \Rightarrow\ [\ce{I2}]=[\ce{Cl2}]=\dfrac{0.080}{2}=0.040\ \mathrm{mol\,dm^{-3}}

[\ce{ICl}]=1.27\ \mathrm{M},\quad[\ce{I2}]=[\ce{Cl2}]=0.040\ \mathrm{M}\ \ (\text{amounts: }2.54,\ 0.080,\ 0.080\ \mathrm{mol})

Worked example: sketching the response to added hydrogen

More \ce{H2} is added at constant temperature to the equilibrium mixture \ce{CO2(g) + H2(g) <=> CO(g) + H2O(g)}. Sketch how the concentrations change with time and explain each part of the sketch.

Before the disturbance every concentration is constant, so each line is horizontal. Draw four lines (the lines for \ce{CO} and \ce{H2O} coincide, because they have equal coefficients).
At time t_1 the \ce{H2} is added. Its concentration jumps up at once, as a vertical line, because nothing has had time to react yet. The other concentrations do not move at this instant.
The system is no longer at equilibrium. By Le Chatelier's principle it shifts to the right to use up some of the added \ce{H2}. More collisions between \ce{H2} and \ce{CO2} make the forward rate greater than the reverse rate.
So [\ce{H2}] and [\ce{CO2}] fall, and [\ce{CO}] and [\ce{H2O}] rise by the same amount (the coefficients are all 1). The curves bend gradually because the net rate drops as the system returns to balance.
When the forward and reverse rates are equal again, all lines become horizontal at new levels. [\ce{H2}] ends below its peak but above its old value, because only part of the added \ce{H2} is used up.
The temperature is unchanged, so K_c has the same value as before. Only the position of equilibrium moved.

Response of the equilibrium \ce{CO2 + H2 <=> CO + H2O} when \ce{H2} is added at t_1. The \ce{H2} curve (accent) jumps up, then falls as the forward reaction uses it. The sketch shows the shape only.

Concept assessment exercise 8.4

K_c for \ce{2ICl(g) <=> Cl2(g) + I2(g)} is 1.0\times10^{-3} at 230^\circ\mathrm{C}. 1.6 moles of \ce{ICl}, 0.05 mole of \ce{I2} and 0.05 mole of \ce{Cl2} are present in the equilibrium mixture in a 2\,\mathrm{dm^3} container at 230^\circ\mathrm{C}. Determine the equilibrium concentrations of \ce{I2}, \ce{Cl2} and \ce{ICl} when the equilibrium is restored after the removal of one mole of \ce{ICl}.

Removing \ce{ICl} shifts the equilibrium in the reverse direction (towards \ce{ICl}). Let y moles each of \ce{I2} and \ce{Cl2} react to give 2y moles of \ce{ICl}. After the removal there are 1.6-1=0.6 mol of \ce{ICl}.
\begin{array}{lccc} & 2\ce{ICl} & \ce{Cl2} & \ce{I2}\\[16pt] \begin{aligned}&\text{Initial}\\ &\text{(after removal)}\end{aligned} & 0.6 & 0.05 & 0.05\\[16pt] \begin{aligned}&\text{New}\\ &\text{equilibrium}\end{aligned} & 0.6+2y & 0.05-y & 0.05-y\end{array}
Write K_c. The volume cancels, so amounts in moles may be used directly. Take the square root of both sides.
\dfrac{(0.05-y)^2}{(0.6+2y)^2}=1.0\times10^{-3}\ \Rightarrow\ \dfrac{0.05-y}{0.6+2y}=0.0316
Cross-multiply and solve for y.
0.05-y=0.0316(0.6+2y)=0.01897+0.0632y\ \Rightarrow\ y=\dfrac{0.03103}{1.0632}=0.0292
Find the new amounts.
\begin{aligned}n_{\ce{I2}}=n_{\ce{Cl2}}&=0.05-0.0292=0.0208\ \mathrm{mol}\\ n_{\ce{ICl}}&=0.6+2(0.0292)=0.658\ \mathrm{mol}\end{aligned}
Divide by the volume 2\,\mathrm{dm^3} for concentrations.
[\ce{I2}]=[\ce{Cl2}]=0.0104\ \mathrm{M},\qquad[\ce{ICl}]=0.329\ \mathrm{M}

Amounts at the new equilibrium: 0.0208 mol of each of I₂ and Cl₂, and 0.658 mol of ICl (concentrations 0.0104 M, 0.0104 M and 0.329 M in the 2 dm³ container).

8.6.2The effect of pressure changeOften examined

Equilibria that contain gases are influenced by changes in pressure. When the pressure on a gaseous system at equilibrium is increased, the system tends to reduce its volume to undo or minimise the effect of the increased pressure. It does this by decreasing the total number of gaseous molecules in the system. This is because, at constant temperature and pressure, the volume of a gas is directly proportional to the total number of gas molecules present.

Consider the equilibrium system below. Which side contains the smaller number of molecules?

\ce{2SO2(g) + O2(g) <=> 2SO3(g)}

Three molecules (two of \ce{SO2} and one of \ce{O2}) react to produce two molecules of \ce{SO3}, which reduces the total number of gaseous molecules. If we suddenly increase the pressure, the system reduces its volume by reducing the number of molecules present, so the reaction shifts to the right. The equilibrium position shifts towards the side with the smaller number of gaseous molecules in the balanced chemical equation.

When the pressure is reduced, the system shifts so as to increase its volume, that is, towards the side with the larger number of gaseous molecules.

Effect of pressure on a gas equilibrium
Pressure change	Shift of equilibrium	Result for \ce{2SO2 + O2 <=> 2SO3}
Increased	Towards the side with fewer gas molecules	Right: more \ce{SO3}
Decreased	Towards the side with more gas molecules	Left: more \ce{SO2} and \ce{O2}

Equal numbers of molecules on both sides

There are equilibrium reactions in which the total number of molecules is the same on either side. For example:

\ce{H2(g) + I2(g) <=> 2HI(g)}

2 moles of gas on the left, 2 moles of gas on the right

When the pressure is changed in such a system, neither the forward nor the reverse reaction is favoured, because the number of molecules is the same on both sides. Changes in pressure or volume do not affect such equilibria.

Worked example: which equilibria respond to pressure?

Predict the effect of increasing the pressure (by decreasing the volume) on each equilibrium: (a) \ce{N2O4(g) <=> 2NO2(g)}, (b) \ce{H2(g) + I2(g) <=> 2HI(g)}, (c) \ce{CaCO3(s) <=> CaO(s) + CO2(g)}.

Rule: higher pressure shifts the equilibrium towards the side with fewer gas molecules. Count gas molecules only: solids and liquids take up almost no volume and are not counted.
(a) Left side: 1 gas molecule. Right side: 2. The left side has fewer, so the equilibrium shifts to the left. More \ce{N2O4} and less \ce{NO2} are present, and the reddish brown colour fades.
1\;\text{mol gas}\ \ \ce{<=>}\ \ 2\;\text{mol gas}\ \Rightarrow\ \text{left}
(b) Left side: 1+1=2 gas molecules. Right side: 2. They are equal, so neither direction reduces the pressure. There is no shift and the amounts stay the same.
2\;\text{mol gas}\ \ \ce{<=>}\ \ 2\;\text{mol gas}\ \Rightarrow\ \text{no shift}
(c) Left side: 0 gas molecules (\ce{CaCO3} is solid). Right side: 1 (\ce{CO2}). The side with fewer gas molecules is the left, so the equilibrium shifts to the left and some \ce{CO2} combines with \ce{CaO}.
0\;\text{mol gas}\ \ \ce{<=>}\ \ 1\;\text{mol gas}\ \Rightarrow\ \text{left}
A common mistake in (c) is to count the solids and say there are 1 on the left and 2 on the right. Only gases matter for a pressure change.

\text{(a) left};\quad\text{(b) no shift};\quad\text{(c) left}

Concept assessment exercise 8.5

The formation of methanol is an important industrial reaction in the processing of new fuels: \ce{CO(g) + 2H2(g) <=> CH3OH(g)}. A student decreases the pressure over the system in an attempt to increase the yield of methanol. Is this approach reasonable? Explain.

Count the gas molecules on each side: 1+2=3 moles of gas on the left (reactants) and 1 mole of gas on the right (methanol).
Decreasing the pressure makes the system shift towards the side with more gas molecules in order to increase its volume. That is the left side, the reactants.
So the equilibrium shifts away from methanol, and the yield of methanol decreases.

No, the approach is not reasonable. To increase the yield of methanol the pressure should be increased, which shifts the equilibrium towards the side with fewer gas molecules (1 mole of \ce{CH3OH} instead of 3 moles of reactants).

8.6.3The effect of change in temperatureOften examined

Chemical reactions that liberate heat are called exothermic and those that absorb heat are called endothermic. In an exothermic reaction heat is placed on the right side of the equation, and in an endothermic reaction it is placed on the left side.

We can use Le Chatelier's principle to predict the direction of change by treating energy (heat) as a reactant or product, in the same way as an actual substance that is added or removed. Therefore an increase in temperature (adding heat) favours the endothermic reaction, and a decrease in temperature (removing heat) favours the exothermic reaction.

An exothermic reaction

Consider the reaction below. Which direction, forward or reverse, is endothermic?

\ce{2SO2(g) + O2(g) <=> 2SO3(g)}\qquad \Delta H^\circ=-198\,\mathrm{kJ}

Because the reaction is exothermic we can write heat as a product:

\ce{2SO2(g) + O2(g) <=> 2SO3(g) + heat}

According to Le Chatelier's principle an increase in temperature shifts the reaction from right to left, to absorb the added heat and oppose the temperature rise. As a result the concentration of \ce{SO3} decreases and the concentrations of \ce{SO2} and \ce{O2} increase. The value of the equilibrium constant therefore decreases, and the equilibrium shifts towards the left.

K_c=\dfrac{[\ce{SO3}]^2\ (\text{decreases})}{[\ce{SO2}]^2[\ce{O2}]\ (\text{increases})}

That is why K_c=2.8\times10^{2} at 1000\,\mathrm{K}, whereas at 298\,\mathrm{K} the value of K_c is 1\times10^{26} (the values given in the book). The equilibrium production of \ce{SO3} is favoured at the lower temperature, because K_c is much larger at 298\,\mathrm{K} than at 1000\,\mathrm{K}.

An endothermic reaction

Now consider the reaction:

\ce{N2O4(g) <=> 2NO2(g)}\qquad\Delta H^\circ=+57.2\,\mathrm{kJ}

Because the reaction is endothermic we can write heat as a reactant:

\ce{N2O4(g) + heat <=> 2NO2(g)}

When the temperature is increased, heat enters the system and the reaction shifts from left to right. As a result the concentration of \ce{NO2} increases and that of \ce{N2O4} decreases. This increases the value of K_c, and the equilibrium shifts in the forward direction.

K_c=\dfrac{[\ce{NO2}]^2\ (\text{increases})}{[\ce{N2O4}]\ (\text{decreases})}

That is why K_c for this reaction is 7.7\times10^{-5} at 0^\circ\mathrm{C} and 0.4 at 100^\circ\mathrm{C}.

Effect of temperature on the position of equilibrium
Reaction type	Temperature raised	Temperature lowered	Value of K_c when T rises
Exothermic (\Delta H<0)	Shifts to the left (reactants)	Shifts to the right (products)	Decreases
Endothermic (\Delta H>0)	Shifts to the right (products)	Shifts to the left (reactants)	Increases

Concept assessment exercise 8.6

1. Consider the equilibrium \ce{2I(g) <=> I2(g)}. What would be the effect on the position of equilibrium when the temperature is decreased?

Two iodine atoms join to form an \ce{I2} molecule, so a bond is formed. Bond formation releases energy, so the forward reaction is exothermic. In the book's form, heat is a product.
\ce{2I(g) <=> I2(g) + heat}
Decreasing the temperature removes heat, so the system shifts in the direction that produces heat, that is, the exothermic forward reaction.

The equilibrium shifts to the right, towards \ce{I2}, and the amount of \ce{I2} increases (the value of K_c increases).

2. Predict the effect of increasing the temperature on the amount of product in the exothermic reaction \ce{CO(g) + 2H2(g) <=> CH3OH(g)}.

The reaction is exothermic, so heat is a product.
\ce{CO(g) + 2H2(g) <=> CH3OH(g) + heat}
Increasing the temperature adds heat, so the system shifts in the direction that absorbs it, the endothermic reverse reaction. The value of K_c falls.

The equilibrium shifts to the left, so the amount of the product, \ce{CH3OH}, decreases.

8.6.4The effect of addition of catalyst

A catalyst added to a reaction mixture speeds up both the forward and the reverse reaction to the same degree. Thus the catalyst has no effect on the equilibrium concentrations of the reaction mixture. However, the catalyst is important in enhancing the rate at which equilibrium is established.

Summary of the factors
Factor	Shifts the position?	Changes K_c?
Concentration	Yes	No
Pressure (when \Delta n\neq0)	Yes	No
Temperature	Yes	Yes
Catalyst	No	No (equilibrium is reached sooner)

Worked example: a catalyst and the position of equilibrium

Sketch the concentration of \ce{SO3} against time for \ce{2SO2(g) + O2(g) <=> 2SO3(g)} with and without a \ce{V2O5} catalyst, starting from \ce{SO2} and \ce{O2} at the same temperature. Explain the sketch and say what the catalyst does to K_c.

Concentration of \ce{SO3} against time for \ce{2SO2 + O2 <=> 2SO3}, with and without \ce{V2O5}. Both curves level off at the same equilibrium concentration (dashed). The catalyst only shortens the time.

Without a catalyst the concentration of \ce{SO3} rises slowly and levels off when the forward and reverse rates become equal. Draw this curve rising towards a horizontal line.
The catalyst speeds up the forward and the reverse reaction to the same degree. So the equilibrium level is reached sooner (a steeper curve), but the level is the same: both curves end on the same dashed line.
Because the equilibrium concentrations are unchanged, the value of K_c=\dfrac{[\ce{SO3}]^2}{[\ce{SO2}]^2[\ce{O2}]} is unchanged. Only a change of temperature changes K_c.
A catalyst therefore gives the same yield in a shorter time. It does not move the position of equilibrium. This is why a catalyst is used together with a compromise temperature in the contact process: it helps the rate and leaves the yield to temperature, pressure and concentration.

K_c\ \text{unchanged; the same equilibrium level is reached faster}

8.7Industrial application of Le Chatelier's principleOften examined

For industrial processes it is important to maximise the concentration of the desired product and minimise the leftover reactants. Le Chatelier's principle and reaction kinetics can both be used to design the best conditions to give the highest possible yield of the product in an economic way.

Two ideas used together
Idea	What it gives
Le Chatelier's principle	Which conditions of temperature, pressure and concentration shift the equilibrium towards the product (a high yield)
Reaction kinetics	Which conditions (higher temperature, catalyst) make the reaction fast enough to be economical

8.7.1Synthesis of ammonia by Haber's processOften examined

The manufacture of ammonia by Haber's process is represented by the equation:

\ce{N2(g) + 3H2(g) <=> 2NH3(g)}\qquad\Delta H^\circ=-92.46\,\mathrm{kJ}

This equation provides the following information.

The reaction is exothermic.
The reaction proceeds with a decrease in the number of molecules (moles): 4 moles of gas on the left and 2 moles on the right.

Le Chatelier's principle suggests three ways to get the maximum yield of ammonia.

i. Low temperature

The forward reaction is exothermic, therefore a low temperature favours the formation of ammonia. The suitable temperature is 400^\circ\mathrm{C}.

ii. High pressure

Four molecules (one of \ce{N2} and three of \ce{H2}) react to produce two molecules of \ce{NH3}. Thus high pressure shifts the equilibrium to the right side, that is, towards the formation of \ce{NH3}. The most suitable pressure is 200 to 300\,\mathrm{atm}.

The optimum condition for the equilibrium production of ammonia is therefore a low temperature and a high pressure. Although the yield of ammonia is high at a low temperature, the rate of its formation is so slow that the process becomes uneconomical. Therefore a catalyst is used to increase the rate of reaction. Usually a piece of iron with other metal oxides is used as the catalyst. The equilibrium mixture contains 35\% \ce{NH3} by volume.

iii. Continual removal of ammonia

A final factor that greatly increases the production of ammonia is the continuous removal of ammonia as it is formed. This is done by liquefying the ammonia: the equilibrium mixture is cooled by refrigeration coils until ammonia condenses at -33.4^\circ\mathrm{C} and is removed. \ce{N2} and \ce{H2}, which do not liquefy at this temperature, are recycled into the reaction chamber.

The stress caused by the continual removal of ammonia shifts the equilibrium towards the production of more ammonia. In fact the mixture need not be allowed to come to equilibrium at all. In this way practically 100\% conversion of \ce{N2} and \ce{H2} to \ce{NH3} is possible.

Haber's process: ammonia is liquefied and removed in the cooler, which shifts the equilibrium to the right. The unreacted gases are recycled.

Conditions of Haber's process
Condition	Reason from Le Chatelier's principle	Value used
Low temperature	Forward reaction is exothermic	400^\circ\mathrm{C}
High pressure	4 moles of gas give 2 moles of gas	200 to 300\,\mathrm{atm}
Catalyst (iron with metal oxides)	Makes the rate fast enough; does not change the position	Iron with other metal oxides
Removal of \ce{NH3} by liquefying	Removing a product shifts the equilibrium to the right	-33.4^\circ\mathrm{C}; \ce{N2} and \ce{H2} recycled

8.7.2Synthesis of sulphuric acid by contact processOften examined

The contact process is an industrial method used to produce sulphuric acid. The manufacture of sulphuric acid is represented by the following chemical reactions.

i. Formation of sulphur dioxide

Sulphur dioxide is produced by burning sulphur.

\ce{S(s) + O2(g) -> SO2(g)}

ii. Oxidation of sulphur dioxide

\ce{SO2} is oxidised into \ce{SO3} using vanadium pentoxide (\ce{V2O5}).

\ce{2SO2(g) + O2(g) <=> 2SO3(g)}\qquad\Delta H^\circ=-198\,\mathrm{kJ}

iii. Conversion of sulphur trioxide into sulphuric acid

The oxidation of sulphur dioxide to sulphur trioxide is the most important step in the production of sulphuric acid. This reaction provides the following information.

The reaction is exothermic.
The reaction occurs with a decrease in the number of moles (3 moles of gas give 2 moles of gas).

Le Chatelier's principle suggests three ways to maximise the yield of sulphuric acid.

a) Low temperature

Since the forward reaction is exothermic, a low temperature favours the formation of sulphur trioxide, and so of sulphuric acid. However, too low a temperature slows down the reaction speed. So a compromise temperature of 450 to 500^\circ\mathrm{C} is used.

b) High pressure

Since the forward reaction produces fewer moles, high pressure shifts the equilibrium to the right. But maintaining a high pressure is very expensive. So a compromise pressure of 2\,\mathrm{atm} is used.

c) Adding excess oxygen

According to Le Chatelier's principle an increase in the concentration of \ce{O2} at equilibrium shifts the equilibrium to the right. However, adding too much \ce{O2} decreases the concentration of \ce{SO3}, which slows down the reaction rate. Thus a compromise 1:1 ratio of \ce{SO2} and \ce{O2} is used to obtain high yields.

The contact process. The contact chamber, where \ce{SO2} is oxidised to \ce{SO3} over \ce{V2O5}, is the most important step.

Conditions of the contact process
Condition	Ideal for yield	Problem	Value used (compromise)
Temperature	Low (reaction is exothermic)	Too low a temperature slows the reaction	450 to 500^\circ\mathrm{C}
Pressure	High (3 moles of gas give 2 moles)	Maintaining high pressure is very expensive	2\,\mathrm{atm}
\ce{O2} concentration	Excess \ce{O2} shifts the equilibrium to the right	Too much \ce{O2} lowers the concentration of \ce{SO3} and slows the reaction	\ce{SO2}:\ce{O2}=1:1

Formula sheet

Equilibrium constant (concentrations)

K_c=\dfrac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b}

Concentrations in \mathrm{mol\,dm^{-3}} at equilibrium for \ce{aA + bB <=> cC + dD}; products over reactants, each raised to its coefficient. Pure solids and liquids are left out.

Unit: depends on \Delta n

Equilibrium constant from rate constants

K_c=\dfrac{k_f}{k_r}

k_f and k_r are the rate constants of the forward and reverse reactions.

Unit: depends on the reaction

Equilibrium constant (partial pressures)

K_p=\dfrac{P_C^{\,c}\,P_D^{\,d}}{P_A^{\,a}\,P_B^{\,b}}

P_A, P_B, P_C, P_D are the partial pressures of the gases at equilibrium.

Unit: depends on \Delta n (for example \mathrm{atm^{-1}} when \Delta n=-1)

Change in the number of moles of gas

\Delta n=(c+d)-(a+b)

Moles of gaseous products minus moles of gaseous reactants in the balanced equation.

Unit: mol

Relation between K_p and K_c

K_p=K_c\,(RT)^{\Delta n}

R=0.0821\,\mathrm{dm^3\,atm\,K^{-1}\,mol^{-1}}; T is the temperature in kelvin.

Units: as for K_p and K_c

Equilibrium constant (moles)

K_n=\dfrac{n_C^{\,c}\,n_D^{\,d}}{n_A^{\,a}\,n_B^{\,b}}

n_A, n_B, n_C, n_D are the moles of each substance at equilibrium.

Unit: depends on \Delta n

Relation between K_p and K_n

K_p=K_n\left(\dfrac{P}{n}\right)^{\Delta n}

P is the pressure of the mixture at equilibrium; n is the total number of moles in the balanced equation.

Unit: as for K_p

Equilibrium constant (mole fractions)

K_x=\dfrac{x_C^{\,c}\,x_D^{\,d}}{x_A^{\,a}\,x_B^{\,b}}

x_A, x_B, x_C, x_D are the mole fractions of A, B, C and D at equilibrium.

Unit: none when \Delta n=0

Relation between K_p and K_x

K_p=K_x\,(P)^{\Delta n}

P is the pressure of the equilibrium mixture.

Unit: as for K_p

Key terms

Irreversible reaction: A reaction in which the reactants are completely consumed and converted into products; it stops when the limiting reactant is used up.
Reversible reaction: A reaction in which the products, once formed, can react to form the reactants again, so that it takes place in both the forward and the backward directions.
Chemical equilibrium: The state of a reversible reaction at which the composition of the reaction mixture does not change; the rates of the forward and reverse reactions are equal.
Dynamic equilibrium: An equilibrium in which individual molecules keep reacting but the rates of the forward and reverse reactions are the same, so the concentrations stay constant.
Macroscopic events: Phenomena that can be observed with the naked eye, such as a change in colour, heat, precipitate, volume or pressure.
Microscopic events: Phenomena that cannot be observed with the naked eye, such as collisions between molecules, breaking and forming of bonds and rearrangement of atoms.
Law of mass action: The rate of a chemical reaction is proportional to the product of the active masses (molar concentrations, each raised to its coefficient) of the reacting substances.
Active mass: The concentration of a reactant or product in \mathrm{mol\,dm^{-3}} for a dilute solution.
Equilibrium constant (K_c): The ratio k_f/k_r, equal to the product concentrations over the reactant concentrations at equilibrium, each raised to its coefficient; it depends only on temperature.
Equilibrium constant expression: The expression for K_c written from the balanced equation, with products in the numerator and reactants in the denominator.
Equilibrium position: A particular set of equilibrium concentrations that satisfies the equilibrium expression; a reaction has infinitely many positions but only one K_c at a given temperature.
K_p: The equilibrium constant written in terms of the partial pressures of the gases; K_p=K_c(RT)^{\Delta n}.
K_n: The equilibrium constant written in terms of the number of moles at equilibrium.
K_x: The equilibrium constant written in terms of mole fractions; K_p=K_x(P)^{\Delta n}.
Homogeneous equilibrium: An equilibrium in which all the reactants and products are in the same phase.
Heterogeneous equilibrium: An equilibrium involving more than one phase; pure solids and pure liquids are left out of the expression.
Spectrometric method: A physical method that follows the concentration of a substance by measuring how much ultraviolet, visible or infrared radiation it absorbs.
Le Chatelier's principle: If a change is imposed on a system at equilibrium, the position of the equilibrium shifts in a direction which tends to reduce that change.
Exothermic reaction: A reaction that liberates heat (\Delta H negative); heat is written on the product side.
Endothermic reaction: A reaction that absorbs heat (\Delta H positive); heat is written on the reactant side.
Catalyst: A substance that speeds up the forward and reverse reactions equally; it does not change the equilibrium concentrations but shortens the time needed to reach equilibrium.
Haber's process: The industrial synthesis of ammonia, \ce{N2 + 3H2 <=> 2NH3}, using about 400^\circ\mathrm{C}, 200 to 300\,\mathrm{atm} and an iron catalyst, with ammonia liquefied and removed.
Contact process: The industrial manufacture of sulphuric acid, in which \ce{SO2} is oxidised to \ce{SO3} over \ce{V2O5} at 450 to 500^\circ\mathrm{C} and 2\,\mathrm{atm}.

Multiple choice questionsp. 183–184

Choose the correct answer. The book prints no answer key for these questions; the keys below were worked out from the book's teaching pages.

K_c is independent of;

ATemperature
BPressureAnswer
CBoth temperature and pressure
DK_p

Why

K_c depends only on temperature. For a given reaction at a fixed temperature it has one value, and changing the pressure of the mixture moves the equilibrium position without changing K_c. Option C is wrong because temperature does change K_c. Option D is another equilibrium constant, not a factor on which K_c depends.

For which of the following reactions, K_c has no units of concentration?

A\ce{2A(g) <=> B(g)}
B\ce{A(g) <=> B(g)}Answer
C\ce{A(g) <=> 2B(g)}
D\ce{3A(g) <=> 2C(g)}

Why

The units of K_c are (\mathrm{mol\,dm^{-3}})^{\Delta n}, where \Delta n is the number of moles of products minus the number of moles of reactants (from the coefficients). Units cancel only when \Delta n=0. Option B gives \Delta n=1-1=0. Option A (\Delta n=-1), option C (\Delta n=+1) and option D (\Delta n=2-3=-1) all leave units.

iii

For the following reaction \ce{2A(g) + B(g) <=> 3C(g)}, we can write?

AK_c > K_p
BK_c < K_p
CK_p - K_c = 0Answer
DK_p - K_c = -1

Why

Here \Delta n = 3-(2+1)=0. Since K_p = K_c(RT)^{\Delta n} and (RT)^0=1, K_p=K_c, so K_p-K_c=0. Options A and B are wrong because the two constants are equal, not different. Option D is wrong because K_p and K_c differ only by the factor (RT)^{\Delta n}, which is 1 here, so their difference cannot be -1.

What is the effect of addition of catalyst on an equilibrium mixture?

AReverse reaction
BForward reaction
CEnthalpy of reaction
DIt has no effectAnswer

Why

A catalyst speeds up the forward and the reverse reaction to the same degree, so the equilibrium concentrations and K_c do not change. It only reduces the time needed to reach equilibrium. Options A and B wrongly suggest that one direction is favoured, and option C is a thermodynamic quantity that a catalyst does not change.

Note. The book prints the last option as "It has not effect"; it is read here as "It has no effect".

NaCl can be purified by passing HCl gas through the ____ solution of NaCl.

ADilute
BConcentratedAnswer
CHot
DCold

Why

HCl ionises to give Cl^- ions. In a concentrated (saturated) NaCl solution, adding Cl^- (a common ion) shifts the equilibrium \ce{NaCl(s) <=> Na+(aq) + Cl-(aq)} to the left, so pure NaCl crystallises and is separated from the impurities. A dilute solution is not saturated, so nothing precipitates. Temperature is not the factor used in this purification.

K_c = K_p when \Delta n is equal to;

AZeroAnswer
B+1
C-1
D-2

Why

K_p = K_c(RT)^{\Delta n}. When \Delta n=0, (RT)^0=1 and K_p=K_c. For \Delta n=+1, -1 or -2, the factor RT (about 24.5\,\mathrm{dm^3\,atm\,mol^{-1}} at 298 K) raises or lowers the value, so the two constants differ.

vii

Consider the following reaction: \ce{2SO2(g) + O2(g) <=> 2SO3(g)}, \Delta H^\circ=-198\,\mathrm{kJ}. Yield of sulphur trioxide can be increased by: (a) increasing pressure (b) increasing temperature (c) adding catalyst (d) increasing concentration of oxygen.

Aa, b
Ba, b, c
Ca, b, c, d
Da, dAnswer

Why

(a) Correct: 3 mol of gas on the left and 2 mol on the right, so high pressure shifts the equilibrium to the right. (b) Wrong: the forward reaction is exothermic, so a higher temperature shifts the equilibrium to the left and lowers the yield. (c) Wrong: a catalyst does not change the equilibrium position. (d) Correct: adding a reactant (oxygen) shifts the equilibrium to the right. Only a and d are correct, which is option 4 in the book.

Note. The book numbers these four combinations 1 to 4; they are keyed here as A to D. Option 4 of the book is the key.

Short questionsp. 184–187

23 marks

Define chemical equilibrium.

Solution

Chemical equilibrium is the state of a reversible reaction in which the composition of the reaction mixture does not change with time. At this stage the rate of the forward reaction equals the rate of the reverse reaction, so the concentrations of the reactants and products stay constant under the existing conditions.

The equilibrium is dynamic. The molecules keep reacting in both directions, but the two opposing processes are balanced, so no further net change is seen unless the system is disturbed.

33 marks

Define and explain the following terms: (i) microscopic events (ii) dynamic equilibrium (iii) Le Chatelier's principle (iv) heterogeneous equilibria (v) homogeneous equilibrium.

Solution

(i) Microscopic events

Microscopic events are changes at the level of particles that cannot be seen with the naked eye, such as collisions between molecules, breaking and forming of bonds, rearrangement of atoms in molecules, and loss or gain of electrons. Macroscopic changes (colour, heat, precipitate, pressure) are the result of many microscopic events happening at the same time.

(ii) Dynamic equilibrium

Dynamic equilibrium is the state of a reversible reaction in which the forward and reverse reactions take place at the same rate. The concentrations of reactants and products stay constant, yet the molecules are constantly reacting. The equilibrium is dynamic because the individual molecules keep reacting while the overall composition does not change.

(iii) Le Chatelier's principle

Le Chatelier's principle states that if a change is imposed on a system at equilibrium (concentration, pressure or temperature), the position of equilibrium shifts in the direction that tends to reduce that change. For example, adding \ce{CO2} to \ce{CO2(g) + H2(g) <=> CO(g) + H2O(g)} shifts the equilibrium to the right, which consumes some of the added \ce{CO2}.

(iv) Heterogeneous equilibria

Heterogeneous equilibria involve more than one phase, for example \ce{CaCO3(s) <=> CaO(s) + CO2(g)} or \ce{C(s) + H2O(g) <=> CO(g) + H2(g)}. The concentrations of pure solids and pure liquids are constant, so they are not included in the equilibrium constant expression.

(v) Homogeneous equilibrium

Homogeneous equilibrium is an equilibrium in which all the reactants and products are in the same phase, for example \ce{N2(g) + 3H2(g) <=> 2NH3(g)}, or the reaction of ethanoic acid with ethanol in the liquid phase, \ce{CH3COOH(l) + C2H5OH(l) <=> CH3COOC2H5(l) + H2O(l)}.

63 marks

Write K_c and K_p expressions for the following reactions.

Solution

Balance each equation first. Gases appear in both expressions; pure solids and liquids do not. The exponent of each concentration or pressure is its coefficient.

(i) \ce{SO2(g) + 1/2 O2(g) <=> SO3(g)}

K_c=\dfrac{[\ce{SO3}]}{[\ce{SO2}]\,[\ce{O2}]^{1/2}}

K_p=\dfrac{P_{\ce{SO3}}}{P_{\ce{SO2}}\,P_{\ce{O2}}^{1/2}}

(ii) \ce{H2O(g) + Cl2O(g) <=> 2HOCl(g)}

K_c=\dfrac{[\ce{HOCl}]^2}{[\ce{H2O}]\,[\ce{Cl2O}]}

K_p=\dfrac{P_{\ce{HOCl}}^2}{P_{\ce{H2O}}\,P_{\ce{Cl2O}}}

(iii) \ce{O3(g) <=> O2(g) + O(g)}

K_c=\dfrac{[\ce{O2}]\,[\ce{O}]}{[\ce{O3}]}

K_p=\dfrac{P_{\ce{O2}}\,P_{\ce{O}}}{P_{\ce{O3}}}

(iv) \ce{O3(g) <=> 3/2 O2(g)}

K_c=\dfrac{[\ce{O2}]^{3/2}}{[\ce{O3}]}

K_p=\dfrac{P_{\ce{O2}}^{3/2}}{P_{\ce{O3}}}

(v) \ce{Fe3O4(s) + H2(g) <=> 3FeO(s) + H2O(g)}

K_c=\dfrac{[\ce{H2O}]}{[\ce{H2}]}

K_p=\dfrac{P_{\ce{H2O}}}{P_{\ce{H2}}}

The solids Fe3O4 and FeO are left out of both expressions.

(vi) \ce{2NO(g) + Cl2(g) <=> 2NOCl(g)}

K_c=\dfrac{[\ce{NOCl}]^2}{[\ce{NO}]^2\,[\ce{Cl2}]}

K_p=\dfrac{P_{\ce{NOCl}}^2}{P_{\ce{NO}}^2\,P_{\ce{Cl2}}}

(vii) \ce{CaCO3(s) <=> CaO(s) + CO2(g)}

K_c=[\ce{CO2}]

K_p=P_{\ce{CO2}}

Only the gas appears; the two solids are left out.

(viii) \ce{C(s) + H2O(g) <=> CO(g) + H2(g)}

K_c=\dfrac{[\ce{CO}]\,[\ce{H2}]}{[\ce{H2O}]}

K_p=\dfrac{P_{\ce{CO}}\,P_{\ce{H2}}}{P_{\ce{H2O}}}

Solid carbon is left out of both expressions.

83 marks

Write the expression for K_c and K_p for the following processes: (a) Blue vitriol is deep blue solid copper(II) sulphate pentahydrate, which is heated to drive off water vapours to form white solid copper(II) sulphate. (b) The decomposition of solid phosphorus pentachloride to gaseous phosphorus trichloride and chlorine gas.

Solution

(a) Blue vitriol, \ce{CuSO4.5H2O(s) <=> CuSO4(s) + 5H2O(g)}

K_c=[\ce{H2O}]^5

K_p=P_{\ce{H2O}}^5

Both solids are left out. Only the water vapour, with coefficient 5, appears in the expressions.

(b) Decomposition of solid phosphorus pentachloride, \ce{PCl5(s) <=> PCl3(g) + Cl2(g)}

K_c=[\ce{PCl3}]\,[\ce{Cl2}]

K_p=P_{\ce{PCl3}}\,P_{\ce{Cl2}}

The solid \ce{PCl5} is left out, so the expressions contain only the two gases.

93 marks

Predict the shift in equilibrium position that will occur for each of the following processes when the volume is reduced.

Solution

(i) \ce{PCl3(g) + 3NH3(g) <=> P(NH2)3(g) + 3HCl(g)}

Count the gas molecules on each side.
\text{left}=1+3=4,\qquad \text{right}=1+3=4
The numbers are equal, so a change in pressure or volume does not favour either side. There is no shift in the equilibrium position.

(ii) \ce{2NO(g) + O2(g) <=> 2NO2(g)}

Count the gas molecules.
\text{left}=2+1=3,\qquad \text{right}=2
Reducing the volume raises the pressure. The system moves to the side with fewer molecules, which is the right-hand side. The equilibrium shifts towards the formation of \ce{NO2}.

(iii) \ce{4NH3(g) + 5O2(g) <=> 4NO(g) + 6H2O(g)}

Count the gas molecules.
\text{left}=4+5=9,\qquad \text{right}=4+6=10
Reducing the volume raises the pressure, so the system moves to the side with fewer molecules, which is the left-hand side. The equilibrium shifts towards \ce{NH3} and \ce{O2}.

103 marks

For each of the following reactions, predict how the value of K_c changes as the temperature is increased.

Solution

(a) \ce{N2(g) + O2(g) <=> 2NO(g)}, \Delta H^\circ=+180\,\mathrm{kJ}

The reaction is endothermic (\Delta H^\circ>0), so heat is treated as a reactant. Raising the temperature shifts the equilibrium to the right, and K_c increases.

(b) \ce{2SO2(g) + O2(g) <=> 2SO3(g)}, \Delta H^\circ=-198\,\mathrm{kJ}

The reaction is exothermic (\Delta H^\circ<0), so heat is a product. Raising the temperature shifts the equilibrium to the left, the [\ce{SO3}] falls and K_c decreases.

The reaction is endothermic. Raising the temperature shifts the equilibrium to the right, so K_c increases.

(d) \ce{CH4(g) + H2O(g) <=> CO(g) + 3H2(g)}, \Delta H^\circ=+256\,\mathrm{kJ}

The reaction is endothermic. Raising the temperature shifts the equilibrium to the right, so K_c increases.

Note. In part (b) the book prints SO3 on both sides of the equation. It is read here as 2SO2 + O2 <=> 2SO3, which is the reaction the book uses at p. 179 with the same enthalpy. Part (c) gives 58 kJ, while the text on p. 180 gives 57.2 kJ for the same reaction.

113 marks

What is the difference between an equilibrium with a K_c value larger than one compared with an equilibrium that has a K_c smaller than one?

Solution

The magnitude of K_c shows the position of equilibrium. When K_c>1, the numerator (the product concentrations) is larger than the denominator (the reactant concentrations) at equilibrium, so the products are present in greater concentration than the reactants. The equilibrium lies towards the products.

When K_c<1, the denominator is larger than the numerator, so the reactants are present in greater concentration than the products at equilibrium. The equilibrium lies towards the reactants. In both cases the equilibrium constant gives the position, and it does not depend on the starting concentrations.

122025 Annual I, Set B, Q.4 (i) OR2025 Annual I, Set B, Q.4 (iii) OR3 marks

Describe the behaviour of the following equilibria with the stated changes: (a) increasing pressure on \ce{C3H8(g) + 5O2(g) <=> 3CO2(g) + 4H2O(g)}; (b) adding \ce{I2(g)} to \ce{2HI(g) <=> I2(g) + H2(g)}; (c) removing heat from \ce{CO2(g) <=> CO(g) + 1/2 O2(g)}, \Delta H^\circ=+284\,\mathrm{kJ}; (d) decreasing pressure on \ce{C2H6(g) <=> C2H4(g) + H2(g)}.

Solution

(a) Increasing pressure on \ce{C3H8(g) + 5O2(g) <=> 3CO2(g) + 4H2O(g)}

Count the gas molecules.
\text{left}=1+5=6,\qquad \text{right}=3+4=7
Increasing the pressure favours the side with fewer molecules, which is the left. The equilibrium shifts to the left, towards \ce{C3H8} and \ce{O2}. K_c does not change.

(b) Adding \ce{I2(g)} to \ce{2HI(g) <=> I2(g) + H2(g)}

Adding a product raises its concentration, so the system moves to consume the added \ce{I2}. The equilibrium shifts to the left, and more \ce{HI} is formed at the expense of \ce{H2} and \ce{I2}.

The reaction is endothermic, so heat is a reactant. Removing heat is like removing a reactant, so the equilibrium shifts to the left, towards \ce{CO2}. The concentrations of \ce{CO} and \ce{O2} fall.

(d) Decreasing pressure on \ce{C2H6(g) <=> C2H4(g) + H2(g)}

Count the gas molecules.
\text{left}=1,\qquad \text{right}=1+1=2
Decreasing the pressure favours the side with more molecules, which is the right. The equilibrium shifts to the right, towards \ce{C2H4} and \ce{H2}.

133 marks

Consider the following gas phase reaction \ce{SO2(g) + Cl2(g) <=> SO2Cl2(g) + Heat}. Describe four changes that would drive the equilibrium to the left.

Solution

The forward reaction is exothermic (heat is on the right), and it has 2 mol of gas on the left and 1 mol on the right. Four changes shift the equilibrium to the left:

Increase the temperature. Heat added to the system is absorbed by the endothermic (reverse) reaction, which shifts the equilibrium to the left.
**Add more \ce{SO2Cl2}.** The added product is consumed when the equilibrium shifts to the left.
**Remove \ce{SO2} or \ce{Cl2}.** Decreasing a reactant concentration shifts the equilibrium towards the reactants, which is the left.
Decrease the pressure (increase the volume). The left side has 2 mol of gas and the right side has 1 mol, so the system moves to the side with more molecules, which is the left.

Removing \ce{SO2Cl2}, adding \ce{SO2} or \ce{Cl2}, cooling, or increasing the pressure would all shift the equilibrium to the right.

143 marks

How would you change the volume of the following reactions to increase the yield of products? (i) \ce{Cl2(g) + H2(g) <=> 2HCl(g)}; (ii) \ce{2NO2(g) <=> N2O4(g)}.

Solution

(i) \ce{Cl2(g) + H2(g) <=> 2HCl(g)}

Count the gas molecules.
\text{left}=1+1=2,\qquad \text{right}=2
The numbers of gas molecules are equal on both sides, so a change of volume (and of pressure) does not shift the equilibrium. No change in volume will increase the yield of \ce{HCl}.

(ii) \ce{2NO2(g) <=> N2O4(g)}

Count the gas molecules.
\text{left}=2,\qquad \text{right}=1
The product side has fewer molecules. Reducing the volume raises the pressure, and the equilibrium moves to the side with fewer molecules, which is the right-hand side. So the volume should be reduced (the pressure increased) to increase the yield of \ce{N2O4}.

153 marks

Potassium dichromate solution has a beautiful clear orange colour. This is due to the colour of the dichromate ion, \ce{Cr2O7^2-}. When the salt is dissolved in water, the following equilibrium is set up, on heating the solution: \ce{Cr2O7^2-(aq) + H2O(l) <=> 2CrO4^2-(aq) + 2H+(aq)} (orange; yellow). What will happen if: (i) dilute sodium hydroxide is added to this solution; (ii) dilute hydrochloric acid is added to this solution?

Solution

(i) Dilute sodium hydroxide is added

Sodium hydroxide supplies \ce{OH-} ions, which react with \ce{H+} ions to form water, \ce{H+ + OH- -> H2O}. The concentration of \ce{H+}, a product, falls. The equilibrium shifts to the right to replace the removed \ce{H+}, so more \ce{CrO4^2-} forms and the solution turns yellow.

(ii) Dilute hydrochloric acid is added

Hydrochloric acid adds \ce{H+} ions, a product. The equilibrium shifts to the left to consume the added \ce{H+}, so more \ce{Cr2O7^2-} forms and the solution stays orange (or becomes more orange).

163 marks

For the reaction between hydrogen and iodine to form hydrogen iodide, \ce{H2(g) + I2(g) <=> 2HI(g)}, the value of K_c is 794 at 298 K but 54 at 700 K. What can you deduce from this information?

Solution

Only a change in temperature changes the value of K_c. Here K_c falls from 794 to 54 as the temperature rises from 298 K to 700 K. The equilibrium therefore moves towards the reactants when the temperature is raised.

Values of K_c for \ce{H2 + I2 <=> 2HI} at two temperatures.
Temperature	K_c	Position of equilibrium
298 K	794	Far to the right, [\ce{HI}] much greater than [\ce{H2}] and [\ce{I2}]
700 K	54	Still to the right, but less \ce{HI} is present

A fall in K_c with rising temperature means that heat is a product of the forward reaction. The forward reaction is therefore exothermic (\Delta H^\circ<0), and the reverse reaction is endothermic. Because K_c is large at 298 K, the reaction gives a high yield of \ce{HI} at low temperature.

173 marks

A propose a set of conditions (temperature, pressure and concentration) for a reversible reaction where the enthalpy change is positive.

Solution

A reaction with a positive enthalpy change is endothermic, so heat is a reactant. The example used here is \ce{N2O4(g) <=> 2NO2(g)} with \Delta H^\circ=+57.2\,\mathrm{kJ}, the reaction in the book's text.

Temperature: high

Raising the temperature supplies heat, which the endothermic forward reaction absorbs. The equilibrium shifts to the right and K_c increases. In the book's example K_c rises from 7.7\times10^{-5} at 0\,^\circ\mathrm{C} to 0.4 at 100\,^\circ\mathrm{C}.

Pressure: low

The forward reaction gives two gas molecules from one, so it increases the number of molecules. Lowering the pressure (increasing the volume) shifts the equilibrium to the side with more molecules, which favours \ce{NO2}.

Concentration: high reactant, product removed

Keeping the concentration of \ce{N2O4} high, or removing \ce{NO2} as it forms, shifts the equilibrium to the right. Adding a reactant or removing a product both favour the forward reaction.

Note. The question begins with "A propose"; it is read as "Propose".

Long questionsp. 184

42026 Annual I, Set D, Q.4 OR6 marks

Explain the industrial application of Le Chatelier's principle using Haber's process as an example.

Solution

Equation and information

\ce{N2(g) + 3H2(g) <=> 2NH3(g)} \qquad \Delta H^\circ=-92.46\,\mathrm{kJ}

The reaction is exothermic.
The number of gas molecules decreases: 4 mol (1 \ce{N2} and 3 \ce{H2}) give 2 mol of \ce{NH3}.

Le Chatelier's principle suggests the conditions that give the maximum yield of ammonia. Each condition is chosen to shift the equilibrium towards \ce{NH3}, while the rate is kept high enough for the process to be economic.

Low temperature

The forward reaction is exothermic, so a low temperature favours the formation of ammonia. However, at a low temperature the rate of reaction is very slow and the process becomes uneconomic. A compromise temperature of about 400\,^\circ\mathrm{C} is therefore used.

High pressure

Four molecules of gas react to give two molecules of ammonia. High pressure shifts the equilibrium to the right, towards \ce{NH3}. Very high pressure is expensive to maintain, so a pressure of 200 to 300\,\mathrm{atm} is used.

Catalyst

A piece of iron with other metal oxides is used as the catalyst. The catalyst does not change the equilibrium position; it increases the rate at which equilibrium is reached. The equilibrium mixture still contains about 35% \ce{NH3} by volume.

Continual removal of ammonia

The mixture is cooled by refrigeration coils until ammonia condenses at -33.4\,^\circ\mathrm{C} and is removed as a liquid. The stress of removing \ce{NH3} shifts the equilibrium towards more ammonia. The unreacted \ce{N2} and \ce{H2}, which do not liquefy at this temperature, are recycled into the reaction chamber. In this way practically 100% conversion of \ce{N2} and \ce{H2} to \ce{NH3} is possible.

Summary

Factor	Condition used	Effect
Temperature	About 400\,^\circ\mathrm{C}	Low temperature favours \ce{NH3}, but the rate is too slow, so a compromise is used
Pressure	200 to 300\,\mathrm{atm}	High pressure shifts the equilibrium to the right (4 mol to 2 mol)
Catalyst	Iron with other metal oxides	Speeds up the attainment of equilibrium; no shift in position
Removal of \ce{NH3}	Liquefied at -33.4\,^\circ\mathrm{C}; \ce{N2} and \ce{H2} recycled	Shifts the equilibrium to the right; near 100% conversion

56 marks

Propose microscopic events that account for the observed macroscopic changes that take place during a shift in equilibrium.

Solution

Macroscopic change

A macroscopic change is one that is seen without looking at the particles: a change in colour, in the concentration of a substance, or in pressure. When a system at equilibrium is disturbed, these properties change for a time and then become constant again at a new equilibrium position.

Microscopic events

Molecules are always colliding. The rate of a reaction depends on the number of effective collisions, which are the collisions that lead to the formation of new bonds between the particles.
At equilibrium the number of effective collisions for the forward reaction equals the number for the reverse reaction.
A change in concentration, pressure or temperature changes the frequency or energy of collisions, so one rate becomes larger than the other. The equilibrium then shifts until the two rates are equal again.

Example: adding CO2 to CO2 + H2 <=> CO + H2O

At equilibrium the forward and reverse effective collisions are equal.
\ce{CO2(g) + H2(g) <=> CO(g) + H2O(g)}
Adding \ce{CO2} raises its concentration. More \ce{CO2} and \ce{H2} molecules collide per second, so the number of effective collisions for the forward reaction increases.
More \ce{CO} and \ce{H2O} are formed. As their concentrations rise, the reverse reaction also gains effective collisions.
As \ce{CO2} and \ce{H2} are used up, the forward rate falls, while the reverse rate increases. When the two rates are equal again, equilibrium is re-established with higher concentrations of \ce{CO} and \ce{H2O}.

The macroscopic observation, a higher concentration of \ce{CO} and \ce{H2O} at the new equilibrium, is therefore the result of this changing balance of effective collisions.

Example: change of pressure

Reducing the volume of a gaseous mixture brings the molecules closer together, so collisions become more frequent. The reaction moves to the side with fewer molecules, because the effective collisions on that side are then more frequent. The equilibrium is re-established when the forward and reverse effective collisions are once again equal.

Summary

Every macroscopic shift in equilibrium is the result of a temporary imbalance in the number of effective collisions for the two opposing reactions. Once the two rates are equal, the macroscopic properties become constant again.

Numerical problemsp. 185

At a particular temperature K_p = 0.133\,\mathrm{atm}. Which of the following conditions corresponds to equilibrium position for the reaction \ce{A(g) <=> 2B(g)}? (a) P_B=0.175\,\mathrm{atm}, P_A=0.102\,\mathrm{atm}; (b) P_B=0.064\,\mathrm{atm}, P_A=0.0308\,\mathrm{atm}; (c) P_B=0.144\,\mathrm{atm}, P_A=0.156\,\mathrm{atm}.

Solution

Given

K_p = 0.133\,\mathrm{atm}

To find

which of the conditions (a), (b), (c) is the equilibrium position

A mixture is at equilibrium when the reaction quotient Q_p, calculated from the given partial pressures, equals K_p. For \ce{A(g) <=> 2B(g)} the expression is Q_p=\dfrac{P_B^2}{P_A}.

(a) P_B=0.175\,\mathrm{atm}, P_A=0.102\,\mathrm{atm}

Substitute the pressures into Q_p.
Q_p=\dfrac{(0.175)^2}{0.102}=\dfrac{0.030625}{0.102}=0.300\,\mathrm{atm}
Compare with K_p=0.133\,\mathrm{atm}. Q_p>K_p, so the mixture is not at equilibrium; the reverse reaction would take place until Q_p=K_p.
0.300\neq0.133

\text{Not at equilibrium}\ (Q_p=0.300\,\mathrm{atm})

(b) P_B=0.064\,\mathrm{atm}, P_A=0.0308\,\mathrm{atm}

Substitute the pressures into Q_p.
Q_p=\dfrac{(0.064)^2}{0.0308}=\dfrac{0.004096}{0.0308}=0.133\,\mathrm{atm}
Q_p equals K_p within the rounding of the data, so the mixture is at equilibrium.
Q_p=K_p=0.133\,\mathrm{atm}

\text{At equilibrium}\ (Q_p=0.133\,\mathrm{atm})

Substitute the pressures into Q_p.
Q_p=\dfrac{(0.144)^2}{0.156}=\dfrac{0.020736}{0.156}=0.133\,\mathrm{atm}
Q_p=K_p to three significant figures, so the mixture is at equilibrium.
Q_p=K_p=0.133\,\mathrm{atm}

\text{At equilibrium}\ (Q_p=0.133\,\mathrm{atm})

\text{Conditions (b) and (c) correspond to equilibrium.}

Book answer. (b) and (c)

Projectp. 187

Project 1

Create a presentation where students explain the equilibrium concept, and discuss the factors that influence equilibrium position.

Solution

This is an open project, so the answer is a plan for the presentation, built from the book's own content.

Reversible reactions and dynamic equilibrium: equal forward and reverse rates, constant concentrations (the rate plot of the unit).
Law of mass action: write K_c from a balanced equation, and what the value of K_c tells about the position.
Le Chatelier's principle with one example each: concentration (\ce{CO2 + H2 <=> CO + H2O}), pressure (\ce{2SO2 + O2 <=> 2SO3}), temperature (\ce{N2O4 <=> 2NO2}) and catalyst (no shift, faster approach to equilibrium).
Industrial applications: Haber's process and the contact process, with the conditions chosen for each.
A short quiz using the multiple choice questions of this unit.

Project 2

Investigate examples of chemical equilibrium in natural systems and prepare a report on it.

Solution

The book gives no list of natural examples and no answer key, so the answer is a method for the report.

Introduce chemical equilibrium as a dynamic state of a reversible reaction with constant composition.
Choose two or three natural systems and write each as a reversible reaction, with its balanced equation and K_c expression (pure solids and liquids are left out).
For each system, state which factor (concentration, pressure, temperature) would shift its position, and in which direction, using Le Chatelier's principle.
State your sources for each example. Check that every equation is balanced and that each factor is applied correctly.

Note. Open-ended research task. The answer depends on the sources the student uses.

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2025 Annual I, Set B, MCQ 61 mark

The following equilibrium will NOT be shifted in the forward direction by: \ce{N2 + 3H2 <=> 2NH3 + \Delta}

AIncreasing temperatureAnswer
BIncreasing pressure
CAdding \ce{H2} gas
DRemoving \ce{NH3}

Why

The forward reaction of \ce{N2 + 3H2 <=> 2NH3 + \Delta} gives out heat, so it is exothermic. Raising the temperature adds heat, and the system opposes this by shifting to the left, the endothermic direction, so the forward direction is not favoured.

The others all push the equilibrium forward: higher pressure favours the side with fewer gas molecules (4 mol to 2 mol), extra \ce{H2} is used up, and removing \ce{NH3} makes more of it form.

2026 Annual I, Set B, Q.2 (viii) OR3 marks

State the necessary conditions for equilibrium.

Solution

Chemical equilibrium is reached by a reversible reaction when the following conditions are met.

A reversible reaction. The products can react to give back the reactants, so the reaction goes in the forward and the reverse directions at the same time (shown by \rightleftharpoons).
A closed system at constant temperature. Nothing enters or leaves the reaction mixture, and the temperature is not changed, because K_c has only one value at a given temperature.
Equal rates. The rate of the forward reaction becomes equal to the rate of the reverse reaction.

R_f=R_r\quad\Rightarrow\quad k_f[\mathrm{A}]^a[\mathrm{B}]^b=k_r[\mathrm{C}]^c[\mathrm{D}]^d

When these conditions are met, the concentrations of all the reactants and products stay constant (the colour, pressure and composition no longer change), and the ratio of the concentrations gives the constant K_c.

K_c=\dfrac{k_f}{k_r}=\dfrac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b}

Note. The book does not list 'conditions for equilibrium' as a separate heading; the answer is built from its sections on dynamic equilibrium and the law of mass action.

2025 Annual I, Set B, Q.2 (xiii)3 marks

Calculate K_c for the following reaction at equilibrium at 100 ^{\circ}C: \ce{2SO2(g) + O2(g) <=> 2SO3(g)} with [\ce{SO2}]=0.59\ \mathrm{M}, [\ce{O2}]=0.05\ \mathrm{M}, [\ce{SO3}]=0.259\ \mathrm{M}

Solution

Given

[\ce{SO2}] = 0.59\,\mathrm{M}
[\ce{O2}] = 0.05\,\mathrm{M}
[\ce{SO3}] = 0.259\,\mathrm{M}
T = 100^\circ\mathrm{C}

To find

Write the equilibrium constant expression. The product \ce{SO3} has coefficient 2 and the reactant \ce{SO2} has coefficient 2, so both are squared.
K_c=\dfrac{[\ce{SO3}]^2}{[\ce{SO2}]^2[\ce{O2}]}
Substitute the equilibrium concentrations.
K_c=\dfrac{(0.259)^2}{(0.59)^2(0.05)}
Evaluate the top and the bottom separately.
K_c=\dfrac{0.0671}{0.3481\times0.05}=\dfrac{0.0671}{0.01741}
Divide. The units are (\mathrm{mol\,dm^{-3}})^2 on top and (\mathrm{mol\,dm^{-3}})^3 below, so one factor of concentration remains in the denominator.
K_c=3.85\ \mathrm{dm^3\,mol^{-1}}

K_c=3.85\ \mathrm{dm^3\,mol^{-1}}

2026 Annual I, Set B, Q.4 OR2+2+2 = 6 marks

Formation of methanol is an important industrial reaction in the processing of new fuels. \ce{CO + 2H2 <=> CH3OH} \Delta H=-90.7\ \mathrm{kJ\,mol^{-1}}. Explain three factors that can increase the yield of methanol in this reaction.

Solution

Information from the equation

\ce{CO(g) + 2H2(g) <=> CH3OH(g)}\qquad \Delta H=-90.7\,\mathrm{kJ\,mol^{-1}}

The forward reaction is exothermic (heat is a product).
The number of gas molecules decreases: 1+2=3 mol of gas on the left and 1 mol on the right.

By Le Chatelier's principle, three changes shift the equilibrium towards methanol and so increase its yield.

Factor 1: high pressure (2 marks)

Three molecules of gas (\ce{CO} and 2\,\ce{H2}) give one molecule of \ce{CH3OH}. Increasing the pressure makes the system reduce its volume by moving to the side with fewer gas molecules, so the equilibrium shifts to the right and more methanol is formed. (The book's Concept assessment 8.5 says the same: decreasing the pressure would lower the yield.)

Factor 2: low temperature (2 marks)

The forward reaction is exothermic, so heat can be written as a product.

\ce{CO(g) + 2H2(g) <=> CH3OH(g) + heat}

Removing heat (lowering the temperature) makes the system produce heat, so the equilibrium shifts to the right and the value of K_c increases. Too low a temperature makes the reaction too slow, so a moderate temperature with a catalyst is used in practice; the catalyst only makes equilibrium come sooner and does not change the yield.

Factor 3: concentration (2 marks)

Adding more \ce{CO} or \ce{H2}, or continually removing the methanol as it is formed, shifts the equilibrium to the right. Removing the product keeps the reaction moving towards methanol. K_c does not change by a change of concentration.

Factor	Change	Shift	Result
Pressure	Increase	Towards 1 mol of gas (right)	More \ce{CH3OH}
Temperature	Decrease (exothermic reaction)	Right, K_c increases	More \ce{CH3OH}
Concentration	Add \ce{CO} or \ce{H2}, or remove \ce{CH3OH}	Right	More \ce{CH3OH}

2025 Annual I, Set B, Q.4 (i) OR2 marks

Discuss the effect of increasing pressure on the equilibrium \ce{C3H8 + 5O2 <=> 3CO2 + 4H2O}

Solution

Count the gas molecules on each side of the equation.
\text{left}=1+5=6,\qquad\text{right}=3+4=7
Increasing the pressure favours the side with the fewer gas molecules, which is the left side.

The equilibrium shifts to the left, towards \ce{C3H8} and \ce{O2}, so less \ce{CO2} and \ce{H2O} is present. The value of K_c does not change, because only temperature changes K_c.

2025 Annual I, Set B, Q.4 (ii) OR2 marks

Discuss the effect of removing \ce{SO3} on the equilibrium \ce{2SO2 + O2 <=> 2SO3}

Solution

Removing a product lowers its concentration, so the system tries to replace it. The forward reaction is favoured, and the equilibrium shifts to the right (towards \ce{SO3}).

\ce{2SO2 + O2 <=> 2SO3}\quad(\ce{SO3}\text{ removed: shift to the right})

More \ce{SO2} and \ce{O2} are used up and more \ce{SO3} is formed, until equilibrium is re-established. K_c does not change.

2025 Annual I, Set B, Q.4 (iii) OR2 marks

Discuss the effect of decreasing temperature on the equilibrium \ce{CO2 + \Delta <=> CO + 1/2 O2}

Solution

The reaction is endothermic (heat \Delta is on the reactant side), so heat is a reactant.

\ce{CO2 + \Delta <=> CO + 1/2 O2}

Decreasing the temperature removes heat, and the system opposes this by favouring the heat-producing (exothermic) direction, the reverse reaction. The equilibrium shifts to the left, towards \ce{CO2}, and the concentrations of \ce{CO} and \ce{O2} fall. The value of K_c decreases.

2025 Annual I, Set D, MCQ 71 mark

Which of the following reaction at equilibrium will NOT be affected by change in pressure?

A\ce{N2O4 <=> 2NO2}
B\ce{CO + 2H2 <=> CH3OH}
C\ce{N2 + O2 <=> 2NO}Answer
D\ce{N2 + 3H2 <=> 2NH3}

Why

Pressure changes the position of an equilibrium only when the number of gas molecules is different on the two sides (\Delta n\neq0). In \ce{N2 + O2 <=> 2NO} there are 2 mol of gas on each side, so a change in pressure favours neither direction.

The other three have unequal numbers of gas molecules: \ce{N2O4 <=> 2NO2} (1 to 2), \ce{CO + 2H2 <=> CH3OH} (3 to 1) and \ce{N2 + 3H2 <=> 2NH3} (4 to 2), so each is affected.

2026 Annual I, Set D, Q.2 (viii) OR1.5+1.5 = 3 marks

Give two possible ways to recognize equilibrium.

Solution

At equilibrium the concentrations of reactants and products stop changing, so any measurable property of the mixture that depends on them also stays constant. The book gives two ways to recognise this.

1. Physical method (1.5 marks)

A physical property of the mixture is measured during the reaction without removing a sample. For example, in \ce{N2O4(g) <=> 2NO2(g)} colourless \ce{N2O4} changes to reddish brown \ce{NO2}. A spectrometer measures the absorbance of radiation, which is proportional to [\ce{NO2}]. When the absorbance stops changing and stays at a constant value, the system is at equilibrium.

2. Chemical method (1.5 marks)

The amount of a substance is found by a chemical reaction. In \ce{CH3COOH + C2H5OH <=> CH3COOC2H5 + H2O} a small portion of the mixture is withdrawn at regular intervals and the acetic acid in it is titrated against standard \ce{NaOH} with phenolphthalein. The concentration of acetic acid falls and then becomes constant; this shows that equilibrium has been reached.

2025 Annual I, Set D, Q.43+3 = 6 marks

Explain one physical and one chemical method to recognize a system at equilibrium.

Solution

Both methods follow how a property of the reaction mixture changes with time. When it stops changing, equilibrium has been reached.

Physical method: the spectrometric method (3 marks)

A physical property of the mixture is measured during the reaction without removing a sample. The reactant or product absorbs ultraviolet, visible or infrared radiation, and its concentration is found from the amount of radiation absorbed.

\ce{N2O4(g) <=> 2NO2(g)}

\ce{N2O4} is a colourless gas and \ce{NO2} is a reddish brown gas. The absorbance, which is proportional to [\ce{NO2}], is measured at regular intervals. It rises as \ce{NO2} forms and then stays at a constant value, which shows that equilibrium has been reached.

Absorbance during the reaction (schematic)

Absorbance during the reaction (schematic): it rises and then stays constant at equilibrium.

Chemical method: titration (3 marks)

The amount of a reactant or product is found by a suitable chemical reaction. Take the reversible reaction between acetic acid and ethanol.

\ce{CH3COOH(l) + C2H5OH(l) <=> CH3COOC2H5(l) + H2O(l)}

Mix a mol of acetic acid and b mol of ethanol in a stoppered flask at room temperature, and add a little mineral acid as a catalyst.
At regular intervals withdraw a small portion of the mixture.
Titrate the acetic acid in the portion against a standard solution of \ce{NaOH}, using phenolphthalein as the indicator.
The concentration of acetic acid falls until it becomes constant. A constant titre means equilibrium has been reached.

Point	Physical method	Chemical method
Measured	Absorbance of radiation	Amount of acetic acid by titration
Sample	Not removed	A portion is withdrawn each time
Sign of equilibrium	Constant absorbance	Constant concentration of acetic acid

2026 Annual I, Set D, Q.4 OR2+2+2 = 6 marks

Manufacturing of ammonia by Haber's process is represented by the following equation. \ce{N2 + 3H2 <=> 2NH3} \Delta H=-92.46\ \mathrm{kJ\,mol^{-1}}. Explain three factors that can increase the yield of ammonia by this reaction.

Solution

Information from the equation

\ce{N2(g) + 3H2(g) <=> 2NH3(g)}\qquad \Delta H=-92.46\,\mathrm{kJ\,mol^{-1}}

The reaction is exothermic.
The number of gas molecules decreases: 1+3=4 mol of gas give 2 mol of \ce{NH3}.

Le Chatelier's principle suggests three ways to increase the yield of ammonia.

Factor 1: low temperature (2 marks)

The forward reaction is exothermic, so a low temperature shifts the equilibrium to the right and favours the formation of \ce{NH3}. At a very low temperature the rate is so slow that the process is uneconomic, so a compromise temperature of about 400\,^\circ\mathrm{C} is used together with an iron catalyst (with other metal oxides), which raises the rate but does not change the position of equilibrium.

Factor 2: high pressure (2 marks)

Four molecules of gas give two molecules of \ce{NH3}. High pressure shifts the equilibrium to the side with fewer molecules, that is to the right, towards \ce{NH3}. A pressure of 200 to 300\,\mathrm{atm} is used, because a still higher pressure is too expensive to maintain.

Factor 3: continual removal of ammonia (2 marks)

The equilibrium mixture is cooled by refrigeration coils until ammonia condenses at -33.4\,^\circ\mathrm{C} and is removed as a liquid. Removing a product shifts the equilibrium to the right, so more \ce{NH3} is formed. The unliquefied \ce{N2} and \ce{H2} are recycled into the reaction chamber, and practically 100\% conversion is possible.

Haber's process: ammonia is liquefied and removed in the cooler, and the unreacted gases are recycled.

Factor	Condition used	Effect
Temperature	About 400\,^\circ\mathrm{C}	Low temperature favours \ce{NH3} (exothermic); compromise for the rate
Pressure	200 to 300\,\mathrm{atm}	Shifts to the right (4 mol to 2 mol)
Removal of \ce{NH3}	Liquefied at -33.4\,^\circ\mathrm{C}	Shifts to the right; \ce{N2} and \ce{H2} recycled

Exam focus

Explain how the yield of a product can be increased by changing temperature, pressure and concentration for a given reaction (for example methanol from \ce{CO + 2H2}, or ammonia by Haber's process). (long)
A 6-mark Section C question in both 2026 sets (Set B Q.4 OR: methanol; Set D Q.4 OR: Haber's process), and book long question 1.
Go to answer
Predict the shift of an equilibrium when pressure, temperature or concentration is changed (increase pressure on \ce{C3H8 + 5O2 <=> 3CO2 + 4H2O}, remove \ce{SO3}, lower the temperature of an endothermic reaction). (short)
Asked as three 2-mark parts in 2025 (Set B Q.4 (i) to (iii) OR), and in book short question 8 and short questions 5, 6, 9 and 10.
Go to answer
Say which equilibria are, and which are not, affected by a change in pressure (compare gas molecules on the two sides). (mcq)
Asked in 2025 (Set D MCQ 7 and Set B MCQ 6), and book exercise MCQ 7.
Go to answer
Describe the physical (spectrometric) and chemical (titration) methods to recognise a system at equilibrium. (long)
A 6-mark question in 2025 (Set D Q.4) and a 3-mark one in 2026 (Set D Q.2 (viii) OR: two ways to recognise equilibrium).
Go to answer
Calculate K_c (or K_p) from equilibrium concentrations, with units. (numerical)
Asked in 2025 (Set B Q.2 (xiii): K_c for \ce{2SO2 + O2 <=> 2SO3}); the same data appear in the book's Concept assessment 8.2.
Go to answer
Define dynamic equilibrium and state the conditions for equilibrium (equal rates, constant concentrations). (short)
Asked in 2026 (Set B Q.2 (viii) OR: necessary conditions for equilibrium); the 2025 Set D Q.2 (ii) OR question on dynamic equilibrium is solved in unit 5.
Go to answer
Derive the K_c expression from the law of mass action and write K_c and K_p for homogeneous and heterogeneous reactions. (long)
The base of every numerical in the unit and of the book's short questions 3 and 4; it was not asked as a separate question in the 2025 or 2026 papers.
Go to answer
Explain the contact process for sulphuric acid and the effect of a catalyst on equilibrium. (short)
Book exercise MCQ 4 and MCQ 7 use these ideas; the contact process was not asked directly in the 2025 or 2026 papers.
Go to answer

8.1Macroscopic events and microscopic events

Macroscopic events

Microscopic events

8.2Reversible reactions and dynamic equilibriumOften examined

Approach to equilibrium: steam and carbon monoxide

Why equilibrium is dynamic

8.3The law of mass actionOften examined

Derivation The equilibrium constant expression Kc​

Conditions for equilibrium

Examples of equilibrium constant expressions

Concept assessment exercise 8.1

8.3.1Equilibrium expressions involving partial pressure, number of moles and mole fractionOften examined

Equilibrium constant K_p

Equilibrium constant K_n

Equilibrium constant K_x

Derivation How K_p is related to K_c, K_n and K_x

Concept assessment exercise 8.2

8.4Types of equilibrium

i. Homogeneous equilibria

ii. Heterogeneous equilibria

Concept assessment exercise 8.3

8.5Ways to recognize equilibrium

a) Physical method (spectrometric method)

b) Chemical method

8.6Factors affecting equilibrium (the Le Chatelier's principle)Often examined

8.6.1The effect of a change in concentrationOften examined

The microscopic events

Concept assessment exercise 8.4

8.6.2The effect of pressure changeOften examined

Equal numbers of molecules on both sides

Concept assessment exercise 8.5

8.6.3The effect of change in temperatureOften examined

An exothermic reaction

An endothermic reaction

Concept assessment exercise 8.6

8.6.4The effect of addition of catalyst

8.7Industrial application of Le Chatelier's principleOften examined

8.7.1Synthesis of ammonia by Haber's processOften examined

i. Low temperature

ii. High pressure

iii. Continual removal of ammonia

8.7.2Synthesis of sulphuric acid by contact processOften examined

i. Formation of sulphur dioxide

ii. Oxidation of sulphur dioxide

iii. Conversion of sulphur trioxide into sulphuric acid

a) Low temperature

b) High pressure

c) Adding excess oxygen

Formula sheet

Key terms

Multiple choice questionsp. 183–184

Short questionsp. 184–187

Temperature: high

Pressure: low

Concentration: high reactant, product removed

Long questionsp. 184

Equation and information

Low temperature

High pressure

Catalyst

Continual removal of ammonia

Summary

Macroscopic change

Microscopic events

Example: adding CO2 to CO2 + H2 <=> CO + H2O

Example: change of pressure

Summary

Numerical problemsp. 185

Projectp. 187

FBISE 2025 and 2026 questions

Information from the equation

Factor 1: high pressure (2 marks)

Factor 2: low temperature (2 marks)

Factor 3: concentration (2 marks)

1. Physical method (1.5 marks)

2. Chemical method (1.5 marks)

Physical method: the spectrometric method (3 marks)

Chemical method: titration (3 marks)

Information from the equation

Factor 1: low temperature (2 marks)

Derivation The equilibrium constant expression $K_{c}$