Law of Chemical Equilibrium And Equilibrium Constant

ChemistryEquilibriumFor JEE aspirants

INTRODUCTION

It is an experimental fact that most of the process including chemical reactions, when carried out in a closed vessel, do not go to completion. They proceed to some extent leaving considerable amounts of reactants & products. When such stage is reached in a reaction, it is said that the reaction has attained the state of equilibrium. Equilibrium represents the state of a process in which the properties like temperature, pressure, concentration etc of the system do not show any change with passage of time. In all processes which attain equilibrium, two opposing processes are involved. Equilibrium is attained when the rates of the two opposing processes become equal.

If the opposing processes involve only physical changes, the equilibrium is called Physical Equilibrium. If the opposing processes are chemical reactions, the equilibrium is called Chemical Equilibrium.

Some common physical equilibria are:

Solid $⇌$ Liquid

Liquid $⇌$ Gas

Gas $⇌$ Solution

Generally, a chemical equilibrium is represented as

aA + bB $⇌$ xX + yY

Where A, B are reactants and X, Y are products.

Note: The double arrow between the left hand part and right hand part shows that changes is taking place in both the directions.

On the basis of extent of reaction, before equilibrium is attained chemical reactions may be classified into three categories.

(a) Those reactions which proceed to almost completion.

(b) Those reactions which proceed to almost only upto little extent.

(c) Those reactions which proceed to such an extent, that the concentrations of reactants and products at equilibrium are comparable.

EQUILIBRIUM IN PHYSICAL PROCESS

The different types of physical Equilibrium are briefly described below

(a) Solid – liquid Equilibrium

The equilibrium that exist between ice and water is an example of solid – liquid equilibrium. In a close system, at 0ºC ice and water attain equilibrium. At that point rate of melting of ice is equal to rate of freezing of water. The equilibrium is represented as

$H_{2} O_{(s)} ⇌ H_{2} O_{(ℓ)}$

(b) Liquid-Gas Equilibrium

Evaporation of water in a closed vessel is an example of liquid – gas equilibrium. Where rate of evaporation is equal to rate of condensation. The equilibrium is represented as

$H_{2} O_{(ℓ)} ⇌ H_{2} O_{(g)}$

(c) Solid – solution equilibrium

If you add more and more salt in water taken in a container of a glass and stirred with a glass rod, after dissolving of some amount. You will find out no further salt is going to the solution and it settles down at the bottom. The solution is now said to be saturated and in a state of equilibrium. At this stage, many molecule of salt from the undissolved salt go into the solution (dissolution) and same amount of dissolved salt are deposited back (Precipitation).

Thus, at equilibrium rate of dissolution is equal to rate of precipitation.

$S a l t (S o l i d) ⇌ S a l t (in so l u t i o n)$

(d) Gas –Solution equilibrium

Dissolution of a gas in a liquid under pressure in a closed vessel established a gas – liquid equilibrium. The best example of this type of equilibrium is cold drink bottles. The equilibrium that exists with in the bottle is

$C O_{2}_{(g)} ⇌ C O_{2} (in so l u t i o n)$

Equilibrium in Chemical Process (Reversible and irreversible reactions)

A reaction in which not only the reactants react to form the products under certain conditions but also the products react to form reactants under the same conditions is called a reversible reaction.

Examples are

(i) 3Fe_(s) + 4H₂O_(g) $⇌$ Fe₃O_4(s) + 4H_2(g)

(ii) CaCO_3(s) $⇌$ CaO_(s) + CO_2(g)

(iii) N_2(g) + 3H_2(g) $⇌$ 2NH_3(g)

If a reaction cannot take place in the reverse direction, i.e. the products formed do not react to give back the reactants under the same condition it is called an irreversible reaction.

Examples are:

(i) $A g N O_{3 (a q)} + N a C l_{(a q)} \to A g C l_{(s)} + N a N O_{3 (g)}$

(ii) $2 M_{(g)} + O_{2 (g)} \to 2 M g O_{(s)}$

Note: If any of the product will be removed from the system, reversible reaction will become irreversible one.

CONCEPT OF CHEMICAL EQUILIBRIUM

Let we have a general reversible reaction,

at time t = 0 $a_{0} A + b_{0} B ⇌ 0 C + 0 D$

as time will pass, A and B will be converted to C and D. As soon as C and D will be formed, reaction will start in back direction also. A time will come when rate of decomposition of A and B will be equal to the rate of formation of A & B, i.e. the rate of forward reaction will be equal to the rate of reverse reaction. At this stage the reaction is said to be in a state of Chemical Equilibrium.

If we see variation of reactants and products with time on graph we will be having following graph.

Diagram being restored — will be back shortly

When the equilibrium is reacted, the concentration of reactant as well as product remains constant. At this stage reaction seems to be stopped but in actual the reaction is going in forward as well as reverse direction with same rate and hence this equilibrium is called as dynamic equilibrium.

Characteristics of Chemical Equilibrium

$∙$ The equilibrium is dynamic i.e. the reaction continues in both forward and reverse directions.

$∙$ The rate of forward reaction equals to the rate of reverse reaction.

$∙$ The observable properties of the system such as pressure, concentration, density remains invariant with time.

$∙$ The chemical equilibrium can be approached from either side.

$∙$ A catalyst can hasten the approach of equilibrium but does not alter the state of equilibrium.

Types of Equilibria

There are mainly two types of equilibria:

(a) Homogeneous: Equilibrium is said to be homogeneous if reactants and products are in same phase.

H₂(g) + I₂(g) $⇌$ HI(g)

N₂(g) + 3H₂(g) $⇌$ 2NH₃(g)

N₂O₄(g) $⇌$ 2NO₂(g)

CH₃COOH(l) + C₂H₅OH(l) $⇌$ CH₃COOC₂H₅(l) + H₂O(l)

(b) Heterogeneous: Equilibrium is said to be heterogeneous if reactants and products are in different phases

CaCO₃(s) $⇌$ CaO(s) + CO₂(g)

NH₄HS(s) $⇌$ NH₃(g) + H₂S(g)

NH₂CO₂NH₄(s) $⇌$ 2NH₃(g) + CO₂(g)

Note: To write the equilibrium constant expression, the concentration of pure liquid and pure solid assumed to be unity, as the concentration of such substances remain constant, i.e.,

concentration = mole/litre $\propto$ density.

LAW OF MASS ACTION

Guldberg and Waage established a relationship between rate of chemical reaction and the concentration of the reactants or, with their partial pressure in the form of law of mass action.

According to this law, "The rate at which a substance reacts is directly proportional to its active mass and rate of a chemical reaction is directly proportional to product of active masses of reactants each raised to a power equal to corresponding stoichiometric coefficient appearing in the balanced chemical equation".

$a A + b B \to c C + d D$

rate of reaction $\propto$ [A]^a.[B]^b

rate of reaction = K[A]^a[B]^b

where K is rate constant or velocity constant of the reaction at that temperature.

Unit of rate constant (K) $= [m o l es / l i t]^{1 - n} t im e^{- 1}$ (where n is order of reaction.)

Note: For unit concentration of reactants rate of the reaction is equal to rate constant or specific reaction rate.

At equilibrium the rate of both forward and backward reactions become equal and after achieving equilibria, the concentration of reactants and products remains constant as shown in figure.

Diagram being restored — will be back shortly

Note: Active mass is the molar concentration of the reacting substances actually participating in the reaction.

Hence

Active mass = $\frac{n u mb er o f m o l es}{v o l u m e in l i t r es}$

Active mass of solid is taken as unity.

Illustration 1. Calculate the partial pressure of each component in the following equilibria

N_2(g) + 3H_2(g) $⇌$ 2NH_3(g)

Solution: N_2(g) + 3H_2(g) $⇌$ 2NH_3(g)

at t = 0 a b 0

at equilibrium a – x b – 3x 2x

n_total = a – x + b – 3x + 2x = (a + b – 2x)

Partial pressure:

$P_{N_{2}} = \frac{n _{N_{2}}}{n _{T}} \times P_{T} = \frac{( a - x )}{( a + b - 2 x )} \times P_{T}$

$P_{H_{2}} = \frac{n _{H_{2}}}{n _{T}} \times P_{T} = \frac{( b - 3 x )}{( a + b - 2 x )} \times P_{T}$

$P_{N H_{3}} = \frac{n _{N H_{3}}}{n _{T}} \times P_{T} = \frac{2 x}{( a + b - 2 x )} \times P_{T}$

Law of Chemical Equilibrium

According to this law, the ratio of product of concentration of products to the product of concentration of reactants, with each concentration term is raised to the power by its coefficient in overall balanced chemical equation, is a constant quantity at a given temperature and it is called equilibrium constant.

Derivation of law of chemical equilibrium

Let us consider for the following equilibrium

aA + bB $⇌$ cC + dD

then, from Law of mass action

rate of forward reaction r₁ $\propto$ [A]^a [B]^b

or r₁ = K_a [A]^a [B]^b and rate of reverse reaction r₂ $\propto$ [C]^c [D]^d

r₂ = K₂ [C]^c [D]^d

at equilibrium, r₁ = r₂

$\Rightarrow$ K₁ [A]^a[B]^b = K₂ [C]^c[D]^d

$\Rightarrow$ $\frac{K _{1}}{K _{2}} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}$

K_c = K₁/K₂, an equilibrium constant in terms of active masses of reacting species.

For the reaction

SO₂Cl₂ $⇌$ SO₂ + Cl₂

at t = 0 a 0 0

at equilibrium a - x x x

equilibrium conc. $\frac{a - x}{V}$ x $\frac{x}{V}$ . $\frac{x}{V}$ (V is volume of container)

So, $[S O_{2} C l_{2}] = \frac{a - x}{V}; [S O_{2}] = \frac{x}{V} = [C l_{2}]$

So, $K_{c} = \frac{( \frac{x}{V} ) \times ( \frac{x}{V} )}{( \frac{a - x}{V} )} = \frac{x ^{2}}{( a - x ) V}$ $\Rightarrow$ $K_{c} = \frac{x ^{2}}{( a - x ) \times V}$

CHARACTERISTICS OF EQUILIBRIUM CONSTANT (K_c)

(i)K_c for a particular reaction at given temperature has a constant value.(ii)Value of K_c always depends on nature of reactants and the temperature, but independent of presence of catalyst or, of inert material.(iii)Its value is always independent of the initial concentration of reactants as well as the products.(iv)The value of K_c indicates the proportion of products/product formed at equilibrium. Large K_c value means large proportions of product.(v)When the reaction is reversed, equilibrium constant for reverse reaction will also be inversed.

K a p os_{c} = \frac{1}{K _{c}}

Let us haveA + B

⇌

C + D

K_{c} = \frac{[ C ] [ D ]}{[ A ] [ B ]}

By reversing the reaction,C + D

⇌

A + B

K a p o s_{C} = \frac{[ A ] [ B ]}{[ C ] [ D ]} = \frac{1}{K _{C}}

(vi)If the coefficients of reactants of products are halved or, doubled then accordingly, value of K¢_c will change.A + B

⇌

C + D

K_{c} = \frac{[ C ] [ D ]}{[ A ] [ B ]}

for2A + 2B

⇌

2C + 2D

K a p os_{c} = \frac{[ C ] ^{2} [ D ] ^{2}}{[ A ] ^{2} [ B ] ^{2}} = {\frac{[ C ] [ D ]}{[ A ] [ B ]}}^{2} = K_{c}^{2}

(vii) When a number of equilibrium reactions are added, the equilibrium constant, for overall reaction is the product of equilibrium constants of respective reactions.N_2(g) + O_2(g)

⇌

2NO

K_{1} = \frac{[ N O ] ^{2}}{[ N _{2} ] [ O _{2} ]}

N₂O_(g)

⇌

N_2(g) + 1/2O_2(g)

K_{2} = \frac{[ N _{2} ] [ O _{2} ] ^{1/2}}{[ N _{2} O ]}

N₂O_(g) + 1/2O_2(g)

⇌

2NO_(g)

K_{3} = \frac{[ N O ] ^{2}}{[ N _{2} O ] [ O _{2} ] ^{1/2}}

\frac{K _{3}}{K _{2}} = \frac{[ N O ] ^{2}}{[ N _{2} O ] [ O _{2} ] ^{1/2}} \times \frac{[ N _{2} O ]}{[ N _{2} ] [ O _{2} ] ^{1/2}}

= \frac{[ N O ] ^{2}}{[ N _{2} ] [ O _{2} ]} = K_{1}

i.e.

\frac{K _{3}}{K _{2}} = K_{1}

Equilibrium constant in terms of partial pressuresLet us consider a general reactionaA + bB

⇌

cC + dDP_AP_BP_CP_DFrom law of mass action

r_{1} \propto (P_{a})^{a} (P_{B})^{b} \Rightarrow r_{1} = K_{1} (P_{A})^{a} (P_{B})^{b}

r_{2} \propto (P_{C})^{c} (P_{D})^{d} \Rightarrow r_{2} = K_{2} (P_{C})^{c} (P_{D})^{d}

at equilibriumr₁ = r₂or K₁ (P_A)^a (P_B)^b = K₂(P_C)^c(P_D)^d

\Rightarrow \frac{K _{1}}{K _{2}} = \frac{( P _{C} ) ^{a} ( P _{D} ) ^{d}}{( P _{A} ) ^{a} ( P _{B} ) ^{b}}

\Rightarrow K_{p} = \frac{K _{1}}{K _{2}} = \frac{( P _{C} ) ^{c} ( P _{D} ) ^{d}}{( P _{A} ) ^{a} ( P _{B} ) ^{b}}

RELATION BETWEEN K_P AND K_CFor a general equation,aA + bB

⇌

cC + dDWhere, a , b, c and d are coefficients of the reacting substance

K_{p} = \frac{( P _{C} ) ^{c} ( P _{D} ) ^{d}}{( P _{A} ) ^{a} ( P _{B} ) ^{b}}

From gas equationPV = nRT

\Rightarrow

P = \frac{n}{V} R T

So, P_A = [A]RT’ P_B = [B]RTP_C = [C]RT P_D = [D]RTHence,

K_{p} = \frac{( [ C ] R T ) ^{c} \cdot ( [ D ] R T ) ^{d}}{( [ A ] R T ) ^{a} \cdot ( [ B ] R T ) ^{b}}

or,

K_{p} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}} \times (R T)^{{(c + d) - (a + b)}}

Where

Δ

n_g = {(c + d) – (a + b)} = change in the numbers of gaseous moles.Hence,

K_{p} = K_{c} \cdot (R T)^{Δ n_{g}}

When

Δ

n = 0 K_p = K_c

Δ

n > 0 K_p > K_c

Δ

n < 0 K_p < K_cNote: Similarly we can find equilibrium constant (K_x) interms of mole fraction and can find out its relation with K_p and K_c.Illustration 2.At 27ଌ K_p value for reaction $C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)$ is 0.1 atm, calculate its K_c value. Solution:K_P= K_c(RT)^Dn $Δ$ n = 1K_c =

\frac{K _{P}}{R T} = \frac{0.1}{0.82 \times 300}

= 4x10^-3

SIGNIFICANCE OF THE MAGNITUDE OF EQUILIBRIUM CONSTANT

(i)A very large value of K_C or K_P signifies that the forward reaction goes to completion or very nearly so.(ii)A very small value of K_C or K_P signifies that the forward reaction does not occur to any significant extent.(iii)A reaction is most likely to reach a state of equilibrium in which both reactants and products are present if the numerical value of K_c or K_P is neither very large nor very small.Units of Equilibrium ConstantAs we have learned that numerical value of equilibrium constant is a function of stoichiometric coefficient used for any balanced chemical equation.But, what will happen in the size of units of equilibrium constant when the stoichiometric coefficient of the reaction changes? (If sum of the exponent for product is not equal to reactant side sum)So for the sake of simplicity in the units of K_p for K_c the relative molarity or pressure of reactants and products are used with respect to standard condition. (For solution standard state 1 mole/litre and for gas standard pressure = 1 atm)Now the resulting equilibrium constant becomes unitless by using relative molarity and pressure.Illustration 3.The value of K_p for the reaction 2H₂O(g) + 2Cl₂(g) $⇌$ 4HCl(g) + O₂(g) is 0.035 atm at 400^oC, when the partial pressures are expressed in atmosphere. Calculate K_c for the reaction, $\frac{1}{2} O_{2} (g) + 2 H C l (g)$ $⇌$ Cl₂(g) + H₂O(g)Solution:

K_{P} = K_{C} (R T)^{Δ n}

Dn = moles of product - moles of reactants = 5 - 4 = 1

R = 0.082 L atm/mol K,

T = 400 + 273 = 673 K

$∴$ 0.035 = K_C (0.082 × 673)

K_C = 6.342 x 10^-4 mol l^-1

$∴$ $K_{C}^{^{'}}$ for the reverse reaction would be $\frac{1}{K _{C}} .$

$∴$ $K_{C}^{^{'}}$ = $\frac{1}{6.342 \times 10 ^{- 4}}$ = 1576.8 (mol l^-1)^-1

When a reaction is multiplied by any number n (integer or a fraction) the $K_{C}^{^{'}}$ or $K_{P}^{^{'}}$ becomes (K_C)ⁿ or (K_P)ⁿ of the original reaction.

$∴$ K_C for $\frac{1}{2}$ O₂(g) + 2HCl(g) $⇌$ Cl₂(g) + H₂O(g)

is $1576.8 =$ 39.7 (mol.l^-1)^-½

Dependence of Degree of Dissociation on Density Measurements

The degree of dissociation of a substance is defined as the fraction of its molecules dissociating at a given time. The following is the method of calculating the degree of dissociation of a gas using vapour densities. This method is valid only for reactions whose K_P exist, i.e., reactions having at least one gas and having no solution.

Since PV = nRT

PV = $\frac{w}{M} R T$

M = $\frac{w R T}{P V} = \frac{ρR T}{P}$

$∴$ VD = $\frac{ρR T}{2 P}$

Since P = $\frac{n R T}{V}$

$∴$ VD = $\frac{ρR T}{2 n R T} \times V = \frac{ρ V}{2 n}$

For a reaction at equilibrium V is a constant and r is a constant. $∴$ vapour Density $α \frac{1}{n}$

$∴$ $\frac{T o t a l m o l es a t e q u i l ib r i u m}{I ni t ia l t o t a l m o l es} = \frac{v a p o u r d e n s i t y ini t ia l}{v a p o u r d e n s i t y a t e q u i l ib r i u m} = \frac{D}{d}$ .= $\frac{M}{m}$

( $∴$ molecular weight = 2 x V.D)

Here M = molecular weight initial

m = molecular weight at equilibrium

Let us take a reaction

PCl₅ $⇌$ PCl₃ + Cl₂

Initial moles C 0 0

At eqb. C(1- $α$ ) C $α$ C $α$

$∴ \frac{C ( 1 + α )}{C} = \frac{D}{d}; 1 + α = \frac{D}{d} = \frac{M}{M}$

Knowing D and d, $α$ can be calculated and so for M and m.

Illustration 4. When PCl₅ is heated it dissociates into PCl₃ and Cl₂. The density of the gas mixture at 200^oC and at 250^oC is 70.2 and 57.9 respectively. Find the degree of dissociation at 200^oC and 250^oC.

Solution: PCl₅(g) $⇌$ PCl₃(g) + Cl₂(g)

We are given the vapour densities at equilibrium at 200^oC and 250^oC.

The initial vapour density will be the same at both the temperatures as it would be $\frac{M _{P C l_{5}}}{2} .$

$∴$ Initial vapour density = $\frac{( 31 + 5 \times 35.5 )}{2} = 104.25$

Vapour density at equilibrium at 200^oC = 70.2

$∴$ $\frac{T o t a l m o l es a t e q u i l ib r i u m}{T o t a l m o l es ini t ia l} = 1 + α = \frac{V a p o u r d e n s i t y ini t ia l}{V a p o u r d e n s i t y a t e q u i l ib r i u m}$

= $\frac{104.25}{70.2} = 1.485$

$∴$ $α$ = 0.485

At 250^oC, 1 + $α$ = $\frac{104.25}{57.9} = 1.8$

$∴$ $α$ = 0.8

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