Samacheer Class 11 Chemistry - Physical and Chemical Equilibrium

At chemical equilibrium the rate of two opposing reactions are equal and the concentration of. reactants and products do not change with time. This condition is not static and is dynamic because both the forward and reverse reactions are still occurring at the same rate and no macroscopic change is observed. So the chemical equilibrium is in a state of dynamic equilibrium.

The relation between K p and K c is K p = K C (RT) ∆n g
K p = equilibrium constant is terms of partial pressure.
K c = equilibrium constant is terms of concentration.
R = gas constant
T = Temperature.
∆n g = Difference between the sum of the number of moles of products and the sum of number of moles of reactants in the gas phase. When ∆n g = 0
K p = K C (RT) 0 = K C i.e., K p = K C
Example: H 2 (g) + I 2 (g) \(\rightleftharpoons\) 2HI(g)
∆n g = 2 – 2 = 0
∴ K p = K C for the synthesis of HI.

K c = \(\frac{[\mathrm{AB}]^{2}}{\left[\mathrm{~A}_{2}\right]\left[\mathrm{B}_{2}\right]}\); A – green, B – blue
Given that ‘V’ is constant(closed system)
At equilbrium,
K c = \(\frac{(-)}{\left(\frac{2}{V}\right)\left(\frac{2}{V}\right)}\) = \(\frac{16}{4}\) = 4
K p = K c (RT) ∆n = 4 (RT) 0 = 4
At stage ‘x’,
Q = \(\frac{\left(\frac{6}{\mathrm{~V}}\right)^{2}}{\left(\frac{2}{\mathrm{~V}}\right)\left(\frac{1}{\mathrm{~V}}\right)}=\frac{36}{2}\) = 18
Q > K c i.e., reverse reaction is favoured.
At stage ‘y’,
Q = \(\frac{\left(\frac{3}{V}\right)^{2}}{\left(\frac{3}{V}\right)\left(\frac{3}{V}\right)}=\frac{9}{3 \times 3}\) = 1
K c > Q i.e., forward reaction is favoured.

a) H 2 (g) + I 2 (g) ⇌ 2HI(g)
In this reaction, there is no effect on changing the volume. ∆ng = 0
b) CaCO 3 (s) ⇌ CaO (s) + CO 2 (g)
In this reaction, increases in volume favours forward reaction.
c) S(s) + 3 F 2 (g) ⇌ SF 6 (g)
In this reaction, decreases in volume favours forward reaction.

The law states that “At any instant, the rate of chemical reaction at a given temperature is directly proportional to the product of the active masses of the reactants at that instant.”

From the knowledge of equilibrium constant, it is possible to predict the di¬rection in which the net reaction is taking place for a given concentration or partial pressure of reactants and products.
Consider a general homogeneous reversible reaction,
xA + yB ⇌ lC + mD
For the above reaction under non-equilibrium conditions, reaction quotient ‘Q’ is defined as the ratio of the product of active masses of reaction products raised to the respective stoichiometric coefficients in the balanced chemical equation to that of the reactants.
Under non – equilibrium conditions, the reaction quotient Q can be calculated using the following expression.
Q = \(\frac{[\mathrm{C}]^{1}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\)
As the reaction proceeds, there is a continuous change in the concentration of reactants and products and also the Q value until the reaction reaches equilibrium. At equilibrium, Q is equal to K c at a particular temperature. Once the equilibrium is attained, there is no change in the Q value. By knowing the Q value, we can predict the direction of the reaction by comparing it with K c.
If Q = K c, the reaction is in equilibrium state.
If Q > K c, the reaction will proceed in the reverse direction, i.e., formation of reactants.
If Q < K c, the reaction will proceed in the forward direction i.e., formation of products.
Predicting the direction of a reaction

Let us consider the formation of ammonia in which, ‘a’ moles nitrogen and h’ moles hydrogen gas are allowed to react in a container of volume V. Let ‘x’ moles of nitrogen react with 3 x moles of hydrogen to give 2x moles of ammonia.
3H 2 (g) + N 2 (g) ⇌ 2NH 3 (g)
Applying law of mass action,
K 2 = \(\frac{\left[\mathrm{NH}_{3}\right]^{2}}{\left[\mathrm{~N}_{2}\right]\left[\mathrm{H}_{2}\right]^{3}}\)

When an inert gas (i.e., a gas which does not react with any other species involved in equilibrium) is added to an equilibrium system at constant volume, the total number of moles. of gases present in the container increases, that is, the total pressure of gases increases, the partial pressure of the reactants and the products are unchanged. Hence at constant volume, the addition of inert gas has no effect on the equilibrium.

Let us consider the general reaction in which all reactants and products are ideal gases.
xA + yB ⇌ lC + mD
The equilibrium constant, K c is,
K c = \(\frac{\left[\mathrm{Cl}^{l}[\mathrm{D}]^{\mathrm{m}}\right.}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\) ……………..(1)
and K p is,
K p = \(\frac{p_{C}^{l} \times p_{D}^{m}}{p_{A}^{x} \times p_{B}^{y}}\) ………….(2)
The ideal gas equation is,
PV = nRT
P = \(\frac{n}{V}\) RT
since, Active mass = molar concentration = n/V
P = active mass × RT
Based on the above expression the partial pressure of the reactants and products can be expresssed as,
P X A = [A] x [RT] x
P Y B = [A] y [RT] y
P l C = [A] l [RT] l
P m D = [A] m [RT] m
On substituting in Eqn.(2).,
By comparing equation (1) and (4),
we get
K p ∆n g =K c (RT) …………..(5)
Where,
∆n g is the difference between the sum of number of moles of products and the sum of number of moles of reactants in the gas phase.
The following relations become immediately obvious.
when ∆n g = 0.
K p = K c (RT) 0 = K c
Example:
H 2 (g) + I 2 (g) ⇌ 2HI(g)
N 2 + O 2 ⇌ 2NO(g)
when ∆n g = +ve
K p = K c (RT) +ve
K p > K c
Example:
2NH 3 (g) ⇌ N 2 (g) + 3H 2 (g)
PCl 5 (g) ⇌ PCl 3 (g) + Cl 2 (g)
When ∆n g = -ve
K p = K c (RT) -ve
K p < K c
Example:
2H 2 (g) + O 2 (g) ⇌ 2H 2 O(g)
2SO 2 (g)+ O 2 (g) ⇌ 2SO 3 (g)

Given that,
[PCl 5 ] initial = \(\frac{1 \text { mole }}{1 \mathrm{dm}^{3}}\)
[Cl 2 ] eq = 0.6 mole dm -3
PCl 5 ⇌ PCl 3 + Cl 2
[PCl 3 ] eq = 0.6 mole dm -3
[PCl 5 ] eq = 0.4 mole dm -3
∴ K c = \(\frac{\left[\mathrm{PCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]}{\left[\mathrm{PCl}_{5}\right]}=\frac{0.6 \times 0.6}{0.4}\)
= 0.9 mole dm -3

:
For the reaction,
SrCO 3 (s) ⇌ SrO(s) + CO 2 (g)
∆n g = 1 – 0 = 1
∴ K p = K c (RT)
2.2 × 10 -4 = K c (0.0821) (1002)
K c = \(\frac{2.2 \times 10^{-4}}{0.0821 \times 1002}\) = 2.674 × 10 -6

:
V = 3 L
[HI] initial = \(\frac{0.3 \mathrm{~mol}}{3 \mathrm{~L}}\) 0.1 M
[HI] eq = 0.05 M
2 HI(g) ⇌ H 2 (g) + I 2 (g)
K c = \(\frac{\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]}{[\mathrm{HI}]^{2}}=\frac{0.025 \times 0.025}{0.05 \times 0.05}\) = 0.25
K p = K c (RT) (∆n g )
∆n g = 2 – 2 = 0
K p = 0.25(RT) 0
K p = 0.25.

2 NO(g) + O 2 (g) ⇌ 2NO 2 (g)
K p = \(\frac{\left(P_{N O_{2}}\right)^{2}}{\left(P_{N O}\right)^{2}\left(P_{O}\right)}\)
= \(\frac{0.96 \times 0.96}{0.04 \times 0.04 \times 0.52}\)
K p = 1.017 × 10 3.

:
CH 4 (g) + 2 H 2 S (g) ⇌ CS 2 (g) + 4H 2 (g)
K c = 4 x 10 -2 mol 2 lit -2
Volume = 500 ml = \(\frac{1}{2}\) L
[CH 4 ] in = \(\frac{1 \mathrm{~mol}}{\frac{1}{2} L}\) = 2 mol L -1;
[CH 2 ] in = \(\frac{1 \mathrm{~mol}}{\frac{1}{2} L}\) = 2 mol L -1

:
K c = 4 × 10 -2 for the reaction,
H 2 S ⇌ 2H 2 (g) + S 2 (g)
K c = \(\frac{\left[\mathrm{H}_{2}\right]\left[\mathrm{S}_{2}\right]^{1 / 2}}{\left[\mathrm{H}_{2} \mathrm{~S}\right]}\)
⇒ 4 × 10 -2 = \(\frac{\left[\mathrm{H}_{2}\right]\left[\mathrm{S}_{2}\right]^{1 / 2}}{\left[\mathrm{H}_{2} \mathrm{~S}\right]}\)
(i) 2H 2 S(g) ⇌ 2H 2 + S 2 (g)
For the reaction,
2H 2 S(g) ⇌ 2H 2 + S 2 (g)
K c = \(\frac{\left[\mathrm{H}_{2}\right]^{2}\left[\mathrm{~S}_{2}\right]}{\left[\mathrm{H}_{2} \mathrm{~S}\right]^{2}}\)
= (4 × 10 -2 ) 2 = 16 × 10 -4
(ii) 3H 2 S(g) ⇌ 3H 2 (g) + 3/2 S 2 (g)
For the reaction,
3H 2 S(g) ⇌ 3H 2 (g) + 3/2 S 2 (g)
K c = \(\frac{\left[\mathrm{H}_{2}\right]^{3}\left[\mathrm{~S}_{2}\right]^{3 / 2}}{\left[\mathrm{H}_{2} \mathrm{~S}\right]^{3}}\)
= (4 × 10 -2 ) 3 = 64 × 10 -6

Given m N 2 =28g;
m H 2 = 6g;
V = 1 L.
(n N 2 ) initial = \(\frac{28}{28}\) = 1
(n H 2 ) initial = \(\frac{6}{2}\) = 3
N 2 (g) + 3 H 2 (g) ⇌ 2NH 3 (g)
[NH 3 ] = \(\frac{17}{17}\) = 1 mol
Weight of N 2 = (no. of moles of N 2 ) × (molar mass of N 2 )
= 0.5 × 28 = 14g
Weight of H 2 = (no. of moles of H 2 ) × (molar mass of H 2 )
= 1.5 × 2 = 3g

A 2 (g) + B 2 (g) ⇌ 2 AB (g)
Given that, K p = 1;
\(\frac{4 x^{2}}{(1-x)^{2}}\) = 1
⇒ 4x 2 = (1 – x) 2 = 1
⇒ 4x 2 = 1 + x 2 – 2x
3x 2 + 2x – 1 = 0
x = 0.33 – 1(not possible)
∴ [A 2 ] eq = 1 – x = 1 – 0.33 = 0.67
[B 2 ] eq = 1 – x = 1 – 0.33 = 0.67
[AB 2 ] eq = 2x × 0.33 = 0.66

This equation gives that quantitative temperature dependence of equilibrium constant (K). The relation between standard free energy change (∆G°) and equilibrium constant is
∆G° = – RT In K ………………..(1)
We know that, ∆G° = ∆H° – T∆S° …………(2)
Substituting (1) in equation (2)
– RT In K = ∆H° – T∆S°
Rearranging, In K = \(\frac{\Delta \mathrm{H}^{0}}{\mathrm{RT}}+\frac{\Delta \mathrm{S}^{0}}{\mathrm{RT}}\) ……………(3)
Differentiating equation (3) with respect to temperature,
\(\frac{\mathrm{d}(\ln \mathrm{K})}{\mathrm{dT}}=\frac{\Delta \mathrm{H}^{0}}{\mathrm{RT}^{2}}\) ……………….(4)
Equation (4) is known as differential form of van,t Hoff equation.
On integrating the equation (4), between T 1 and T 2 with their respective equilbrium consatnts K 1 and K 2.
Equation 5 is known as integrated form of Van’t Hoff equation.

:
K p 1 = 8.19 × 10 2;
T 1 = 298 K
K p 1 = 8.19 × 10 2;
T 1 = 298 K
K p 2 = 4.16 × 10 -1;
T 2 = 498 K

P CO 2 = 1.017 × 10 -3 atm
T = 500°C;
K p = P CO 2
∴ K p 1 = 1.017 × 10 -3;
T = 500 + 273 = 773 K
K p 2 = ?
T = 600 + 273 = 873 K
∆H° = 181 KJ mol -1