Chemical Equilibrium
Maharashtra State Board-Class-11-Science-Chemistry-Chapter -12
Solutions
Question 1. Choose the correct option
(A) The equlilibrium , H2O(l) ⇌ H+(aq) + OH(aq) is
(a) dynamic
(b) static
(c) physical
(d) mechanical
(a) dynamic
(B) For the equlibrium, A ⇌ 2B + Heat, the number of ‘A’ molecules increases if
(a) volume is increased
(b) temperature is increased
(c) catalyst is added
(d) concerntration of B is decreased
(b) temperature is increased
(C) For the equilibrium Cl2(g) + 2NO(g) ⇌ 2NOCl(g) the concentration of NOCl will increase if the equlibrium is disturbed by
(a) adding Cl2
(b) removing NO
(c) adding NOCl
(c) removal of Cl2
(a) adding Cl2
(D) The relation between Kc and Kp for the reaction A (g) + B (g) ⇌ 2C(g) + D(g) is
(a) KC = YKP
(b) KP = KC2
(c) KC = \(\frac{1}{\sqrt{K_p}}\)
(d) KP/KC = 1
(a) KC = YKP
(E) When volume of the equilibrium reaction C (g) + H2O (g) ⇌ CO(g) + H2(g) is increased at constant temperature the equilibrium will
(a) shift from left to right
(b) shift from right to left
(c) be unaltered
(d) can not be predicted
(c) be unaltered
Question 2. Answer the following
(A) State Law of Mass action.
The law of mass action states that the rate of a chemical reaction at each instant is proportional to the product of concentrations of all the reactants.
(B) Write an expression for equilibrium constant with respect to concentration.
Consider a reversible reaction,
A(g) + B(g) ⇌ C(g) + D(g)
The equilibrium constant with respect to concentration of reactants and products is,
kC = \(\frac{[C]×[D]}{[A]×[B]}\)
(C) Derive mathematically value of kP for for A(g) + B(g) ⇌ C(g) + D(g)
When the concentrations of reactants and products in gaseous reactions are expressed in terms of their partial pressure, then the equilibrium constant is represented as Kp.
∴ For the reaction,
A(g) + B(g) ⇌ C(g) + D(g)
the equilibrium constant (Kp) can be expressed using partial pressure as:
Kp = \(\frac{P_C×P_D}{P_A×P_B}\)
Where PA, PB, PC and PD are equilibrium partial pressures of A, B, C and D respectively.
(D) Write expressions of Kc for following chemical reactions
(i) 2SO2(g) + O2(g) ⇌ 2SO3(g)
(ii) N2O4(g) ⇌ 2NO2(g)
(i) 2SO2(g) + O2(g) ⇌ 2SO3(g)
Kp = \(\frac{P^2_{SO_3}}{P^2_{SO_2}×P^2_{O_2}}\)
Kp = Kc(RT)Δn
∴ Kc = \(\frac{K_p}{(RT)^{Δn}}\)
∴ Kc = Kp × (RT)−Δn
Δn = 2 − (2 + 1) = −1 mol
∴ Kc = Kp × (RT)−(−1) = Kp × RT
(ii) N2O4(g) ⇌ 2NO2(g)
Kp = \(\frac{P^2_{NO_2}}{P^2_{N_2O_4}}\)
Kp = Kc(RT)Δn
Δn = 2 − (1) = 1 mol
∴ Kp = Kc × RT
∴ Kc = \(\frac{K_p}{RT}\)
(E) Mention various applications of equilibrium constant.
Various applications of the equilibrium constant:
- Prediction of the direction of the reaction
- To know the extent of the reaction
- To calculate equilibrium concentrations
- The link between chemical equilibrium and chemical kinetics.
(F) How does the change of pressure affect the value of equilibrium constant ?
A change in pressure of gaseous substances at equilibrium may shift the equilibrium forward or reverse but equilibrium constant will remain unchanged at constant temperature.
(J) Differentiate irreversible and reversible reaction.
Irreversible reaction | Reversible reaction |
It is a unidirectional reaction. | The reaction takes place in forward and reverse directions. |
Rate of forward reaction only is considered. | Rates of forward and backward reaction are considered. |
The rate constant of forward direction is only considered. | The rate constants of forward and backward directions are considered. |
A reaction goes to completion. | A reaction never goes to completion. |
A catalyst accelerates the rate of the forward reaction only. | A catalyst accelerates the forward and backward rates to the same extent. |
Rate constant depends on temperature. | Equilibrium constant depends on temperature. |
(K) Write suitable conditions of concentration, temperature and pressure used during manufacture of ammonia by Haber process.
The Haber process is the process of synthesis of ammonia gas by reacting together
hydrogen gas and nitrogen gas in a particular stoichiometric ratio by volumes and at selected optimum temperature.
- Iron (Fe) (containing a small quantity of molybdenum) is used as catalyst.
(i) Effect of concentration : By increasing the concentration of the reactants N2(g) and H2(g) the reaction proceeds forward increasing the yield of NH3(g).
(ii) Effect of temperature : Since the formation of NH3(g) is an exothermic reaction, it is carried out at high temperature. The optimum temperature is 773 K.
(iii) Effect of pressure :
For this reversible reaction,
Δn = (n2)product — (n1)reactant = 2 − (1 + 3) = −2 mol
Since during the formation of ammonia, there is a decrease in number of moles, the reaction is carried out at high pressure. The optimum pressure required is about 500 atm.
(L) Relate the terms reversible reactions and dynamic equilibrium.
- Reversible reactions are the reactions which do not go to completion and occur in both directions simultaneously.
- If such a reaction is allowed to take place for a long time, so that the concentrations of the reactants and products do not vary with time, then the reaction will attain equilibrium.
- Since, both the forward and backward reactions continue to take place in opposite directions in the same speed, the equilibrium achieved is dynamic in nature. Thus, if the reaction is not reversible then it cannot attain dynamic equilibrium.
(M) For the equilibrium. BaSO4(s) ⇌ Ba2+(aq) + SO42−(aq) state the effect of
(a) Addition of Ba2+ ion.
(b) Removal of SO42− ion
(c) Addition of BaSO4(s) on the equilibrium.
(a) Addition of Ba2+ ion will shift the equilibrium in reverse direction.
(b) Removal of SO42− ion will shift the reaction in forward direction.
(c) Addition of BaSO4(s) will not affect the equilibrium.
Question 3. Explain :
(A) Dynamic nature of chemical equilibrium with suitable example.
- At chemical equilibrium state, the rates of the forward and the backward reactions are equal and the concentrations of the reactants and the products reach constant values.
- At chemical equilibrium the reaction still continues to take place in both the directions with the same rates.
- The reactants continue to form the products while the products also continue to form the reactants.
- Because of such a situation at the equilibrium state in the reaction system, the chemical equilibrium is called a dynamic equilibrium.
Example :
In the reaction between H2 and I2 to form H1, the colour of the reaction mixture becomes constant because the concentrations of H2, I2 and H1 become constant at equilibrium.
H2 + I2 ⇌ 2HI
Thus, when equilibrium is reached, the reaction appears to have stopped. However, this is not the case. The reaction is still going on in the forward and backward direction but the rate of forwarding reaction is equal to the rate of backward
reaction. Hence, chemical equilibrium is dynamic in nature and not static.
(B) Relation between KC and KP.
Consider the equilibrium constants designated as KP and KC for the concentrations
expressed in pressures and molar concentrations respectively.
Consider the following gaseous reversible reaction,
aA(g) + bB(g) ⇌ cC(g) + dD(g)
Then, KP = \(\frac{[P_C]^c×[P_D]^d}{[P_A]^a×[P_B]^b}\)
where PA, PB, PC and PD are the equilibrium partial pressures of A, B, C and D respectively.
The partial pressure, p of each component is directly proportional to its molar concentration.
If gaseous equilibrium mixture contains nA, nB, nC and nD moles of A, B, C and D respectively in a vessel of volume V, then the molar concentrations are, [A] = nA/V [B] = nB/V, [C] = nC/V, [D] = nD/V mol L-1
By the general gas equation,
pAV = nART,
∴ pA = \(\frac{n_A}{V}RT\) = [A]RT
Similarly,
pB = [B]RT, pC = [C]RT and pD = [D]RT
Since,
KP = \(\frac{[p_C]^c×[p_D]^d}{[p_A]^a×[p_B]^b}\) = \(\frac{([C]RT)^c×([D]RT)^d}{([A]RT)^a×([B]RT)^b}\)
= \(\frac{[C]^c×[D]^d}{[A]^a×[B]^b}×\frac{(RT)^{c+d}}{(RT)^{a+b}}\)\)
= \(\frac{[C]^c×[D]^d}{[A]^a×[B]^b}× (RT)^{(c+d)-(a+b)}\)
= \(\frac{[C]^c×[D]^d}{[A]^a×[B]^b}× (RT)^{Δn}\)
Since,
KC = \(\frac{[C]^c×[D]^d}{[A]^a×[B]^b}\)
∴ KP = KC (RT)Δn
where, Δn = (c + d) — (a + b)
= [Number of moles of gaseous products]—[Number of moles of gaseous reactants]
= n(g)(products) — n(g)(reactants)
(C) State and explain Le Chatelier’s principle with reference to
(1) change in temperature
(2) change in concentration.
(1) Le Chatelier’s principle with reference to change in temperature :
(i) For endothermic reaction equilibrium :
Consider the following endothermic reaction,
N2(g) + O2(g) ⇌ 2NO(g) —79.9 kJ
- In this equilibrium, forward reaction is endothermic while backward reaction is
- By increasing the temperature, according to Le Chatelier’s principle, the equilibrium will be shifted to the forward endothermic direction giving more products. Hence, the endothermic reaction is favoured at high temperature.
(ii) For exothermic reaction equilibrium :
Consider the following exothermic reaction,
PCl5(g) ⇌ PCl3(g) + Cl2(g) +95.5 kJ
- In this equilibrium, the forward reaction is exothermic, while backward reaction is
- By increasing the temperature, heat is added to a reaction system at equilibrium. This applies a stress to the system. According to Le Chatelier’s principle, the equilibrium is shifted to the backward endothermic reaction absorbing the heat which decreases a stress. But decrease in temperature will shift the equilibrium forward exothermic reaction yielding more products PCl3(g) and Cl2(g).
- Hence exothermic reaction is favoured at low temperature.
(2) Le Chatelier’s principle with reference to change in concentration :
At equilibrium, all the reactants and the products present are at equilibrium concentrations.
Consider the following reaction at the equilibrium state.
Na2(g) + 3H2(g) ⇌ 2NH3(g)
At equilibrium QC = KC
- Addition of the reactants : By the addition of one or more reactants, (H2, N2) a stress is applied of the increase in the concentration of the reactants and QC < KC.
- According to Le Chatelier’s principle, this stress will be reduced by shitting the equilibrium forward by decreasing the concentration of the reactants and increasing the concentration of the products (NH3).
- Addition of products : By addition of the products (NH3), a stress is applied to the chemical equilibrium on the right hand side and QC > Kc. This stress will be reduced by shifting the equilibrium backward thus by decreasing the concentration of the products.
- Removal of the reactants : By removal of the reactants (N2 or H2), the stress produced will be reduced by shifting the equilibrium backward, by decreasing the concentrations of the products.
- Removal of the products : By removal of the products (NH3), the stress produced will be reduced by shifting the equilibrium forward, by converting more reactants into the products.
(D) (a) Reversible reaction (b) Rate of reaction
(a) Reversible reaction : Reactions which do not go to completion and occur in both directions simultaneously are called reversible reactions.
- Reversible reactions proceed in both directions. The direction from reactants to products is the forward reaction, whereas the opposite reaction from products to reactants is called the reverse or backward reaction.
- A reversible reaction is denoted by drawing in between the reactants and product a double arrow, one pointing in the forward direction and other in the reverse direction (⇌).
Example :
(i) H2(g) + I2(g) ⇌ 2HI(g)
(ii) CH3COOH(aq) + H2O(l) ⇌ CH3COO−(aq) + H3O+(aq)
(b) Rate of a reaction : The rate of a decrease in concentration of the reactants per unit time or increase in concentration of products per unit time is called rate of the reaction.
(E) What is the effect of adding chloride on the position of the equilibrium ?
AgCl(s) ⇌ Ag+(aq) + Cl(aq)
On the addition of chloride ion, Cl−(aq), the equilibrium will shift in the reverse direction.
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