Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Chemistry Guide Pdf Chapter 8 Physical and Chemical Equilibrium Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium

### 11th Chemistry Guide Physical and Chemical Equilibrium Text Book Back Questions and Answers

Textbook Evaluation:

I. Choose the best answer:

Question 1.

If K_{b} and K_{f} for a reversible reaction are 0.8 × 10^{-5} and 1.6 × 10^{-4 }respectively, the value of the equilibrium constant is,

a) 20

b) 0.2 × 10^{-1}

c) 0.05

d) None of these

Answer:

a) 20

Question 2.

At a given temperature and pressure, the equilibrium constant values for the equilibria

3A_{2} + B_{2} + 2Cr 2A_{3}BC and

A_{3}BC \(\frac{3}{2}\) [A_{2}] + \(\frac{1}{2}\) B_{2} +C

The relation between K_{1} and K_{2} is

a) K_{1} = \(\frac{1}{\sqrt{\mathrm{K}_{2}}}\)

b)K_{2} = \(\mathrm{K}_{1}^{-1 / 2}\)

C) K_{1}^{2} = 2K_{2}

d) \(\frac{\mathrm{K}_{1}}{2}\) = K_{2}

Answer:

b)K_{2} = \(\mathrm{K}_{1}^{-1 / 2}\)

Question 3.

The equilibrium constant for a reaction at room temperature is K_{1} and that at 700 K is K_{2} If K_{1} > K_{2}, then

a) The forward reaction is exothermic

b) The forward reaction is endothermic

c) The reaction does not attain equilibrium

d) The reverse reaction is exothermic

Answer:

a) The forward reaction is exothermic

Question 4.

The formation of ammonia from N_{2}(g) and H_{2}(g) is a reversible reaction 2NO(g) + O_{2}(g) 2NO_{2}(g) + Heat What is the effect of increase of temperature on this equilibrium reaction

a) equilibrium is unaltered

b) formation of ammonia is favoured

c) equilibrium is shifted to the left

d) reaction rate does not change

Answer:

c) equilibrium is shifted to the left

Question 5.

Solubility of carbon dioxide gas in cold water can be increased by

a) increase in pressure

b) decrease in pressure

c) increase in volume

d) none of these

Answer:

a) increase in pressure

Question 6.

Which one of the following is incorrect statement?

a) for a system at equilibrium, Q is always less than the equilibrium constant

b) equilibrium can be attained from either side of the reaction

c) the presence of catalyst affects both the forward reaction and reverse reaction to the same extent

d) Equilibrium constant varied with temperature

Answer:

a) for a system at equilibrium, Q is always less than the equilibrium constant

Question 7.

K_{1} and K_{2} are the equilibrium constants for the reactions respectively.

N_{2}(g) + O_{2}(g) 2NO(g)

2NO(g) + O_{2}(g) 2NO_{2} (g)

What is the equilibrium constant for the reaction NO_{2}(g) ⇌ 1/2 N_{2}(g) + O_{2}(g)

a) \(\frac{1}{\sqrt{\mathrm{K}_{1} \mathrm{~K}_{2}}}\)

b) \(\left(\mathrm{K}_{1}=\mathrm{K}_{2}\right)^{1 / 2}\)

c) \(\frac{1}{2 \mathrm{~K}_{1} \mathrm{~K}_{2}}\)

d) \(\left(\frac{1}{\mathrm{~K}_{1} \mathrm{~K}_{2}}\right)^{3 / 2}\)

Answer:

a) \(\frac{1}{\sqrt{\mathrm{K}_{1} \mathrm{~K}_{2}}}\)

Question 8.

In the equilibrium, 2A(g) ⇌ 2B(g) + C_{2}(g) the equilibrium concentrations of A, B and C_{2} at 400K are 1 × 10^{-4}M, 2.0 × 10^{-3} M, 1.5 × 10^{-4} M respectively. The value of K_{c} for the equilibrium at 400 K is

a) 0.06

b) 0.09

c) 0.62

d) 3 × 10^{-2}

Answer:

a) 0.06

Question 9.

An equilibrium constant of 3.2 × 10^{-6} for a reaction means, the equilibrium is

a) largely towards forward direction

b) largely towards reverse direction

c) never established

d) none of these

Answer:

b) largely towards reverse direction

Question 10.

\(\frac{\mathrm{K}_{\mathrm{C}}}{\mathrm{K}_{\mathrm{P}}}\) for the reaction,

N_{2}(g) + 3H_{2}(g) ⇌ 2NH_{3} (g) is

a) \(\frac{1}{\mathrm{RT}}\)

b) \(\sqrt{\mathrm{RT}}\)

c) RT

d) (RT)^{2}

Answer:

d) (RT)^{2}

Question 11.

For the reaction AB (g) ⇌ A(g) + B(g), at equilibrium, AB is 20 % dissociated at a total pressure of P, the equilibrium constant K_{p} is related to the total pressure by the expression

a) P = 24 K_{p}

b) P = 8 K_{p}

c) 24 P = K_{p}

d) none of these

Answer:

a) P = 24 K_{p}

Question 12.

In which of the following equilibrium, K_{p} and K_{c} are not equal?

a) 2N0(g) ⇌ N_{2}(g) + O_{2}(g)

b) SO_{2}(g) + NO_{2} ⇌ SO_{3}(g) + NO(g)

c) H_{2}(g) + I_{2}(g) ⇌ 2HI(g)

d) PCl_{5} ⇌ PCl_{3}(g) + Cl_{2}(g)

Answer:

b) SO_{2}(g) + NO_{2} ⇌ SO_{3}(g) + NO(g)

Question 13.

If x is the fraction of PCl_{5} dissociated at equilibrium in the reaction PCl_{5} ⇌ PCl_{3} + Cl_{2} then starting with 0.5 mole of PCl_{5}, the total number of moles of reactants and products at equilibrium is

a) 0.5 – x

b) x + 0.5

c) 2x + 0.5

d) x + 1

Answer:

b) x + 0.5

Question 14.

The values of K_{p1} and K_{p2}; for the reactions, X ⇌ Y + Z, A ⇌ 2B are in the ratio 9 : 1 if degree of dissociation of X and A be equal then total pressure at equilibrium P_{1}, and P_{2} are in the ratio

a) 36 : 1

b) 1 : 1

c) 3 : 1

d) 1 : 9

Answer:

a) 36 : 1

Question 15.

In the reaction

Fe(OH)_{3}(S) ⇌ Fe^{3+} (aq) + 3OH^{–} (aq),

if the concentration of OH^{–} ions is decreased by 1/4 times, then the equilibrium concentration of Fe^{3+} will

a) not changed

b) also decreased by 1/4 times

c) increase by 4 times

d) increase by 64 times

Answer:

d) increase by 64 times

Question 16.

Consider the reaction where K_{p} = 0.5 at a particular temperature PCl_{5}(g) ⇌ PCl_{3} (g) + Cl_{2} (g) if the three gases are mixed in a container so that the partial pressure of each gas is initially 1 atm, then which one of the following is true.

a) more PCl_{3} will be produced

b) more Cl_{2} will be produced

c) more PCl_{5} will be produced

d) none of these

Answer:

c) more PCl_{5} will be produced

Question 17.

Equimolar concentrations of H_{2} and I_{2} are heated to equilibrium in a 1 litre flask. What percentage of the initial concentration of H_{2} has reacted at equilibrium if rate constant for both forward and reverse reactions are equal

a) 33%

b) 66%

c) (33)^{2}%

d) 16.5 %

Answer:

a) 33%

Question 18.

In a chemical equilibrium, the rate constant for the forward reaction is 2.5 × 10^{-2,} and the equilibrium constant is 50. The rate constant for the reverse reaction is,

a) 11.5

b) 50

c) 2 × 10^{2}

d) 2 × 10^{-3}

Answer:

b) 50

Question 19.

Which of the following is not a general characteristic of equilibrium involving physical process

a) Equilibrium is possible only in a closed system at a given temperature

b) The opposing processes occur at the same rate and there is a dynamic but stable condition

c) All the physical processes stop at equilibrium

d) All measurable properties of the system remains constant

Answer:

c) All the physical processes stop at equilibrium

Question 20.

For the formation of Two moles of SO_{3}(g) from SO_{2} and O_{2}, the equilibrium constant is K1. The equilibrium constant for the dissociation of one mole of SO_{3} into SO_{2} and O_{2} is

a) \(1 / \mathrm{K}_{1}\)

b) K_{1}^{2}

c) \(\left(\frac{1}{\mathrm{~K}_{1}}\right)^{1 / 2}\)

d) \(\frac{\mathrm{K}_{1}}{2}\)

Answer:

c) \(\left(\frac{1}{\mathrm{~K}_{1}}\right)^{1 / 2}\)

Question 21.

Match the equilibria with the corresponding conditions:

i) Liquid ⇌ Vapour

ii) Solid ⇌ Liquid

iii) Solid ⇌ Vapour

iv) Solute(s) ⇌ Solute (Solution)

1) Melting point

2) Saturated solution

3) Boiling point

4) Sublimation point

5) Unsaturated solution

Answer:

b) 3 1 4 2

Question 22.

Consider the following reversible reaction at equilibrium, A + B ⇌ C, If the concentration of the reactants A and B are doubled, then the equilibrium constant will

a) be doubled

b) become one fourth

c) be halved

d) remain the same

Answer:

d) remain the same

Question 23.

[Co(H_{2}O)_{6}]^{2+} (aq) (pink) + 4Cl^{–} (aq) ⇌ [CoCl_{4}]^{2-}(aq) (blue) + 6 H_{2}O (l)

In the above reaction at equilibrium, the reaction mixture is blue in colour at room temperature. On cooling this mixture, it becomes pink in color. On the basis of this information, which one of the following is true?

a) ∆H > 0 for the forward reaction

b) ∆H = 0 for the reverse reaction

c) ∆H < 0 for the forward reaction

d) Sign of the ∆H cannot be predicted based on this information

Answer:

a) ∆H > 0 for the forward reaction

Question 24.

The equilibrium constants of the following reactions are:

N_{2} + 3H_{2} ⇌ 2NH_{3}; K_{1}

N_{2} + O_{2} ⇌ 2NO; K_{2}

H_{2} + 1/2O_{2} ⇌ H_{2}O; K_{3}

The equilbrium constant (K) for the reaction ;

2NH_{3} + 5/2 O_{2} ⇌ 2NO + 3H_{2}O, will be

a) K_{2}^{3} K_{3}/K_{1}

b) K_{1} K_{3}^{3}/K_{2}

c) K_{2}K_{3}^{3}/K_{1}

d) K_{2}K_{3}/K_{1}

Answer:

c) K_{2}K_{3}^{3}/K_{1}

Question 25.

A 20 litre container at 400 K contains CO_{2} (g) at pressure 0.4 atm and an excess (neglect the volume of solid SrO). The volume of the container is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when the pressure of CO_{2} attains its maximum value will be: Given that: SrCO_{3}(S) ≅ SrO + CO_{2} (g) [K_{p} = 1.6 atm]

a) 2 litre

b) 5 litre

c) 10 litre

d) 4 litre

Answer:

b) 5 litre

II. Write brief answer to the following questions:

Question 26.

If there is no change in concentration, why is the equilibrium state considered dynamic?

Answer:

Rate of forward reaction = Rate of backward reaction

Question 27.

For a given reaction at a particular temperature, the equilibrium constant has constant value. Is the value of Q also constant? Explain.

Answer:

K_{c} and Q_{c} are constant at equilibrium, both are temperature dependent. When K_{c} is constant at given temperature, Q_{c} also constant.

Question 28.

What is the relation between K_{p} and K_{c}? Given one example for which K_{p} is equal to K_{c}.

Answer:

K_{p} = K_{c}(RT)^{∆ng}

H_{2}(g) + I_{2}(g) ≅ 2HI (g) ; ∆n_{g} = 0 ;

K_{p} = K_{c}.

Question 29.

For a gaseous homogeneous reaction at equilibrium, number of moles of products are greater than the number of moles of reactants. Is K_{c} is larger or smaller than K_{p}.

Answer:

K_{p} >K_{c}

n_{p} > n_{R}

Question 30.

When the numerical value of the reaction quotient (Q) is greater than the equilibrium constant, in which direction does the reaction proceed to reach equilibrium?

Answer:

Q > K_{c}

Reverse reaction is favoured.

Question 31.

For the reaction, A_{2}(g) + B_{2}(g) ⇌ 2AB(g); ∆H is -ve.

the following molecular scenes represent differenr reaction mixture.(A-green, B-blue)

i) Calculate the equilibrium constant K_{p} and (K_{c}).

ii) For the reaction mixture represented by scene (x), (y) the reaction proceed in which directions?

iii) What is the effect of increase in pressure for the mixture at equilibrium.

Answer:

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.

Question 32.

State Le – Chateller principle.

Answer:

“If a system at equilibrium is distributed, then the system shifts itself in a direction that nullifies the effect of that disturbance”.

Question 33.

Consider the following reactions,

a) H_{2}(g) + I_{2}(g) ⇌ 2HI(g)

b) CaCO_{3}(s) ⇌ CaO (s) + CO_{2}(g)

c) S(s) + 3 F_{2}(g) ⇌ SF_{6}(g)

In each of the above reactions find out whether you have to increase (or) decrease the volume to increase the yield of the product.

Answer:

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.

Question 34.

State law of mass action.

Answer:

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

Rate ∝ [Reactant]x

Active mass = \(\left(\frac{\mathrm{n}}{\mathrm{V}}\right)\) mol dm^{-3} (or) mol L^{-1}

Question 35.

Explain how will you predict the direction of an equilibrium reaction.

Answer:

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 stiochiometric 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

Question 36.

Derive a general expression for the equilibrium constant Kp and Kc for the reaction, 3H_{2}(g) + N_{2}(g) ⇌ 2NH_{3}(g).

Answer:

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} = \(\frac{\left[\mathrm{NH}_{3}\right]^{2}}{\left[\mathrm{~N}_{2}\right]\left[\mathrm{H}_{2}\right]^{3}}\)

Question 37.

Write the balanced chemical equation for an equilibrium reaction for which the equilibrium constant is given by expression.

Answer:

K_{c} = \(\frac{\left[\mathrm{NH}_{3}\right]^{4}\left[\mathrm{O}_{2}\right]^{7}}{\left.[\mathrm{NO}]^{4} \mid \mathrm{H}_{2} \mathrm{O}\right]^{6}}\)

Balanced equation is:

4NO + 6H_{2}O ⇌ 4NH_{3} + 7O_{2}

Question 38.

What is the effect of added Inert gas on the reaction at equilibrium?

Answer:

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.

Question 39.

Derive the relation between K_{p} and K_{c}.

Answer:

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}_{∆ng} =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)

Question 40.

One mole of PCl_{5} is heated in one litre closed container. If 0.6 mole of chlorine is found at equilibrium, Calculate the value of equilibrium constant.

Solution:

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}

Question 41.

For the reaction SrCO_{3} (s) ⇌ SrO(s) + CO_{2}(g) the value of equilibrium constant K_{p} = 2.2 × 10^{-4} at 1002 K. Calculate K_{c} for the reaction.

Solution :

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}

Question 42.

To study the decomposition of hydrogen iodide, a student fills an evacuated 3 litre flask with 0.3 mol of HI gas and allows the reaction to proceed at 500°C. At equilibrium he found the concentration of HI which is equal to 0.05 M. Calculate K_{c} and K_{p}.

Solution :

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)^{(∆ng})

∆n_{g} = 2 – 2 = 0

K_{p} = 0.25(RT)^{0}

K_{p} = 0.25.

Question 43.

Oxidation of nitrogen monoxide was studied at 200 with initial pressures of 1 atm NO and 1 atm of O_{2}. At equilibrium partial pressure of oxygen is found to be 0. 5 atm calculate K_{p} value.

Solution:

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}.

Question 44.

1 mol of CH_{4}, 1 mole of CS_{2} and 2 mol of H_{2}S are 2 mol of H_{2} are mixed in a 500 ml flask. The equilibrium constant for the reaction K_{c} = 4 x 10^{-2} mol^{2} lit^{-2}. In which direction will the reaction proceed to reach equilibrium ?

Solution :

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}

Question 45.

At particular temperature K_{c} = 4 × 10^{-2} for the reaction, H_{2}S ⇌ 2H_{2}(g) + S_{2}(g). Calculate the K_{c} for each of the following reaction.

i) 2H_{2}S(g) ⇌ 2H_{2} + S_{2}(g)

ii) 3H_{2}S(g) ⇌ 3H_{2}(g) + 3/2 S_{2}(g)

Solution :

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}

Question 46.

28 g of Nitrogen and 6 g of hydrogen were mIed In a 1 litre closed container. At equIlibrium 17 g NH3 was produced. Calculate the weight of nitrogen, hydrogen at equilibrium.

Solution:

Given m_{N2} =28g;

m_{H2} = 6g;

V = 1 L.

(n_{N2})_{initial} = \(\frac{28}{28}\) = 1

(n_{H2})_{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

Question 47.

The equilibrium for the dissociation of XY_{2} is given as,

2 XY_{2}(g) ⇌ 2 XY(g) + Y_{2}(g)

if the degree of dissociation x is so small compared to one. Show that 2 K_{p} = PX^{3} where P is the total pressure and K_{p} is the dissociation equilibrium constant of XY_{2}.

Answer:

2 XY_{2}(g) ⇌ 2 XY(g) + Y_{2}(g)

Question 48.

A sealed container was filled with 0.3 mol H_{2}(g), 0.4 mol I_{2}(g) and 0.2 mol HI(g) at 800 K and total pressure 1.00 bar. Calculate the amounts of the components in the mixture at equilibrium given that K = 870 for the reaction, H_{2}(g) + I_{2}(g) ⇌ 2 HI (g).

Solution:

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

Question 49.

Deduce the Vant Hoff equation.

Answer:

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.

Question 50.

The equilibrium constant K_{p} for the reaction N_{2}(g) + 3H_{2}(g) ⇌ 2NH_{3}(g) is 8.19 × 10^{2} at 298 K and 4.6 × 10^{-1} at 498 K. Calculate ∆H° for the reaction.

Solution :

K_{p1} = 8.19 × 10^{2};

T_{1} = 298 K

K_{p1} = 8.19 × 10^{2};

T_{1} = 298 K

K_{p2} = 4.16 × 10^{-1};

T_{2} = 498 K

Question 51.

The partial pressure of carbon dioxide in the reaction CaCO_{3}(s) ⇌ CaO(s) + CO_{2}(g) is 1.017 × 10^{-3} atm at 500°C. Calculate K_{p} at 600°C for the reaction. H for the reaction is 181 KJ mol^{-1} and does not change in the given range of temperature.

Solution:

P_{CO2} = 1.017 × 10^{-3} atm

T = 500°C;

K_{p} = P_{CO2}

∴ K_{p1} = 1.017 × 10^{-3};

T = 500 + 273 = 773 K

K_{p2} = ?

T = 600 + 273 = 873 K

∆H° = 181 KJ mol^{-1}

### 11th Chemistry Guide Physical and Chemical Equilibrium Additional Questions and Answers

I.Choose the best answer:

Question 1.

For which of the following K_{p} is less than K_{c}?

a) N_{2}O_{4} ⇌ 2NO_{2}

b) N_{2} + 3H_{2} ⇌ 2NH_{3}

c) H_{2} + I_{2} ⇌ 2HI

d) CO + H_{2}O ⇌ CO_{2} + H_{2}

Answer:

b) N_{2} + 3H_{2} ⇌ 2NH_{3}

Question 2.

In which of the following reaction, the value of K_{p} will be equal to K_{c}?

a) H_{2} + I_{2} ⇌ 2HI

b) PCl_{5} ⇌ PCl_{3} + Cl_{2}

c) 2NH_{3} ⇌ N_{2} + 3H_{2}

d) 2SO_{2} +O_{2} ⇌ 2SO_{3}

Answer:

a) H_{2} + I_{2} ⇌ 2HI

Question 3.

For homogeneous gas reaction 4NH_{3} + 5O_{2} ⇌ 4NO + 6H_{2}O. The equilibrium constant K_{c} has the unit

a) (concentration)^{1}

b) (concentration)^{1}

c) (concentration)^{9}

d) (concentration)^{10}

Answer:

a) (concentration)^{1}

Question 4.

The reaction, 2SO_{2}(g) + O_{2}(g) ⇌ 2SO_{3}(g) is carried out in a 1 dm3 vessel and 2 dm^{3} vessel separately. The ratio of the reaction velocities will be

a) 1 : 8

b) 1 : 4

c) 4 : 1

d) 8 : 1

Answer:

d) 8 : 1

Question 5.

K_{p} for the following reaction at 700 K is 1.3 × 10^{-3} atm^{-1}. The K_{c} same temperature for the reaction. 2SO_{2} + O_{2} ⇌ 2SO_{3} will be

a) 1.1 × 10^{-2}

b) 3.1 × 10^{-2}

c) 5.2 × 10^{-2}

d) 7.4 × 10^{-2}

Answer:

d) 7.4 × 10^{-2}

Question 6.

The equilibrium constant expression for the equilibrium :

2NH_{2}(g) + 2O_{2}(g) ⇌ N_{2}O(g) + 3H_{2}O(g) is

a) K_{c} = \(\frac{\left[\mathrm{N}_{2} \mathrm{O}\right]\left[\mathrm{H}_{2} \mathrm{O}\right]^{3}}{\left[\mathrm{NH}_{3}\right]\left[\mathrm{O}_{2}\right]}\)

b) K_{c} = \(\frac{\left[\mathrm{H}_{2} \mathrm{O}\right]^{3}\left[\mathrm{~N}_{2} \mathrm{O}\right]}{\left[\mathrm{NH}_{3}\right]^{2}\left[\mathrm{O}_{2}\right]^{2}}\)

c) K_{c} = \(\frac{\left[\mathrm{NH}_{3}\right]^{2}\left[\mathrm{O}_{2}\right]^{2}}{\left[\mathrm{~N}_{2} \mathrm{O}\right]\left[\mathrm{H}_{2} \mathrm{O}\right]^{3}}\)

d) K_{c} = \(\frac{\left[\mathrm{NH}_{3}\right]\left[\mathrm{O}_{2}\right]}{\left[\mathrm{N}_{2} \mathrm{O}\right]\left[\mathrm{H}_{2} \mathrm{O}\right]}\)

Answer:

b) K_{c} = \(\frac{\left[\mathrm{H}_{2} \mathrm{O}\right]^{3}\left[\mathrm{~N}_{2} \mathrm{O}\right]}{\left[\mathrm{NH}_{3}\right]^{2}\left[\mathrm{O}_{2}\right]^{2}}\)

Question 7.

For the system 3A + 2B ⇌ C, the expression for equilibrium constant is

a) \(\frac{[3 \mathrm{~A}][2 \mathrm{~B}]}{[\mathrm{C}]}\)

b) \(\frac{[\mathrm{C}]}{[3 \mathrm{~A}][2 \mathrm{~B}]}\)

c) \(\frac{[\mathrm{A}]^{3}[\mathrm{~B}]^{2}}{[\mathrm{C}]}\)

d) \(\frac{[\mathrm{C}]}{[\mathrm{A}]^{3}[\mathrm{~B}]^{2}}\)

Answer:

d) \(\frac{[\mathrm{C}]}{[\mathrm{A}]^{3}[\mathrm{~B}]^{2}}\)

Question 8.

For the reaction PCl_{3}(g) + Cl_{2}(g) ⇌ PCl_{5}(g) at 250°C, the value of K_{c} is 26, then the value of K_{p} at the same temperature will be

a) 0.61

b) 0.57

c) 0.83

d) 0.46

Answer:

a) 0.61

Question 9.

The equilibrium constant for the reaction N_{2}(g) + O_{2}(g) ⇌ 2NO(g) at temperature T is 4 × 10^{-4}. The value of K_{c} for the reaction NO(g) ⇌ 1/2 N_{2}(g) + 1/2 O_{2}(g) at the same temperature is ___________.

a) 4 × 10^{-4}

b) 50

c) 2.5 × 10^{2}

d) 0.02

Answer:

b) 50

Question 10.

In which of the following equilibria, the value of K_{p} is less than K_{c}?

a) H_{2} + I_{2} ⇌ 2HI

b) N_{2} + 3 H_{2} ⇌ 2NH_{3}

c) N_{2} + O_{2} ⇌ 2NO

d) CO + H_{2}O ⇌ CO_{2} + H_{2}

Answer:

b) N_{2} + 3 H_{2} ⇌ 2NH_{3}

Question 11.

In the reversible reaction A + B ⇌ C + D. The concentration of each C and D at equilibrium was 0.8-mole lit, then the equilibrium constant will be

a) 6.4

b) 0.64

c) 1.6

d) 16.0

Answer:

d) 16.0

Question 12.

The K_{c} for H_{2}(g) + I_{2}(g) ⇌ 2 HI(g), is 64. If the volume of the container is reduced to one – half of its original volume, the value of the equilibrium constant will be

a) +28

b) 64

c) 32

d) 16

Answer:

b) 64

Question 13.

For the reaction C(s) + CO_{2}(g) ⇌ 2CO (g), the partial pressure of CO_{2} and CO are 2.0 and 4.0 atm respectively at equilibrium. The K_{p} for the reaction is

a) 0.5

b) 4.0

c) 8.0

d) 32.0

Answer:

c) 8.0

Question 14.

When the rate of forward reaction becomes equal to backward reaction, this state is termed as

a) chemical equilibrium

b) Reversible state

c) Equilibrium

d) All of these

Answer:

d) All of these

Question 15.

The equilibrium constant of the reaction H_{2}(g) + I_{2}(g) ⇌ 2 HI(g) is 64. If the volume of the container is reduced to one – fourth of its original volume, the value of the equilibrium constant will be

a) 16

b) 32

c) 64

d) 128

Answer:

c) 64

Question 16.

For the reaction PCl_{3}(g) + Cl_{2}(g) ⇌ PCl_{5}(g), the position of equilibrium can be

shifted to the right by _______.

a) increasing the temperature

b) Doubling the volume

c) Addition of Cl_{2} at constant volume

d) Addition of equimolar quantities of PCl_{3} and PCl_{5}

Answer:

b) Doubling the volume

Question 17.

For the reaction 2 NOCl(g) ⇌ 2 NO(g) + Cl_{2}(g), K_{c} at 427 is 3 × 10^{-6} L mol^{-1}. The value of K_{p} is nearly

a) 7.50 × 10^{-5}

b) 2.50 × 10^{-5}

c) 2.50 × 10^{-4}

d) 1.72 × 10^{-4}

Answer:

d) 1.72 × 10^{-4}

Question 18.

Rate of reaction curve for equilibrium can be like [r_{f} = forward rate, r_{b} = backward rate]

Answer:

a)

Question 19.

For a reaction, N_{2} + 3H_{2} ⇌ 2NH_{3}(g), the value of K_{c} does not depend upon

A) initial concentration of the reactants

B) pressure

C) temperature

D) Catalyst

a) only C

b) A, B, C

c) A, B, C

d) A, B, C, D

Answer:

c) A, B, C

Question 20.

2NO_{2} ⇌ 2NO + O_{2}, K = 1.6 × 10^{-12} NO + \(\frac{1}{2}\)O_{2} ⇌ NO_{2}; K’ = ?

a) K’ = \(\frac{1}{\mathrm{~K}^{2}}\)

b) K’ = \(\frac{1}{\mathrm{~K}}\)

c) K’ = \(\frac{1}{\sqrt{\mathrm{K}}}\)

d) None of these

Answer:

c) K’ = \(\frac{1}{\sqrt{\mathrm{K}}}\)

Question 21.

At 1000 K, the value of Kp for the reaction, A(g) + 2B(g) ⇌ 3C (g) + D(g) is 0.05 atm. The value of Kc in terms of R would be

a) 20000 R

b) 0.02 R

c) 5 × 10^{-5} R

d) 5 × 10^{-5} × R^{-1}

Answer:

d) 5 × 10^{-5} × R^{-1}

Question 22.

In a chemical equilibrium, the rate constant for the backward reaction is 7.5 x 10^{-4} and the equilibrium constant is 1.5. The rate constant for the forward reaction

a) 2 × 10^{-3}

b) 5 × 10^{-4}

c) 1.12 × 10^{-3}

d) 9.0 × 10^{-4}

Answer:

c) 1.12 × 10^{-3}

Question 23.

A reversible reaction is one which

a) Proceeds in one direction

b) proceeds in both direction

c) proceeds spontaneously

d) All the statements are wrong

Answer:

b) proceeds in both direction

Question 24.

The equilibrium constant in a reversible reaction at a given temperature

a) depends on the initial concentration of the reactants

b) depends on the concentration of the products at equilibrium

c) does not depend on the initial concentrations

d) It is not characteristic of the reaction

Answer:

c) does not depend on the initial concentrations

Question 25.

For the system A(g) + 2B(g) ⇌C (g), the equilibrium concentrations are (A) 0.06 mole / lit (B) 0.12 mole / lit and (C) 0.216 mole / lit. The K_{eq} for the reaction is

a) 250

b) 416

c) 4 × 10^{-3}

d) 125

Answer:

a) 250

Question 26.

A chemical reaction is at equilibrium when

a) Reactants are completely transformed into products

b) The rates of forward and backward reactions are equal

c) Formation of products is minimized

d) Equal amounts of reactants and products are present

Answer:

b) The rates of forward and backward reactions are equal

Question 27.

Partial pressures of A, B, C and D on the basis of gaseous system A + 2B ⇌ C + 3D are A = 0.20, B = 0.10, C = 0.30 and D = 0.50 atm. The numerical value of equilibrium constant is _______.

a) 11.25

b) 18.75

c) 5

d) 3.75

Answer:

b) 18.75

Question 28.

Molar concentration of 96 g of O_{2} contained in a 2 L vessel is ______.

a) 16 mol / L

b) 1.5 mol / L

c) 4 mol / L

d) 24 mol / L

Answer:

b) 1.5 mol / L

Question 29.

According to law of mass action rate of a chemical reaction is proportional to

a) concentration of reactants

b) molar concentration of reactants

c) concentration of products

d) molar concentration of products

Answer:

b) molar concentration of reactants

Question 30.

The equilibrium constant of the reaction SO_{2}(g) + \(\frac{1}{2}\) O_{2}(g) ⇌ SO_{3} (g) is 4 × 10^{-3} atm^{-1/2}. The equilibrium constant of the reaction 2SO_{3}(g) ⇌ 2SO_{2}(g) + O_{2}(g) would be

a) 250 atm

b) 4 × 10^{3} atm

c) 0.25 × 10^{4} atm

d) 6.25 × 10^{4} atm

Answer:

d) 6.25 × 10^{4} atm

Question 31.

The rate constant for forward and backward reactions of the hydrolysis of ester are 1.1 × 10^{-2} and 1.5 × 10^{-3} per minute respectively. The equilibrium constant for the reaction is ___________.

a) 4.33

b) 5.33

c) 6.33

d) 7.33

Answer:

d) 7.33

Question 32.

The active mass of 64 g of HI is a two litre of flask would be

a) 2

b) 1

c) 5

d) 0.25

Answer:

d) 0.25

Question 33.

At a certain temperature, 2HI ⇌ H_{2} + I_{2}. Only 50 % HI is dissociated at equilibrium. The equilibrium constant is

a) 0.25

b) 1.0

c) 3.0

d) 0.50

Answer:

a) 0.25

Question 34.

Unit of equilibrium constant for the reversible reaction, H_{2} + I_{2} ⇌ 2HI is ________.

a) mol^{-1} litre

b) mol^{-2} litre

c) mol litre^{-1}

d) none of these

Answer:

d) none of these

Question 35.

For the reaction

CO(g) + \(\frac{1}{2}\)O_{2}(g) ⇌ CO_{2}(g), K_{p}/K_{c} is _______.

a) RT

b) (RT)^{-1}

c) (RT)^{-1/2}

d) (RT)^{1/2}

Answer:

c) (RT)^{-1/2}

Question 36.

The following equilibrium are given

N_{2} +3H_{2} ⇌ 2NH_{2} ………K_{1}

N_{2} + O_{2} ⇌ 2N0 ………….K_{2}

H_{2} + 1/2 O_{2} ⇌ H_{2}O …….. K_{3}

The equilibrium constant of the reaction.

2 NH_{3} + 5/2 O_{2} ⇌ 2NO + 3H_{2}O, in terms of K_{1}, K_{2} and K_{3} is _________.

a) \(\frac{\mathbf{K}_{1} \mathbf{K}_{2}}{\mathbf{K}_{3}}\)

b) \(\frac{\mathrm{K}_{1} \mathrm{~K}_{3}^{2}}{\mathrm{~K}_{2}}\)

c) \(\frac{\mathrm{K}_{2} \mathrm{~K}_{3}^{2}}{\mathrm{~K}_{1}}\)

d) K_{1} K_{2} K_{3}

Answer:

c) \(\frac{\mathrm{K}_{2} \mathrm{~K}_{3}^{2}}{\mathrm{~K}_{1}}\)

Question 37.

For the reaction

2NO_{2}(g) ⇌ 2NO(g) + O_{2}(g) K_{c} = 1.8 × 10^{-6} at 184°C and R = 0.083 Jk^{-1} mol^{-1}. When K_{p} and are compared at 184C, it

is found that

a) K_{p} > K_{c}

b) K_{p} < K_{c}

c) K_{p} – K_{c}

d) K_{p} × K_{c}( depends upon pressure of gases)

Answer:

a) K_{p} > K_{c}

Question 38.

The rate of forward reaction is two times that of reverse reaction at a given temperature and identical concentration. K equilibrium is

a) 2.5

b) 2.0

c) 0.5

d) 1.5

Answer:

b) 2.0

Question 39.

In a reaction A+B ⇌ C+D,the concentrations of A, B, C and D [in mole/lit] are 0.5, 0.8, 0.4 & 1.0 respectively. Th. equilibrium constant is

a) 0.1

b) 1.0

c) 10

d) ∞

Answer:

b) 1.0

Question 40.

For the following thre. reactions 1, 2 and 3 equilibrium constants are given:

1. CO(g) + H_{2}O(g) ⇌ CO(g) + H_{2}(g); K_{1}

2.CH_{4}(g) + H_{2}O(g) ⇌ CO + 3H(g); K_{2}

3. CH_{4}(g) + 2H_{2}O(g) ⇌ CO(g) + 4H_{2}(g); K_{3}

Which of the following relations is correct?

a) K_{1}√K_{2} = K_{3}

b) K_{2}K_{3} = K_{1}

c) K_{3} = K_{1}K_{2}

d) K_{3} = K_{2}^{3} K_{1}^{2}

Answer:

c) K_{3} = K_{1}K_{2}

Question 41.

4 moles of A is mixed with 4 moles of B. At equilibrium for the reaction A + B ⇌ C + D. 2 moles of C and D is formed. The equilibrium constant for the reaction will be

a) \(\frac{1}{4}\)

b) \(\frac{1}{2}\)

c) 1

d) 4

Answer:

c) 1

Question 42.

In a reaction PCl_{5} ⇌ PCl_{3} + Cl_{2} degree of dissociation is 30%. If Initial moles of PCl_{5} is one then total moles at equilibrium is

a) 1.3

b) 0.7

c) 1.6

d) 1.0

Answer:

a) 1.3

Question 43.

When 3 moles of A and 1 mole of B are mixed in 1 lit vessel, the following reaction takes place A(g) + B(g) ⇌ 2C(g), 1.5 moles of ‘C’ are formed. The equilibrium constant for the reaction is

a) 0.12

b) 0.25

c) 0.50

d) 4.0

Answer:

d) 4.0

Question 44.

In which of the following, the reaction proceeds towards completion

a) K = 10^{3}

b) K = 10^{-2}

c) K = 10

d) K = 1

Answer:

a) K = 10^{3}

Question 45.

Two moles of NH_{3} when put into a previously evacuated vessel (1L) partially dissociate into N_{2} and H_{2}. If at equilibrium one mole of NH_{3} is present, the equilibrium constant is ________.

a) \(\frac{3}{4}\) mol^{2} lit^{-2}

b) \(\frac{27}{64}\) mol^{2} lit^{-2}

c) \(\frac{27}{32}\) mol^{2} lit^{-2}

d) \(\frac{27}{16}\) mol^{2} lit^{-2}

Answer:

d) \(\frac{27}{16}\) mol^{2} lit^{-2}

Question 46.

For the reaction N_{2}(g) + O_{2}(g) ⇌ 2NO(g) the value of K_{c} at 800°C is 0.1. When the equilibrium concentration of both the reactions is 0.5 mol, what is the value of K_{p} at the same temperature?

a) 0.5

b) 0.1

c) 0.01

d) 0.025

Answer:

b) 0.1

Question 47.

28 g of N_{2}(g) and 6g of H_{2}(g) were mixed at equilibrium 17 g NH_{3} was produced. The weight of N_{2} and H_{2} at equilibrium are respectively.

a) 11g, 0g

b) 1g, 3g

c) 14g, 3g

d) 11g, 3g

Answer:

c) 14g, 3g

Question 48.

2SO_{3} ⇌ 2SO_{2} + O_{2} If K_{c} = 100, α = 1, half of the reaction is completed, the concentration of SO_{3} and SO_{2} are equal, the concentration of O_{2} is _________.

a) 0.001 M

b) \(\frac{1}{2}\) SO_{2}

c) 2 times of SO_{2}

d) Data in complete

Answer:

d) Data in complete

Question 49.

For reaction HI ⇌ \(\frac{1}{2}\)H_{2} + \(\frac{1}{2}\)I_{2} value of K_{c} is \(\frac{1}{8}\), then value of K is \(\frac{1}{8}\) for H_{2} + I_{2} ⇌ 2HI.

a) \(\frac{1}{64}\)

b) 64

c) \(\frac{1}{8}\)

d) 8

Answer:

b) 64

Question 50.

2 moles of PCl_{5} were heated in a closed vessel of 2 liter capacity. At equilibrium, 40% of PCl_{5} is dissociated into PCl_{3} and Cl_{2}. The value of equilibrium constant is

a) 0.266

b) 0.53

c) 2.66

d) 5.3

Answer:

a) 0.266

Question 51.

For a reaction H_{2} + I_{2} ⇌ 2HI at 721 k, the value of equilibrium constant is 50. If 0.5 mol each of H_{2} and I_{2} is added to the system the value of equilibrium constant will be

a) 40

b) 60

c) 50

d) 30

Answer:

c) 50

Question 52.

In a reaction the rate of reaction is proportional to its active mass, this statement is known as

a) Law of mass action

b) Le – Chatelier principle

c) Faradays’s law of electrolysis

d) law of constant proportion

Answer:

b) Le – Chatelier principle

Question 53.

In a chemical equilibrium A + B ⇌ C + D, when one mole each of the two reactants are mixed, 0.6 mole each of the products are formed. The equilibrium constant calculated is

a) 1

b) 0.36

c) 2.25

d) \(\frac{4}{9}\)

Answer:

c) 2.25

Question 54.

Under a given set of experimental conditions, with an increase in the concentration of the reactants, the rate of a chemical reaction.

a) Decreases

b) Increases

c) Remains unaltered

d) First decreases and then increases

Answer:

c) Remains unaltered

Question 55.

A + B ⇌ C + D, If finally the concentrations of A and B are both equal but at equilibrium concentration of D will be twice of that of A then what will be the equilibrium constant of reaction.

a) \(\frac{4}{9}\)

b) \(\frac{9}{4}\)

c) \(\frac{1}{9}\)

d) 4

Answer:

d) 4

Question 56.

Theory of ‘active mass’ indicates that the rate of chemical reaction is directly proportional to the ___________.

a) Equilibrium constant

b) Properties of reactants

c) Volume of apparatus

d) Concentration of reactants

Answer:

d) Concentration of reactants

Question 57.

In the reaction, A + B ⇌ 2C, at equilibrium the concentration of A and B is 0.20 mol L^{-1} each and that of C was found to be 0.60 mol L^{-1}. The equilibrium constant of the reaction is

a) 2.4

b) 18

c) 4.8

d) 9

Answer:

d) 9

Question 58.

The rate at which substances react depends on their _______.

a) Atomic weight

b) Molecular weight

c) Equivalent weight

d) Active mass

Answer:

c) Equivalent weight

Question 59.

If in the reaction N_{2}O_{4} ⇌ 2NO_{2}, α is that part of N_{2}O_{4} which dissociates, then the number of moles at equilibrium will be

a) 3

b) 1

c) (1- α)^{2}

d) (1 + α)

Answer:

d) (1 + α)

Question 60.

On decomposition of NH_{4}HS, the following equilibrium is established NH_{4}HS(s) ⇌ NH_{3}(g) + H_{2}S(g). If the total pressure is P atm, then the equilibrium constant Kp is equal to _______.

a) P atm

b) P^{2} atm^{2}

c) \(\frac{\mathrm{P}^{2}}{4}\) atm^{2}

d) 2P atm

Answer:

c) \(\frac{\mathrm{P}^{2}}{4}\) atm^{2}

Question 61.

Some gaseous equilibrium are given below:

i) CO + H_{2}O ⇌ CO_{2} + H_{2}

ii) 2CO + O_{2} ⇌ 2CO

iii) 2H_{2} + O_{2} ⇌ 2H_{3}O

find out the relation between equilibrium constants.

a) K = K_{1}K_{2}

b) K = (K_{1}K_{2})^{2}

c) K = (K_{1}K_{2})^{–\(\frac{1}{2}\)}

d) K = \(\left(\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}\right)^{\frac{1}{2}}\)

Answer:

d) K = \(\left(\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}\right)^{\frac{1}{2}}\)

Question 62.

K_{c} = 9 for the reaction A + B ⇌ C + D. If A and B are taken In equal amounts, then amount of ‘C’ in equilibrium is

a) 1

b) 0.25

c) 0.75

d) none of these

Answer:

c) 0.75

Question 63.

In the reaction C(S) + CO_{2}(g) ⇌ 2CO(g) the equilibrium pressure is 12 atm. If 50 % of CO_{2} reacts, then K_{p} will be ________.

a) 12 atm

b) 16 atm

c) 20 atm

d) 24 atm

Answer:

b) 16 atm

Question 64.

For the following gases equilibrium, N_{2}O_{4}(g) ⇌ 2NO_{2}(g) K_{p} is found to be equal to K_{c}. This is attained when,

a) 0°C

b) 273 k

c) 1 k

d) 12.19 k

Answer:

d) 12.19 k

Question 65.

Consider the following reversible gaseous reactions {At 298 K)

a) N_{2}O_{4} ⇌ 2NO_{2}

b) 2SO_{2} + O_{2} ⇌ 2SO_{3}

c) 2HI ⇌ H_{2} + I_{2}

d) X + Y ⇌ 4Z

Answer:

a) N_{2}O_{4} ⇌ 2NO_{2}

Question 66.

For the reaction A + 2B ⇌ 2C at equilibrium [C] = 1.4 M, [A]_{0} = 1 M, [B]_{0} = 2M, [C]_{0} = 3 M. The value of K_{c} is .

a) 0.084

b) 8.4

c) 84

d) 840

Answer:

a) 0.084

Question 67.

For the reaction H_{2}(g) + I_{2}(g) ⇌ 2HI(g). K_{c} = 66.9 at 350°C and K_{c} = 50.0 at 448°C. The reaction has

a) ∆H = +ve

b) ∆H = -ve

c) ∆H = zero

d) ∆H = not found the signs.

Answer:

b) ∆H = -ve

Question 68.

In an equilibrium reaction H_{2}(g) + I_{2}(g) ⇌ 2HI(g), ∆H = -3000 calories, which factor favours dissociation of HI

a) Low temperature

b) High pressure

c) High temperature

d) Low pressure

Answer:

c) High temperature

Question 69.

In an equilibrium reaction for which ∆G°= 0, the equilibrium constant K Is

a) 0

b) 1

c) 2

d) 10

Answer:

b) 1

Question 70.

Consider the following reversible reaction at equilibrium, 2H_{2}O(g) ⇌ 2H_{2}(g) + O_{2}(g). Which one of the following ∆H = 241. 7 kJ changes in a conditions will lead to maximum decomposition of H_{2}O(g)?

a) Increasing both temperature and pressure

b) Decreasing temperature and increasing pressure

c) Increasing temperature and decreasing pressure

d) Increasing temperature at constant pressure.

Answer:

c) Increasing temperature and decreasing pressure

Question 71.

In which of the following system doubling the volume of the container causes a shift to the right.

a) H_{2}(g) + Cl_{2}(g) ⇌ 2HCl(g)

b) 2CO(g) + O_{2}(g) ⇌ 2CO_{2}(g)

c) N_{2}(g) + 3H_{2}(g) ⇌ 2NH_{3}(g)

d) PCl_{5}(g) ⇌ PCl_{3}(g) + Cl_{2}(g)

Answer:

d) PCl_{5}(g) ⇌ PCl_{3}(g) + Cl_{2}(g)

Question 72.

∆G° (HI, g) ≈ 1.7 kJ. What is the equilibrium constant at 25° C for 2HI(g) ⇌ H_{2}(g) + I_{2}(g)

a) 24.0

b) 3.9

c) 2.0

d) 0.5

Answer:

d) 0.5

Question 73.

In the reaction A_{2}(g) + 4B_{2}(g) ⇌ 2AB_{4} ∆H < 0 the formation of AB_{4} is will, be favoured at

a) Low temperature, high pressure

b) High temperature, low pressure

c) Low temperature, low pressure

d) High temperature, high pressure

Answer:

a) Low temperature, high pressure

Question 74.

N_{2} + 3H_{2} ⇌ 2NH_{3}. If temperature of following equilibrium reaction increases then the reaction.

a) Shifts Right side

b) Shifts left side

C) Remains unchanged

d) No change

Answer:

b) Shifts left side

Question 75.

Consider the equilibrium N_{2}(g) + 3H_{2}(g) ⇌ 2NH_{3} ∆H = -93.6 kJ. The maximum yield of ammonia is obtained by

a) Decrease of temperature and increase of pressure.

b) Increase of temperature and decreases of pressure.

c) Decrease of both the temperature and pressure

d) Increase of both the temperature and pressure

Answer:

a) Decrease of temperature and increase of pressure.

Question 76.

H_{2}(g) + I_{2}(g) ⇌ 2HI(g) ∆H = +q cal, then formation of HI

a) Is favoured by lowering the temperature

b) Is favoured by increasing the pressure

c) Is unaffected by change in pressure

d) Is unaffected by change in temperature

Answer:

c) Is unaffected by change in pressure

Question 77.

The formation of SO_{3} takes place according to the following reaction, 2SO_{2} + O_{2} ⇌ 2SO_{3}; ∆H = 45.2 k cal. The formation of SO_{3} is favoured by

a) Increasing in temperature

b) Removal of oxygen

c) Increase of volume

d) Increase of pressure

Answer:

d) Increase of pressure

Question 78.

Which of the following equilibrium is not shifted by an increase in the pressure?

a) H_{2}(g) + 3H_{2}(g) ⇌ 2HI(g)

b) N_{2}(g) + 3H_{2}(g) ⇌ 2NH_{3}(g)

c) 2CO(g) + O_{2}(g) ⇌ 2CO_{2}(g)

d) 2C(g) + O_{2}(g) ⇌ 2CO(g)

Answer:

a) H_{2}(g) + 3H_{2}(g) ⇌ 2HI(g)

Question 79.

Consider the heterogeneous equilibrium in a closed container. NH_{4}HS(s) ⇌ NH_{3}(g) + H_{2}S(g) if more NH_{4}HS is added to the equilibrium

a) Partial pressure of NH_{3} increases

b) Partial pressure of H_{2}S increases

c) Total pressure in the container increases

d) No effect on partial pressure of NH_{3} and H_{2}S

Answer:

d) No effect on partial pressure of NH_{3} and H_{2}S

Question 80.

According to Le – Chatelier’s principle adding heat to a solid and liquid in equilibrium with endothermic nature will cause the ____________.

a) Temperature to rise

b) Temperature to fall

c) Amount of solid to decrease

d) Amount of liquid to decrease

Answer:

c) Amount of solid to decrease

Question 81.

According to Le-chatelier’s principle, adding heat to a solid to liquid in equilibrium will causes the

a) temperature to increase

b) temperature to decrease

c) amount of liquid to increase

d) amount of solid to increase

Answer:

c) amount of liquid to increase

Question 82.

For the reaction A + B + Q ⇌ C + D, if the temperature is increased, then concentration of the products will

a) increase

b) decrease

c) Remain same

d) become zero

Answer:

a) increase

Question 83.

H_{2}(g) + I_{2}(g) ⇌ 2HI(g). In this reaction when pressure increases, the reaction direction

a) does not change

b) is forward

c) is backward

d) decreases

Answer:

a) does not change

Question 84.

Which of the following factor is shifted the reaction PCl_{3} + Cl_{2} ⇌ PCl_{5} at the left side?

a) Adding PCl_{5}

b) increase pressure

c) constant temperature

d) catalyst

Answer:

a) Adding PCl_{5}

Question 85.

The rate of reaction of which of the following is not affected by pressure?

a) PCl_{3} + Cl_{2} ⇌ PCl_{5}

b) N_{2} + 3 H_{2 } ⇌ 2NH_{3}

c) N_{2} + O_{2} ⇌ 2NO

d) 2 SO_{2} + O_{2} ⇌ 2 SO_{3}

Answer:

c) N_{2} + O_{2} ⇌ 2NO

Question 86.

Which reaction is not affected by change in pressure?

a) H_{2} + I_{2} ⇌ 2HI

b) 2 C + O_{2} ⇌ 2CO

c) N_{2} + 3 H_{2} ⇌ 2NH_{3}

d) PCl_{5} ⇌ PCl_{3} + Cl_{2}

Answer:

a) H_{2} + I_{2} ⇌ 2HI

Question 87.

Which of the following reactions proceeds at low pressure?

a) N_{2} + 3H_{2 }⇌ 2NH_{3}

b) H_{2} + I_{2} ⇌ 2HI

c) PCl_{5} ⇌ PCl_{3} + Cl_{2}

d) N_{2} + O_{2} ⇌ 2NO

Answer:

c) PCl_{5} ⇌ PCl_{3} + Cl_{2}

Question 88.

In the following reaction PCl_{5} ⇌ PCl_{3}(g) + Cl_{2}(g) at constant temperature, rate of backward reaction, is increased by

a) inert gas mixed at constant volume

b) Cl_{2} gas mixed at constant volume

c) inert gas mixed at constant pressure

d) PCl_{5} mixed in constant volume

Answer:

b) Cl_{2} gas mixed at constant volume

Question 89.

On cooling of following system at equilibrium CO_{2}(s) ⇌ CO_{2}(g)

a) There is no effect on the equilibrium state

b) more gas is formed

c) more gas solidifies

d) None

Answer:

c) more gas solidifies

Question 90.

On the velocity in a reversible reaction the correct explanation of the effect of catalyst is _______.

a) it provides a new reaction path of low activation energy

b It increases a kinetic energy of the reacting molecules

c) it displaces the equilibrium state on right side

d) it decreases the velocity of backward reaction

Answer:

a) it provides a new reaction path of low activation energy

Question 91.

According to Le – Chatelier’s principle, if heat is given to solid – liquid system, then

a) Quantity of solid will reduce

b) Quantity of liquid will reduce

c) increase in temperature

d) decrease in temperature

Answer:

a) Quantity of solid will reduce

Question 92.

In a gives system, water and ice are in equilibrium, if pressure is applied to the system then .

a) more of ice is formed

b) amount of ice and water will remain same

c) more of ice is melted

d) either (a) or (c)

Answer:

c) more of ice is melted

Question 93.

Answer:

b)

Question 94.

The value of ∆G° for a reaction in aqueous phase having K_{c} = 1, would be

a) – RT

b) -1

c) 0

d) + RT

Answer:

c) 0

Question 95.

The equilibrium constants for the reaction, Br_{2} ⇌ 2 Br at 500 K and 700 K are 1 × 10^{-10} and 1 × 10^{-5} respectively. The reaction is

a) endothermic

b) exothermic

c) fast

d) slow

Answer:

a) endothermic

Question 96.

In the reaction A_{2}(g) + 4 B_{2}(g) ⇌ 2AB_{4}(g), ∆H < 0, the decomposition of AB_{4}(g) will be favoured at

a) low temperature and high pressure

b) high temperature and low pressure

c) low temperature and low pressure

d) high temperature and high pressure

Answer:

c) low temperature and low pressure

Question 97.

The equilibrium constant for the reaction N_{2}(g) + O_{2}(g) ⇌ 2NO (g) is 4 × 10^{-4} at 200 K. In the presence of a catalyst the equilibrium is attained 10 times faster. Therefore the equilibrium constant in pressure of the catalyst at 200 K is

a) 4 × 10^{-3}

b) 4 × 10^{-4}

c) 4 × 10^{-5}

d) None

Answer:

b) 4 × 10^{-4}

Question 98.

Change in volume of the system does not alter the number of moles in which of the following equilibrium.

a) N_{2}(g) + O_{2}(g) ⇌ 2 NO (g)

b) PCl_{5}(g) ⇌ PCl_{3}(g) + Cl_{2}(g)

c) N_{2}(g) + 3H_{2}(g) ⇌ 2 NH_{3}(g)

d) SO_{2}Cl_{2}(g) ⇌ SO_{2}(g) + Cl_{2}(g)

Answer:

a) N_{2}(g) + O_{2}(g) ⇌ 2NO_{2} (g)

Question 99.

In which of the following equilibrium reactions the equilibrium would shift to the right, if total pressure is increased

a) N_{2} + 3H_{2} ⇌ 2 NH_{3}

b) H_{2} + I_{2} ⇌ 2HI

c) H_{2} + Cl_{2} ⇌ 2 HCl

d) N_{2}O_{4} ⇌ 2NO

Answer:

a) N_{2} + 3H_{2} ⇌ 2 NH_{3}

Question 100.

The graph in K (eq) vs \(\frac{1}{T}\) relates for a reaction must be

a) Exothermic

b) Endothermic

c) ∆H is negligible

d) High spontaneous at ordinary temperature

Answer:

a) Exothermic

Question 101.

Which of the following statements is cor¬rect for a reversible process in state of equilibrium?

a) ∆G° = -2.30 RT log K

b) ∆G° = 2.30 RT log K

c) ∆G = -2.30 RT log K

d) ∆G = 2.30 RT log K

Answer:

a) ∆G° = -2.30 RT log K

Question 102.

Which of the following is a characteristic of a reversible reaction?

a) Number of moles of reactants and products are equal

b) it can be influenced by a catalyst

c) it can never proceed to completion

d) none of the above

Answer:

c) it can never proceed to completion

Question 103.

The equilibrium constant in a reversible reaction at a given temperature

a) depends on the initial concentration of the reactants

b) depends on the concentration of the products at equilibrium

c) does not depend on the initial concentration

d) it is not characteristic of the reaction

Answer:

c) does not depend on the initial concentration

II. Very short question and answers(2 Marks):

Question 1.

What are reversible reactions?

Answer:

Reactions in which the products can react to give back reactants under suitable conditions are called reversible reactions.

Question 2.

What are irreversible reactions?

Answer:

Reactions in which the products formed do not react to give back reactants under normal conditions are called irreversible reactions.

Question 3.

Why the chemical equilibrium is referred to as Dynamic Equilibrium?

Answer:

At equilibrium the forward and the backward reactions are proceeding at the same rate and no macroscopic change is observed. So chemical equilibrium is often called as dynamic equilibrium.

Question 4.

What is equilibrium constant?

Answer:

The equilibrium constant K_{c} may be defined as the ratio of product of the active masses of the products to that of the reactants.

Question 5.

Write the Applications of equilibrium constant?

Answer:

The knowledge of equilibrium constant helps us in

- predicting the extent of a reaction, predicting the direction in which the net reaction is taking place and
- calculation of equilibrium constants and concentration of the reactants and products.

Question 6.

What is Le – Chatelier’s principle?

Answer:

It states that “If a system at equilibrium is subjected to a change of concentration, pressure or temperature, then the equilibrium is shifted in such a ways to nullify the effect of that disturbance.

Question 7.

For the H_{2}(g) + I_{2}(g) ⇌ 2HI(g) reaction the value of K_{c} at 717 K is 48. If at a particular instant, the concentration of H_{2}, I_{2,} and HI is 0.2 mol lit^{-1} and 0.6 mol lit^{-1} respectively, then calculate Q and predict the direction of reaction.

Solution:

Q = \(\frac{\left[\mathrm{H}_{1}\right]^{2}}{\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]}=\frac{0.6 \times 0.6}{0.2 \times 0.2}\) = 9

Q < K_{c}, the reaction will proceed in the forward direction.

Question 8.

Define Degree of dissociation.

Answer:

The fraction of the total number of moles of the reactants are dissociated.

Degree of dissociation ‘x’ =

‘x’ = \(\frac{\text { no.of moles dissociated }}{\text { Total no. of moles present initially }}\)

Question 9.

Dissociation of PCl_{5} decreases in presence of increase in Cl_{2}, Why?

Answer:

- The dissociation equilibrium of PCl
_{5}in gaseous state is written as PCl_{5}(g) ⇌ PCl_{3}(g) + Cl_{2}(g) - In a chemical equilibrium increasing the concentration of the products results in shifting the equilibrium in favour of the reactants. So in the presence of increased Cl
_{2}reversed reaction only favoured. - Hence dissociation of PCl
_{5}decreases in presence of increase in Cl_{2}.

Question 10.

The equilibrium constant for K_{c} for A(g) ⇌ B(g) is 2.5 × 10^{-2} . The rate constant of the forward reaction is 0.05 sec^{-1}. Calculate the rate constant of the reverse reaction.

Answer:

K_{c} = 2.5 × 10^{-2};

K_{f} = 0.05 sec^{-1}

K_{c} = \(\frac{\mathrm{K}_{\mathrm{f}}}{\mathrm{K}_{\mathrm{r}}}\)

= \(\frac{0.05}{2.5 \times 10^{-2}}\) = 2

Rate constant of the reverse reaction,

K_{r} = 2 sec^{-1}.

Question 11.

The equilibrium constant for the reaction 2SO_{3}(g) ⇌ 2SO_{2}(g) + O_{2}(g) is 0.15 mol dm^{-3} at 900K. Calculate the equilibrium constant for the reaction 2SO_{2}(g) + O_{2}(g) ⇌ 2SO_{3}(g) at the same temperature. 2SO_{3}(g) ⇌ 2SO_{2}(g) + O_{2}(g).

Answer:

K_{c} = \(\frac{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]}{\left[\mathrm{SO}_{3}\right]^{2}}\) = 0.15

K’ = \(\frac{1}{\mathrm{~K}_{\mathrm{c}}}=\frac{1}{0.15}\)

= 6.6 mole dm^{-3}

K’ = 6.6 mole dm^{-3}

III. Short question and answers(3 Marks):

Question 1.

What is solid – liquid equilibrium? Give example.

Answer:

- The equilibrium between solid and liquid at its melting point (or freezing point) at normal atmospheric pressure is called solid-liquid equilibrium.
- If some ice-cubes along with water are placed in a thermos flask (at 00 C and 1 atm), then there will be no change in the mass of ice and water.
- At equilibrium,

Rate of melting of ice = Rate of freezing of water

H_{2}O(s) ⇌ H_{2}O(l)

Question 2.

What is liquid – vapour equilibrium?

Answer:

The equilibrium of a liquid with its gas at its boiling point at 1 atm pressure is called liquid – vapour equilibrium.

Rate of evaporation = Rate of condensation

H_{2}O(l) ⇌ H_{2}O(g).

Question 3.

What is Solid – vapour equilibrium?

Answer:

- When solid iodine is placed in a closed transparent vessel, after sometime the vessel gets filled up with violet vapour.
- The intensity of the colour increases and finally becomes constant, i.e., equilibrium is attained.

I_{2}(s) ⇌ I_{2}(g)

Question 4.

What are Homogeneous and Heterogeneous equilibriums?

Answer:

Homogeneous equilibrium:

In a homogeneous equilibrium, all the reactants and products are in the same phase.

Example:

H_{2}(g) + I_{2}(g) ⇌ 2HI (g)

In the above equilibrium, H_{2}, I_{2,} and HI are in the gaseous state.

Similarly, for the following reactions,

CH_{3}COOCH_{3}(aq) + H_{2}O(aq) ⇌ CH_{3}COOH (aq) + CH_{3}OH (aq) all the reactants and products are in homogenous solution phase.

Heterogeneous equilibrium:

If the reactants and products are present in different phases, it is known as heterogeneous equilibrium.

Example:

H_{2}O(l) ⇌ H2O (g)

CaCO_{3}(s) ⇌ CaO(s) + CO_{2}(g).

Question 5.

What is Law of mass action?

Answer:

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

Rate ∝ [Reactants]^{x}

[ ] = \(\frac{\mathrm{n}}{\mathrm{v}}\), {Unit is mol.dm^{-3} or mol.L^{-1}}

where,

n is the number of moles and

V is the volume of the container (dm^{3} or L).

Question 6.

What is the reaction quotient? Write the relation of K & Q.

Answer:

Reaction quotient ‘Q’ is defined as the ratio of the product of active mass of reaction products raised to the respective stoichiometric coefficients in the balanced chemical equation to that of the reactants under non equilibrium.

Reaction quotient Q.

Q = \(\frac{[\mathrm{C}]^{1}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\)

If Q = K_{c}, the reaction is in equilibrium

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.

Question 7.

Explain effect of pressure of formation of HI.

Answer:

Pressure has no effect on the equilibrium when the total number of the moles of the gaseous reactants and the gaseous products are equal. ( ∆n_{g} = 0).

Therefore, pressure has effect only on the equilibrium with ∆n_{g} = 0.

Let us consider the following reactions

H_{2}(g) + I_{2}(g) ⇌ 2HI(g)

N_{2}(g) + O_{2}(g) ⇌ 2NO(g)

Here,

The number of moles on reactants and products are equal. So pressure has no effect on them.

Question 8.

The equilibrium concentrations of NH_{3}, N_{2} and H_{2} are 1.8 × 10^{-2} M, 1.2 × 10^{-2} M and 3 × 10^{-2} M respectively. Calculate the equilibrium constant for the formation of NH_{3} from N_{2} and H_{2}. [Hint: M= mol lit-1]

Solution:

Given data:

[NH_{3}] = 1.8 × 10^{-2 }M ;

[N_{2}] = 1.2 × 10^{-2 }M ;

[H_{2}] = 3 × 10^{-2 }M ;

K_{c} = ?

N_{2}(g) + 3H_{2}(g) ⇌ 2NH_{3}(g)

K_{c} = \(\frac{\left[\mathrm{NH}_{3}\right]^{2}}{\left[\mathrm{~N}_{2}\right]\left[\mathrm{H}_{2}\right]^{3}}\)

K_{c} = \(\frac{1.8 \times 10^{-2} \times 1.8 \times 10^{-2}}{1.2 \times 10^{-2} \times 3 \times 10^{-2} \times 3 \times 10^{-2} \times 3 \times 10^{-2}}\)

K_{c} = 1 × 10^{-2} l^{2} mol^{-2}

Question 9.

One mole of H_{2} and one mole of I_{2} are allowed to attain equilibrium. If the equilibrium mixture contains 0.4 mole of HI. Calculate the equilibrium constant.

Answer:

Given data:

[H_{2}] = 1 mole;

[I_{2}] = 1 mole

At equilibrium, [HI] = 0.4 mole; K_{c} = ?

H_{2}(g) + I_{2} ⇌ 2HI(g)

Question 10.

Relate the three equilibrium constants.

i) N_{2} + O_{2} ⇌ 2NO; K_{1}

ii) 2NO + O_{2} ⇌ 2NO_{2}; K_{2}

iii) N_{2} + 2O_{2} ⇌ 2NO_{2}; K_{3}

Answer:

Question 11.

Write the equilibrium constant for the following:

i)H_{2}O_{2}(g) ⇌ H_{2}O + 1/2O_{2}(g)

ii) CO(g) + H_{2}O(g) ⇌ CO_{2}(g) + H_{2}(g)

iii) N_{2}O_{4}(g) ⇌ 2NO_{2}(g)

Answer:

i)H_{2}O_{2}(g) ⇌ H_{2}O + 1/2O_{2}(g)

ii) CO(g) + H_{2}O(g) ⇌ CO_{2}(g) + H_{2}(g)

iii) N_{2}O_{4}(g) ⇌ 2NO_{2}(g)

Question 12.

Calculate the ∆n_{g}, for the following reactions:

i) H_{2}(g) + I_{2}(g) ⇌ 2HI

ii) 2H_{2}O(g) + 2Cl_{2}(g) ⇌ 4HCl(g) + O_{2}(g)

∆n_{g} = (Total number of moles of gaseous products) – (Total number of moles of gaseous reactants)

Answer:

i) H_{2}(g) + I_{2}(g) ⇌ 2HI

∆n_{g} = 2 – 2 = 0

∆n_{g} = 0

ii) 2H_{2}O(g) + 2Cl_{2}(g) ⇌ 4HCl(g) + O_{2}(g)

∆n_{g} = 5 – 4 = 1

∆n_{g} = 1

Question 13.

For the reaction A+B ⇌ 3C at 25°C, a 3 litre volume reaction vessel contains 1, 2, and 4 moles of A, B, and C respectively at equilibrium, calculate the equilibrium constant K_{c} of the reaction at 25°C.

Answer:

Question 14.

At temperature T_{1} the equilibrium constant of reaction is K_{1} At a higher temperature T_{2}, K_{2} is 10% of K_{1}. Predict whether the equilibrium is endothermic or exothermic.

Answer:

At temperature, T_{1}, the equilibrium constant is K_{1}

At a high temperature, T_{2}, the equilibrium constant K_{2} = 10%KX

K_{2} = \(\frac{10}{100}\) K_{1}

K_{2} = 0.1 K_{1}

K_{1} > K_{2}

Hence, the equilibrium is exothermic.

Question 15.

Calculate the K_{c} when a mixture containing 8.07 moles of H_{2} and 9.08 moles of I_{2} are reacted at 448°C until 13.38 moles of HI was formed at the equilibrIum.

Answer:

The equilibrium reaction is,

H_{2}(g) + I_{2}(g) ⇌ 2HI(g)

K_{c} = \(\frac{(2 x)^{2}}{(a-x)(b-x)}\)

a = 8.07 moles;

b = 9.08 moles

2x = 13.38 moles

x = 6.69 moles

K_{c} = \(\frac{4(6.69)^{2}}{(8.07-6.69)(9.08-6.69)}\)

= \(\frac{4 \times 44.75}{1.38 \times 2.39}\)

K_{c} = 54.29

Question 16.

For the equilibrium 2NOCl(g) ⇌ 2NO(g) + Cl_{2}(g) the value of the equilibrium constant K_{c} is 3.75 × 10^{-6} at 790°C. Calculate K_{p} for this equilibrium at the same temperature. [Hint: K_{p} = K_{c}(RT)^{∆ng}]

Answer:

K_{c} = 3.75 × 10^{-6};

T = 790 + 273 = 1063 K

K_{p} = K_{c} × (RT)^{∆ng}

For the reaction,

2NOCl(g) ⇌ 2NO(g) + Cl_{2}(g)

∆n_{g} = 3 – 2 = 1

K_{p} = 3.75 × 10^{-6} × (8.314 × 1063)

= 33141.6 × 10^{-6}

= 3.31 × 10^{4} × 10^{-6}

K_{p} = 3.31 × 10^{-2}

Question 17.

How much PCl_{5} must be added to one litre volume reaction vessel at 250°C in order to obtain a concentration of 0.1 mole of Cl_{2}, K_{c} for PCl_{5} ⇌ PCl_{3} + Cl_{2} is 0.0414 mol dm^{-3} at 250°C.

Answer:

PCl_{5} ⇌ PCl_{3} + Cl_{2}

Number of moles of PCl_{5} at equilibrium = a – 0.1

Number of moles PCl_{3} at equilibrium = 0.1

Number of moles Cl_{2} at equilibrium = 0.1

K_{c} = \(\frac{\left[\mathrm{PCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]}{\left[\mathrm{PCl}_{5}\right]}\)

= 0.0414

= \(\frac{0.1 \times 0.1}{a-0.1}\)

0.01 = (a – 0.1) 0.0414

0.01 = 0.0414a – 0.00414

0.0414a = 0.01414

a = 0.3415 mole

Question 18.

At 540 K, the equlllbrium constant K_{p} for PCl_{5} dissociation equilibrium at 1.0 atm is 1.77 atm. Calculate equilibrium constant in molar concentration (K_{c}) at same temperature and pressure.

Answer:

K_{p} = 1.77 atm;

T = 540 K;

a = 0.082

PCl_{5} ⇌ PCl_{3} + Cl_{2}

∆n_{g} = 2 – 1 = 1

K_{p} = K_{c}(RT)^{∆ng}

1.77 = K_{c} × (0.0821 × 450)^{1}

K_{c} = \(\frac{1.77}{(0.0821 \times 540)}\)

= 4 × 10^{-2} moles/litres.

III. Long question and answers (5 Marks):

Question 1.

Derive the Equilibrium constants Kp and K for the homogeneous reactions.

Answer:

Let us consider a reversible reaction,

xA + yB ⇌ lC + mD

where A and B are the reactants, C and D are the products, and x, y, l, and m are the coefficients of A, B, C, and D respectively.

Applying Law of mass action, the rate of the forward reaction,

r_{f} α [A]^{x} [B]^{y} (or)

r_{f} = kf [A]^{x} [B]^{y}

Similarly, the rate of the backward reaction,

r_{b} a [C]^{l} [D]^{m} (or)

r_{b} = k, [C]^{l} [D]^{m}

where k_{f} and k_{b} are proportionality constants

At equilibrium, Rate of forward reaction(r_{f}) = Rate of backward reaction(r_{b})

k_{f} α [A]^{x} [B]^{y} = k_{b} a [C]^{l} [D]^{m} (or)

\(\frac{\mathrm{k}_{\mathrm{f}}}{\mathrm{k}_{\mathrm{b}}}=\frac{[\mathrm{C}]^{l}[\mathrm{D}]^{m}}{[\mathrm{~A}]^{x}[\mathrm{~B}]^{y}}\) = K_{c}

where, K_{c} is the equilibrium constant in terms of concentration.

At a given temperature, the ratio of the product of active masses of products to that the reactants is a constant, known as equilibrium constant. Each concentration term is raised to the power which is equal to the respective stoichiometric coefficients in the balanced chemical equation. Later when we study chemical kinetics we will learn that this is only approximately true. If the reactants and products behave as gases, then the equilibrium constant in terms of partial pressures can be written as

\(\frac{\mathrm{P}_{\mathrm{C}}^{l} \times \mathrm{P}_{\mathrm{D}}^{m}}{\mathrm{P}_{\mathrm{A}}^{x} \times \mathrm{P}_{\mathrm{B}}^{y}}\)where, P_{c} is the partial pressure of the gas, C and so on.

Question 2.

Derive K_{p} & K_{c} relation for the formation of HI.

Answer:

Synthesis of HI:

Consider that ‘a’ moles and ‘b’ moles are the initial concentration of H_{2} and I_{2} respectively, taken in a container of volume V. Let ‘x’ moles of each of H_{2} and I_{2} react together to form 2x moles of HI.

H_{2}(g) + I_{2}(g) ⇌ 2HI(g)

The equilibrium constant K_{p} can also be calculated as follows:

Total number of moles at equilibrium = a – x + b – x + 2x = a + b.

If the total pressure of the system at equilibrium be P, then

Partial pressure of H_{2},

p_{H2} = \(\left(\frac{a-x}{a+b}\right)\)P

Partial pressure of I_{2},

p_{I2} = \(\left(\frac{b-x}{a+b}\right)\)P

Partial pressure of HI,

p_{HI} = \(\left(\frac{2x}{a+b}\right)\)P

K_{p} = \(\frac{\left(\frac{2 x}{a+b}\right)^{2} \mathrm{P}^{2}}{\left(\frac{a-x}{a+b}\right) \mathrm{P}\left(\frac{b-x}{a+b}\right) \mathrm{P}}=\frac{4 x^{2}}{(a-x)(b-x)}\)

In the above equilibrium, K_{c} and K_{p} are thus shown to be equal.

Question 3.

Derive the equilibrium constant K_{p} & K_{c} for the Dissociation of PCl_{5}.

Answer:

Dissociation of PCl_{5}:

Consider that ‘a’ moles is the initial concentration of PCl_{5} taken in a container of volume V. Let ‘x’ moles of PCl_{5} be dissociated into x moles each of PCl_{3} and Cl_{2}.

PCl_{5}(g) ⇌ PCl_{3}(g) + PCl_{2}(g)

The equilibrium constant K_{p} can also be calculated as follows:

Total number of moles at equilibrium = a – x + x + x = a + x

If the total pressure of the system at equilibrium be P, then

Partial pressure of PCl_{5},

p_{PCl5} = \(\left(\frac{a-x}{a+x}\right)\)P;

Partial pressure of PCl_{3},

p_{PCl3} = \(\left(\frac{x}{a+x}\right)\)P;

Partial pressure of Cl_{2}

p_{Cl2} = \(\left(\frac{x}{a+x}\right)\)P

K_{p} = \(\frac{\left(\frac{x}{a+x}\right) \mathrm{P}\left(\frac{x}{a+x}\right) \mathrm{P}}{\left(\frac{a-x}{a+x}\right) \mathrm{P}}=\frac{x^{2} \mathrm{P}}{(a-x)(a+x)}\)

Question 4.

Explain the effect of Concentration of formation of HI?

Answer:

Effect of concentration:

At equilibrium, the concentration of the reactants and the products does not change. The addition of more reactants or products to the reacting system at equilibrium causes an increase in their respective concentrations.

According to Le Chatelier’s principle, the effect of increase in concentration of a substance is to shift the equilibrium in a direction that consumes the added substance. Suppose at equilibrium if one of the reactants is added, then the equilibrium will shift in the forward direction and if one of the products is added, then the equilibrium will shift in backward direction.

Let us consider the reaction,

H_{2}(g) + I_{2}(g) ⇌ 2HI(g)

On addition of H_{2} or I_{2} to the equilibrium mixture product HI will tend to increase. In order to minimize the stress, the system shifts the reaction in a direction where H_{2} and I_{2} are consumed i.e., the formation of HI would relieve the stress. Hence, the equilibrium shifts to the right (forward direction).

Similarly, removal of HI (product) also boosts the forward reaction and increases the concentration of products.

If HI is added to the equilibrium mixture, the concentration HI is increased, then the system proceeds in the reverse direction to nullify the stress.

Question 5.

The equilibrium constant at 298 K for a reaction is 100. A + B ⇌ C + D. If the initial concentration of all the four species were 1 M each what will be equilibrium concentration of D (in mol lit^{-1})

Solution:

Given data:

[A] = [B] = [C] = 1 M;

K_{c} = 100; [D] = ?

10(1 – x) = 1 + x

10 – 10x – 1 – x = 0

9 – 11x = 0

11x = 9

x = 9/11 = 0.818

So, [D] at equilibrium = 1 + 0.818 = 1.818 M

Question 6.

What happens when ∆n_{g} = 0, ∆n_{g} = -ve, ∆n_{g}= +ve In a gaseous reaction?

Answer:

(i) When ∆n_{g} = 0, the total number of moles of gaseous products are equal to the total number of moles of gaseous reactants.

Eg:

H_{2}(g) + I_{2}(g) ⇌ 2HI(g)

∆n_{g} = 2 – 2 = 0

K_{p} = K_{c}(RT)^{∆ng}

If ∆n_{g} = 0,

K_{p} = K_{c}(RT)^{0}

∴ K_{p} = K_{c}

(ii) When ∆n_{g} = +ve, the total number of moles of gaseous products are greater than the total number of moles of gaseous reactants.

Eg:

2H_{2}O + 2Cl_{2}(g) ⇌ 4HCl(g) + O_{2}(g)

∆n_{g} = 5 – 4 = 1

K_{p} = K_{c}(RT)^{1}

K_{p} = K_{c}(RT)

∴ K_{p} > K_{c}

(iii) When ∆n_{g} = -ve, the total number of moles of gaseous products are greater than the total number of moles of gaseous reactants.

N_{2}(g) + 3H_{2}(g) ⇌ 2NH_{3}(g)

∆n_{g} = 2 – 4 = -2

K_{p} = K_{c}(RT)^{-2}

K_{p} = \(\frac{K_{c}}{(R T)^{2}}\)

∴ K_{p} < K_{c}

Question 7.

What is the relationship between the dissociation and formation equilibrium constant?

Answer:

In a chemical equilibrium reaction, the equilibrium constant for the dissociation equilibrium reaction which is also known as dissociation constant is found to be the reciprocal value of the equilibrium constant for the formation equilibrium reaction.

Formation of equilibrium reaction,

2SO_{2}(g) + O_{2} ⇌ 2SO_{3}

K_{c} = \(\frac{\left[\mathrm{SO}_{3}\right]^{2}}{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]}\) dm^{3} /mole.

Dissociation equilibrium reaction of SO_{3} the reactants became products and vice versa.

2SO_{3}(g) ⇌ 2SO_{2}(g) + O_{2}(g)

K_{c}‘ = \(\frac{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]}{\left[\mathrm{SO}_{3}\right]^{2}}\) mole/dm^{3}

= \(\frac{1}{\mathrm{~K}_{\mathrm{c}}}\)

K_{c}‘ is considered as the dissociation constant of SO_{3} gas. Usually, the equilibrium constant of the dissociation equilibrium is the reciprocal of the equilibrium constant of the formation equilibrium reaction.

Question 8.

In the equilibrium H_{2} + I_{2} ⇌ 2HI, the number of moles of H_{2}, I_{2,} and HI are 1, 2, 3 moles respectively. Total pressure of the reaction mixture is 60 atm. Calculate the partial pressures of H_{2}, I_{2,} and HI in the mixture.

Solution:

H_{2} + I_{2} ⇌ 2HI

Number of moles of H_{2} at equilibrium = 1 mole

Number of moles of I_{2} at equilibrium = 2 mole

Number of moles of HI at equilibrium = 3 mole

Total number of moles of equilibrium = 1 + 2 + 3 = 6 moles

Mole fraction of H_{2} = 1/6;

Mole fraction of I_{2} = 2/6;

Mole fraction of HI = 3/6

Partial pressure = Mole fraction × Total pressure

P_{H2} = \(\frac{1}{6}\) × 60 = 10 atm;

P_{I2} = \(\frac{2}{6}\) × 60 = 20 atm;

P_{HI} = \(\frac{3}{6}\) × 60 = 30 atm.

Question 9.

In 1 litre volume reaction vessel, the equilibrium constant K of the reaction PCl_{5} ⇌ PCl_{3} + Cl_{2} is 2 × 10^{-4} What will be the degree of dissociation assuming only a small extent of 1 mole of PCl_{5} has dissociated?

Answer:

since ‘x’ is very small compared to 1,

then K_{2} = x^{2}

2 × 10^{-4} = x^{2}

Degree of dissociation x = \(\sqrt{2 \times 10^{-4}}\)

= 1.414 × 10^{-2}

Question 10.

At 35°C, the value of K for the equilibrium reaction N_{2}O_{4} ⇌ 2NO_{2} is 0.3174, Calculate the degree of dissociation when P is 0.2382 atm.

Answer:

N_{2}O_{4} ⇌ 2NO_{2} |
N_{2}O_{4} |
NO_{2} |

Initial number of moles | 1 | 0 |

Number of moles dissociated | x | 2x |

Number of moles equilibrium | 1 – x | 2x |

Total number of moles at equilibrium = 1 – x + 2x = 1 + x

Question 11.

Degree of dissociation of PCl_{5} at 1 atm and 25°C is 0.2. Calculate the pressure at which PCl_{5} is half dissociated at 25°C.

Answer:

For PCl_{5} dissociation equilibrium,

K_{p} = \(\frac{x^{2} p}{1-x^{2}}\) [ P = total pressure = 1 atm]

x = 0.2

K_{p} = \(\frac{(0.2)^{2}(1.0)}{1-(0.2)^{2}}\)

= \(\frac{0.04}{1-0.04}\)

Question 12.

Initially, 0.1 moles each of H_{2} and I_{2} gases and 0.02 moles of HI gas are mixed in a reaction vessel of constant volume at 300K. Predict the direction towards which the reaction proceeds [K = 3.5 x 10^{-2}].

Answer:

The formation of equilibrium is

H_{2}(g) + I_{2}(g) ⇌ 2HI(g)

K_{c} = \(\frac{[\mathrm{HI}]_{\mathrm{e}}^{2}}{\left[\mathrm{H}_{2}\right]_{\mathrm{e}}\left[\mathrm{I}_{2}\right]_{\mathrm{e}}}\)

= 3.5 × 10^{-2} at 300 K.

Under non equilibrium conditions;

Q = \(\frac{[\mathrm{HI}]^{2}}{\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]}\)

= \(\frac{\left(2 \times 10^{-2}\right)^{2}}{0.1 \times 0.1}\)

= 4 × 10^{-2}

Thus Q > K_{c}.

Therefore, the reaction will proceed initially before attaining the equilibrium in a direction such that the Q value is reduced. That is, concentrations of H_{2} and I_{2} should be increases, which is a reverse reaction. The Traction proceeds in the left side of fomiation equilibrium of HI and HI decomposes initially to H_{2} and I_{2} until Q = K_{c}.