UNIT TEST 8

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Questions:

Generic 3

22 x 3 = 66

The partial pressure of carbon dioxide in the reaction
CaCO₃ (s) ⇌ CaO (s) + CO₂(g) is 1.017 × 10^–3 atm at 500⁰C. Calculate K_p at 600⁰C for the reaction. ΔH for the reaction is 181 KJ mol^–1 and does not change in the given range of temperature.

Deduce the Vant Hoff equation.

1 mol of CH₄, 1 mole of CS₂ and 2 mol of H₂S are 2 mol of H₂ are mixed in a 500 ml flask. The equilibrium constant for the reaction K_C = 4 x 10^–2 mol² lit^–2. In which direction will the reaction proceed to reach equilibrium ?

Oxidation of nitrogen monoxide was studied at 200^o C with initial pressures of 1 atm NO and 1 atm of O₂. At equilibrium partial pressure of oxygen is found to be 0.52 atm calculate K_P value.

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^o C. At equilibrium he found the concentration of HI which is equal to 0.05 M. Calculate K_C and K_{P for this reaction.}

For the reaction
SrCO₃ (s) ⇌ SrO (s) + CO₂(g),
the value of equilibrium constant K_P = 2.2 x 10^–4 at 1002 K. Calculate K_C for the reaction.

One mole of PCl₅ is heated in one litre closed container. If 0.6 mole of chlorine is found at equilibrium, calculate the value of equilibrium constant.

What is the effect of added inert gas on the reaction at equilibrium at constant volume.

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

10.

State Le-Chatelier principle.

11.

For the reaction,
A₂(g) + B₂(g) ⇌ 2AB(g) ; ΔH is –ve.
the following molecular scenes represent different 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.

12.

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

13.

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.

14.

What is the relation between K_P and K_C. Give one example for which K_P is equal to K_C.

15.

The equilibrium constant for the following reaction is 0.15 at 298 K and 1 atm pressure.
N₂O₄(g) ⇌ 2NO₂(g);
\(\Delta \mathrm{H}_{\mathrm{f}}^{\circ}=57.32 \mathrm{KJmol}^{-1}\)
The reaction conditions are altered as follows.
a) The reaction temperature is altered to 100^o C keeping the pressure at 1 atm, Calculate the equilibrium constant.

16.

1 mol of PCl₅, kept in a closed container of volume 1 dm³ and was allowed to attain equilibrium at 423 K. Calculate the equilibrium composition of reaction mixture. (The K_c value for PCl₅ dissociation at 423 K is 2)

17.

The following water gas shift reaction is an important industrial process for the production of hydrogen gas.
CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g)
At a given temperature K_p = 2.7. If 0.13 mol of CO, 0.56 mol of water, 0.78 mol of CO₂ and 0.28 mol of H₂ are introduced into a 2 L flask, and find out in which direction must the reaction proceed to reach equilibrium

18.

For an equilibrium reaction K_p = 0.0260 at 25° C ΔH= 32.4 kJmol^-1, calculate Kp at 37° C

19.

The equilibrium concentrations of NH₃, N₂ and H₂ are 1.8 x 10^-2 M, 1.2 x 10^-2 M and 3 x 10^-2M respectively. Calculate the equilibrium constant for the formation of NH₃ from N₂ and H₂. [Hint: M= mol lit^-1]

20.

One mole of H₂ and one mole of I₂ are allowed to attain equilibrium in 1 lit container. If the equilibrium mixture contains 0.4 mole of HI. Calculate the equilibrium constant.

21.

The value of Kc for the reaction
N₂O₂(g) \(\rightleftharpoons \) 2NO₂(g)

22.

The value of K_c for the following reaction at 717 K is 48.

Generic 5

9 x 5 = 45

23.

The equilibrium constant K_P for the reaction
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is 8.19 x 10² at 298 K and 4.6 x 10^–1 at 498 K. Calculate ΔH^o for the reaction

24.

A sealed container was filled with 1 mol of A₂ (g), 1 mol B₂ (g) at 800 K and total pressure 1.00 bar. Calculate the amounts of the components in the mixture at equilibrium given that K = 1 for the reaction
A₂ (g) + B₂ (g) ⇌ 2AB (g)

25.

The equilibrium for the dissociation of XY₂ is given as,
2XY₂ (g) ⇌ 2XY (g) + Y₂(g)
if the degree of dissociation x is so small compared to one. Show that 2 K_P = PX₃ where P is the total pressure and K_P is the dissociation equilibrium constant of XY₂.

26.

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

27.

At particular temperature K_C = 4 x 10^–2 for the reaction
H₂S(g) ⇌ H₂(g) + ½ S₂(g)
Calculate KC for each of the following reaction
i) 2H₂S (g) ⇌ 2H₂ (g) + S₂ (g)
ii) 3H₂S (g) ⇌ 3H₂ (g) + 3/2 S₂(g)

28.

Derive the relation between K_P and K_C.

29.

Write a balanced chemical equation for equilibrium reaction for which the equilibrium constant is given by expression
\(K_c={[NH_3]^4[O_2]^5\over [NO]^4[H_2O]^6}\)

30.

Derive a general expression for the equilibrium constant K_P and K_C for the reaction
3H₂(g) + N₂(g) ⇌ 2NH₃(g).

31.

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

Generic 2

6 x 2 = 12

32.

State law of mass action.

33.

Consider the following reactions,
H₂(g) + I₂(g) ⇌ 2 HI(g)
In each of the above reaction find out whether you have to increase (or) decrease the volume to increase the yield of the product.

34.

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

35.

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

36.

The atmospheric oxidation of NO
2NO(g) + O₂(g) ⇌ 2NO₂(g)
was studied with initial pressure of 1 atm of NO and 1 atm of O₂. At equilibrium, partial pressure of oxygen is 0.52 atm calculate K_p of the reaction.

37.

Consider the following reaction
Fe³⁺(aq) + SCN^–(aq) ⇌ [Fe(SCN)]²⁺(aq)
A solution is made with initial Fe³⁺, SCN^- concentration of 1 x 10^-3 M and 8 x 10^-4 M respectively. At equilibrium [Fe(SCN)]²⁺ concentration is 2 x 10^-4 M. Calculate the value of equilibrium constant.

Multiple Choice Question

25 x 1 = 25

38.

A 20 litre container at 400 K contains CO₂ (g) at pressure 0.4 atm and an excess of SrO (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 pressure of CO₂ attains its maximum value will be: Given that: SrCO₃ (S) ⇌ SrO (S) + CO₂(g)
KP = 1.6 atm

(a)

2 litre

(b)

5 litre

(c)

10 litre

(d)

4 litre

39.

The equilibrium constants of the following reactions are:

N₂ + 3H₂ ⇌ 2NH₃ :	K₁
N₂ + O₂ ⇌ 2NO :	K₂
H₂ + ½O₂ ⇌ H₂O :	K₃

The equilibrium constant (K) for the reaction ;
\({ 2NH }_{ 3 }+5/2{ O }_{ 2 }\overset { K }{ \rightleftharpoons } 2NO+{ 3H }_{ 2 }{ O },\) will be

(a)

\(K_2^3{K_3\over K_1}\)

(b)

\(K_1{K_3^3\over K_2}\)

(c)

\(K_2{K_3^3\over K_1}\)

(d)

\(K_2{K_3\over K_1}\)

40.

[Co(H₂O)₆]²⁺ (aq) (pink) + 4Cl^– (aq) ⇌ [CoCl₄]^2– (aq) (blue) + 6 H₂O (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 colour. On the basis of this information, which one of the following is true?

(a)

ΔH > 0 for the forward reaction

(b)

ΔH = 0 for the reverse reaction

(c)

ΔH < 0 for the forward reaction

(d)

Sign of the ΔH cannot be predicted based on this information

41.

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

42.

Match the equilibria with the corresponding conditions,

i) Liquid ⇌ Vapour	1) melting point
ii) Solid ⇌ Liquid	2) Saturated solution
iii) Solid ⇌ Vapour	3) Boiling point
iv) Solute (s) ⇌ Solute (Solution)	4) Sublimation point
	5) Unsaturated solution

(a)

(i)	(ii)	(iii)	(iv)
1	2	3	4

(b)

(i)	(ii)	(iii)	(iv)
3	1	4	2

(c)

(i)	(ii)	(iii)	(iv)
2	1	3	4

(d)

(i)	(ii)	(iii)	(iv)
3	2	4	5

43.

For the formation of Two moles of SO₃(g) from SO₂ and O₂, the equilibrium constant is K₁. The equilibrium constant for the dissociation of one mole of SO₃ into SO₂ and O₂ is __________

(a)

\(1/K_1\)

(b)

\(K_1^2\)

(c)

\(({1\over K_1})^{1/2}\)

(d)

\({K_1\over 2}\)

44.

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

45.

In a chemical equilibrium, the rate constant for the forward reaction is 2.5 \(\times\)10² and the equilibrium constant is 50. The rate constant for the reverse reaction is ____________

(a)

11.5

(b)

(c)

2 x 10²

(d)

2 x 10^-3

46.

Equimolar concentrations of H₂ and I₂ are heated to equilibrium in a 1 litre flask. What percentage of initial concentration of H₂has reacted at equilibrium if rate constant for both forward and reverse reactions are equal ____________

(a)

33%

(b)

66%

(c)

(33)²%

(d)

16.5%

47.

Consider the reaction where K_P = 0.5 at a particular temperature
PCl₅(g) ⇌ PCl₃ (g) + Cl₂ (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 PCl₃ will be produced

(b)

more Cl₂ will be produced

(c)

more PCl₅ will be produced

(d)

none of these

48.

In the reaction,
Fe (OH)₃ (s) ⇌ Fe³⁺(aq) + 3OH^–(aq),
if the concentration of OH^– ions is decreased by ¼ times, then the equilibrium concentration of Fe³⁺ will

(a)

not changed

(b)

also decreased by ¼ times

(c)

increase by 4 times

(d)

increase by 64 times

49.

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 and initial concentration of X and A be equal then total pressure at equilibrium P₁ and P₂ are in the ratio __________

(a)

36 : 1

(b)

1 : 1

(c)

3 : 1

(d)

1 : 9

50.

If x is the fraction of PCl₅ dissociated at equilibrium in the reaction
PCl₅ ⇌ PCl₃ + Cl₂
then starting with 0.5 mole of PCl₅, 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

51.

In which of the following equilibrium, K_P and K_C are not equal ?

(a)

2 NO(g) ⇌ N₂(g) + O₂(g)

(b)

SO₂ (g) + NO₂ ⇌ SO₃(g) + NO(g)

(c)

H₂(g) + I₂(g) ⇌ 2HI(g)

(d)

PCl₅ (g) ⇌ PCl₃(g) + Cl₂(g)

52.

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

53.

\({K_c\over K_p}\) for the reaction,
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is ___________

(a)

\({1\over RT}\)

(b)

\(\sqrt{RT}\)

(c)

(d)

(RT)²

54.

An equilibrium constant of 3.2\(\times\)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

55.

In the equilibrium,
2A(g) ⇌ 2B(g) + C₂(g)
the equilibrium concentrations of A, B and C₂ at 400 K are 1\(\times\)10^–4 M, 2.0 \(\times\)10^–3 M, 1.5 \(\times\)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 x 10^-2

56.

K₁ and K₂ are the equilibrium constants for the reactions respectively.
\({ N }_{ 2 }(g)+{ O }_{ 2 }(g)\overset { { K }_{ 1 } }{ \rightleftharpoons } 2NO(g)\)
\(2NO(g)+{ O }_{ 2 }(g)\overset { { K }_{ 2 } }{ \rightleftharpoons } { 2NO }_{ 2 }(g)\)
What is the equilibrium constant for the reaction NO₂(g) ⇌ ½N₂(g) + O₂(g)

(a)

\({1\over \sqrt{K_1K_2}}\)

(b)

(K₁ = K₂)^1/2

(c)

\({1\over 2K_1K_2}\)

(d)

\(({1\over K_1K_2})^{3/2}\)

57.

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)

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

(d)

Equilibrium constant varied with temperature

58.

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

59.

The formation of ammonia from N₂(g) and H₂(g) is a reversible reaction
N₂(g) + 3H₂(g) ⇌ 2NH₃(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

60.

The equilibrium constant for a reaction at room temperature is K₁ and that at 700 K is K₂. If K₁ > K₂, 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

61.

At a given temperature and pressure, the equilibrium constant values for the equilibria
\({ 3A }_{ 2 }+{ B }_{ 2 }+2C\overset { { K }_{ 1 } }{ \rightleftharpoons } { 2A }_{ 3 }BC\) and
\({ A }_{ 3 }BC\overset { { K }_{ 2 } }{ \rightleftharpoons } 3/2\left[ { A }_{ 2 } \right] +\frac { 1 }{ 2 } { B }_{ 2 }+C\)
The relation between K₁ and K₂ is ___________

(a)

\(K_1={1\over \sqrt{K_2}}\)

(b)

\(K_2=K_1^{-1/2}\)

(c)

\(K_1^2=2K_2\)

(d)

\({K_1\over 2}=K_2\)

62.

If K_b and K_f for a reversible reactions are 0.8 x 10^–5 and 1.6 x 10^–4 respectively, the value of the equilibrium constant is __________

(a)

(b)

0.2 x 10^-4

(c)

0.05

(d)

none of these

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Answers Key:

Generic 3

22 x 3 = 66

\(P_{CO_2}\) = 1.017 x 10^-3 atm T = 500^oC
K_p = \(P_{CO_2}\)
\(\therefore K_{P_1}\) = 1.017 x 10^-3 T = 500 + 273 = 773 K
\(K_{P_2}\) = ? T = 600 + 273 = 873 K
\(\Delta H^o=181\ KJ\ mol^{-1}\)
\(\log({K_{P_1}\over K_{P_1}})={\Delta H^o\over 2.303\ R}({T_2-T_1\over T_1T_2})\)
\(\log({K_{P_2}\over 1.017\times 10^{-3}})={181\times 10^3\over 2.303\times 8.314}({873-773\over 873\times 773})\)
\(\log({K_{P_2}\over 1.017\times 10^{-3}})={181\times 10^3\times 100\over 2.303\times 8.314\times 873\times 773}\)
\({K_{P_2}\over 1.017\times 10^{-3}}=\) anti log of (1.40)
\({K_{P_2}\over 1.017\times 10^{-3}}=25.12\)
\(\Rightarrow K_{P_2}=\) 25.12 x 1.017 x 10^-3
\(K_{P_2}=\) 25.54 x 10^-3.

This equation gives the quantitative temperature dependence of equilibrium constant (K). The relation between standard free energy change (\(\triangle\)G^O) and equilibrium constant is
\(\Delta { G }^{ 0 }=-RTln\ K\) ...(1)
We know that
\(\Delta { G }^{ 0 }=\Delta { H }^{ 0 }-T\Delta { S }^{ 0 }\)
Substituting (1) in equation (2)
\(-RTln\ K\ =\Delta { H }^{ 0 }-T\Delta { s }^{ 0 }\)
Rearranging
In \(K=\cfrac { -\Delta H^{ 0 } }{ RT } +\cfrac { { \Delta S }^{ 0 } }{ R } \) ...(3)
Differentiating equation (3) with respect to temperature
\(\cfrac { d\left( In\quad K \right) }{ dT } =\cfrac { \Delta { H }^{ 0 } }{ { RT }^{ 2 } } \) ...(4)
Equation 4 is known as differential form of Van't Hoff equation.
On integrating the equation 4, between T₁ and T₂with their respective equilibrium constants K₁ and K_2.
\(\int _{ { k }_{ 1 } }^{ { K }_{ 2 } }{ d\left( In\ K \right) =\cfrac { \Delta { H }^{ 0 } }{ R } \int _{ { T }_{ 2 } }^{ { { T }_{ 2 } } }{ \cfrac { dT }{ { T }^{ 2 } } } } \)
\(\left[ In\quad K \right] _{ { K }_{ 1 } }^{ { K }_{ 2 } }=\cfrac { \Delta { H }^{ 0 } }{ R } \left[ -\cfrac { 1 }{ T } \right] ^{ { T }_{ 2 } }_{ { T }_{ 1 } }\)
\(In\quad { K }_{ 2 }-In\quad { K }_{ 1 }=\cfrac { \Delta { H }^{ 0 } }{ R } -\left[ \cfrac { 1 }{ { T }_{ 2 } } +\cfrac { 1 }{ { T }_{ 2 } } \right] \)
\(In\quad \cfrac { { K }_{ 2 } }{ { K }_{ 1 } } =\cfrac { \Delta { H }^{ 0 } }{ R } \left[ \cfrac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 2 }{ T }_{ 1 } } \right] \)
\(log\quad \cfrac { { K }_{ 2 } }{ { K }_{ 1 } } =\cfrac { \Delta { H }^{ 0 } }{ 2.303R } \left[ \cfrac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 2 }{ T }_{ 1 } } \right] \) ...(5)
Equation (5) is known as integrated form of Van't Hoff equation.

CH₄(g) + 2H₂S(g) ⇌ CS₂(g) + 4H₂(g)
K_C = 4 x 10^–2 mol lit^–2
Volume = 500 ml = 1/2 L
\(\left[\mathrm{CH}_{4}\right]_{\text {in }}=\frac{2 \mathrm{~mol}}{1 / 2 \mathrm{~L}} \)
= 2 mol L^-1
\(\left[\mathrm{CS}_{2}\right]_{\text {in }}=\frac{1 \mathrm{~mol}}{1 / 2 \mathrm{~L}}\)
= 2 mol L^-1
\( {\left[\mathrm{H}_{2} \mathrm{~S}\right]_{\text {in }}=\frac{2 \mathrm{~mol}}{1 / 2 \mathrm{~L}}=4 \mathrm{~mol} \mathrm{~L}^{-1} \quad\left[\mathrm{H}_{2}\right]=\frac{2 \mathrm{~mol}}{1 / 2 \mathrm{~L}}=4 \mathrm{~mol} \mathrm{~L}^{-1}} \)
\(\mathrm{Q}=\frac{\left[\mathrm{CS}_{2}\right]\left[\mathrm{H}_{2}\right]^{4}}{\left[\mathrm{CH}_{4}\right]\left[\mathrm{H}_{2} \mathrm{~S}\right]^{2}}=\frac{2 \times(4)^{4}}{(2) \times(4)^{2}}=16 \)
Q > K_c
\(\therefore \) The reaction will proceed in the reverse direction to reach the equilibrium.

2NO (g) + O₂(g) ⇌ 2NO₂(g)

	NO	O₂	NO₂
Initial pressure	1	1	-
Reacted	0.96	0.48	-
Equilibrium partial pressure	0.04	0.52	0.96

\(K_p={(p_{NO_2})^2\over (P_{NO})^2(P_{o_2})}={0.96\times 0.96\over 0.04\times 0.04\times 0.52}\)
K_p = 1.017 x 10³.

V = 3L
\([HI]_{initial}={0.3\ mol\over 3L}=0.1\ M\)
[HI]_eq = 0.05 M
2HI(g) ⇌ H₂(g) + I₂(g)

	HI(g)	H₂(g)	I₂(g)
Initial Concentration	0.1	-	-
Reacted	0.05	-	-
Equilibrium concentration	0.05	0.025	0.025

\(K_c={[H_2][I_2]\over [HI]^2}\)
\(={0.025\times 0.025\over 0.05\times 0.05}\)
K_c = 0.25
K_P = K_C (RT)^Δng
Δn_g = 2 – 2 = 0
K_P = 0.25 (RT)^o
K_P = 0.25

for the reaction,
SrCO₃ (S) ⇌ SrO(S) + CO₂(S)
Δn_g= 1 – 0 = 1
\(\therefore\) K_P = K_C (RT)
2.2 x 10^–24 = K_C (0.0821) (1002)
\(K_c={2.2\times 10^{-4}\over 0.0821\times 1002}\)
KC = 2.674 x 10^-6

\( \therefore\left[\mathrm{PCl}_{5}\right]_{\mathrm{eq}}=0.4 \mathrm{~mol} \mathrm{dm}^{-3} ; \quad\left[\mathrm{PCl}_{3}\right]_{\mathrm{eq}}=0.6 \mathrm{~mol} \mathrm{dm}^{-3} ; \quad\left[\mathrm{Cl}_{2}\right]_{\mathrm{eq}}=0.6 \mathrm{~mol}^{-3} \)
\(\mathrm{K}_{c}=\frac{\left[\mathrm{PCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]}{\left[\mathrm{PCl}_{5}\right]}=\frac{0.6 \times 0.6}{0.4}\)
K_c = 0.9 mol dm^-3

Addition of an inert gas to a reaction at equilibrium, at constant volume has no effect.

If we know the value of k_cand Q, the reaction quotient, we can predict the direction of a reaction
If Q = Kc; the reaction is in equilibrium state.
If Q > Kc: the reaction will proceed in the reverse direction i.e., formation of reactants.
If Q < Kc: the reaction will proceed in the forward direction i.e., formation of products.

10.

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

11.

\(K_c={[AB]^2\over [A_2][B_2]}\)
A - green
B - blue

Given that 'V' is constant (closed system)
At equilibrium
\(K_c={({4\over V})^2\over ({2\over V})({2\over V})}={16\over 4}=4\)
K_P = K_C (RT)^Δn
K_P = 4(RT)^o = 4
At Stage 'x'
\(Q={({6\over V})^2\over ({2\over V})({1\over V})}={36\over 2}=18\)
Q > K_C ie., reverse reaction is favoured
At Stage 'y'
\(Q={({3\over V})^2\over ({3\over V})({3\over V})}={9\over 3\times 3}=1\)
K_C > Q ie., forward reaction is favoured.

12.

If Q > K_c; the reaction will proceed in the reverse direction (ie) formation of reactants to proceed to reach equilibrium.

13.

\(\Delta n_{g}=\sum n p_{(g)}-\sum n R_{(g)}\)
As \(\Delta n_{p}(g)\) is greater \(\Delta n_{g}=+v e\)
\( \therefore K_{p}=K_{c}(R T)^{+v e} \)
\(\therefore K_{p}>K_{c} \)
So K is smaller than K_p.

14.

i) \(K_{p}=K_{c}(R T)^{\Delta n_{g}}\)
K_p = Equilibrium constant in term of partial Pressures.
K_c = Equilibrium constant in term of concentration.
R = Gas constant; T = Temperature
\(\Delta \mathrm{n}_{\mathrm{g}}\) = Difference between the sum of number of moles of products and the sum of number of moles of reactants in gas phases.
ii) Synthesis of HI:
\( \mathrm{H}_{2(\mathrm{~g})}+\mathrm{I}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{HI}_{(\mathrm{g})} \)
\(\Delta n_{g}=0 \therefore K_{p}=K_{c}(R T) \Delta n_{g}\)
\(K_{p}=K_{c}(R T)^{\circ} \)
\(K_{p}=K_{c} \text {. }\)

15.

N₂O₄(g) ⇌ 2NO₂(g)
T₁ = 298 K K_P1 = 0.15
T₂ = 100o C = 100 + 273 = 373 K ;
K_P2 = ?
\(\log({K_2\over K_1})={\Delta H^o\over 2.303R}[{T_2-T_1\over T_1T_2}]\)
K₂ > K₁ and T₂ > T₁
\(\log({K_{p_2}\over 0.15})={57.2KJ\ mol^{-1}\over 2.303\times 8.314JK^{-1}\ mol^{-1}}[{373-298\over 373\times 298}]\)
\(\log({K_{P_2}\over 0.15})={57.2\times 10^{+3}\times 75\over 2.303\times 8.314\times 373\times 298}\)
\(\log({K_{p_2}\over 0.15})=2.02\)
\({K_{p_2}\over 0.15}=104.7\)
\(K_{p_2}=104.7\times 0.15\)
\(K_{p_2}=15.705\)

16.

PCl₅ ⇌ PCl₃ + Cl₂
Given that [PCl₅]_initial = 1 mol; V = 1 dm³; K_C = 2

	PCl₅	PCl₃	Cl₂
Initial no.of moles	1	-	-
No.of moles	x	-	-
No.of moles at equilibrium	1 - x	x	x
Equilibrium concentration	\({1-x\over 1}\)	\({x\over 1}\)	\({x\over 1}\)

\(K_c={[PCl_3][Cl_2]\over [PCl_5]}\)
\(2={x\times x\over (1-x)}\)
2 - 2x = x²|
x² + 2x - 2 = 0
Solution for a quadratic equation
\(a x^{2}+b x+c=0 \text { are }, x=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}\)
a- = 1 b = 2 c = -2
\(x=\frac{-2-\sqrt{4-4 \times 1 \times-2}}{2 \times 1} \)
\(x=\frac{-2-\sqrt{12}}{2}=\frac{-2-\sqrt{4 \times 3}}{2}\)
\( x =\frac{-2-2 \sqrt{3}}{2} \)
\(=\frac{-2+2 \sqrt{3}}{2}, \frac{-2-2 \sqrt{3}}{2} \)
\(x =-1+\sqrt{3} ;-1-\sqrt{3} \)
\(=-1-\sqrt{3} \text { not possible } \)
Since x is + ve,
x = -1 + 1.732
x = 0.732
Equilibrium concentration of
\( {\left[\mathrm{PCl}_{5}\right]_{\mathrm{eq}}=\frac{1-x}{1}=1-0.732=0.268 \mathrm{M}} \)
\({\left[\mathrm{PCl}_{3}\right]_{\mathrm{eq}}=\frac{x}{1}=\frac{0.732}{1}=0.732} \)
\({\left[\mathrm{Cl}_{2}\right]_{\mathrm{eq}}=\frac{x}{1}=\frac{0.732}{1}=0.732} \)

17.

CO(g) + H₂O(g) ⇌ CO₂ (g) + H₂(g) Given K_P = 2.7
[CO] = 0.13, [H₂O] = 0.56
[CO₂] = 0.78 ; [H₂] = 0.28
V = 2L
K_P = K_C (RT)
2.7 = K_C (RT)^o
K_C = 2.7
\(Q_c={[CO_2][H_2]\over [CO][H_2O]}\)\(={({0.78\over 2})({0.28\over 2})\over ({0.13\over 2})({0.56\over 2})}\)
Q = 3
Q > K_c, Hence the reaction proceed in the reverse direction.

18.

T₁= 25 + 273 = 298 K
T₂ = 37 + 273 = 310 K
ΔH = 32.4 KJmol^-1 = 32400 Jmol^-1
R = 8.314 JK^-1 mol^-1
K_P1= 0.0260
K_p2 = ?
\(\log {K_2\over K_1}={\Delta H^o\over 2.303R}[{T_2-T_1\over T_2T_1}]\)
\(\log {K_2\over K_1}={32400\over 2.303\times 8.314}({310-298\over 310\times 298})\)
\(={32400\times 10\over 2.303\times 8.314\times 310\times 298}\)
= 0.2198
\({K_2\over K_1}=\) antilog 0.2198 = 1.6588
K₂ = 1.6588 x 0.026 = 0.0431

19.

Given data:
[NH₃] = 1.8 x 10^-2 M
[N₂] = 1.2 x 10^-2 M
[H₂] = 3 x 10^-2 M
K_c = ?
Solution:
N_2(g) + 3H_2(g) \(\rightleftharpoons \) 2NH_3(g)
\(K_c={[NH_3]^2\over[N_2][H_2]^3}={1.8\times 10^{-2}\times 1.8\times 10^{-2}\over 1.2\times 10^{-2}\times 3\times 10^{-2}\times 3\times 10^{-2}\times 3\times 10^{-2}}\)
= 1 x 10³l²mol^-2

20.

H_2(g) + I₂_(g) \(\rightleftharpoons \) 2HI_(g)
At equilibrium, [HI] = 0.4 mol L^-1 Kc= ?

	H₂	I₂	HI
Initial number of moles	1	1	-
Number of moles at equilibrium	1 - x	1 - x	2x = 0.4 x = 0.2
	0.8	0.8	0.4

\(\therefore K_c={[HI]^2\over [H_2[I_2]]}={0.4\times 0.4\over 0.8\times 0.8}=0.25\)

21.

N₂O₂(g) \(\rightleftharpoons \) 2NO₂(g)
Kc = 0.21 at 373 K. The concentrations N₂O₄ and NO₂ are found to be 0.125 mol dm^-3 and 0.5 mol dm^-3 respectively at a given time. From the above information we can predict the direction of reaction as follows.
\(Q={[NO_2]^2\over [N_2O_4]}={0.5\times 0.5\over 0.125}=2\)
The Q value is greater than K_c. Hence, the reaction will proceed in the reverse direction until the Q value reaches 0.21.

22.

H₂(g) + I₂(g) \(\rightleftharpoons \) 2HI(g)
At a particular instant, the concentration of H₂, I₂and HI are found to be 0.2 mol L^-1, 0.2 mol L^-1 and 0.6 mol L^-1respectively. From the above information we can predict the direction of reaction as follows.
\(Q={[HI]^2\over[H_2][I_2]}={0.6\times 0.6\over 0.2\times 0.2}=9\)
Since Q < K_c, the reaction will proceed in the forward direction.

Generic 5

9 x 5 = 45

23.

K_p1 = 8.19 x 10² T₁ = 298K
K_p2 = 4.6 x 10^-1 T₂ = 498K
\(\log({K_{p_2}\over K_{P_1}})={\Delta H^o\over 2.303\times 8.314}({T_2-T_1\over T_1T_2})\)
\(\log({4.6\times 10^{-1}\over 8.19\times 10^2})={\Delta H^o\over 2.303\times 8.314}({498-298\over 498\times 298})\)
\({-3.2505\times 2.303\times 8.314\times 498\times 298\over 200}=\Delta H^o\)
\(\Delta H^o=\) -46181 J mol^-1
\(\Delta H^o=\) -46.18 J mol^-1.

24.

A₂(g) + B₂(g) ⇌ 2AB(g)

	A₂	B₂	AB
Initial Concentration	1	1	-
No.of moles reacted	x	x	-
No.of moles at equilibrium	1 - x	1 - x	2x

Total no. of moles = 1 – x + 1 – x + 2x = 2
\(K_p={(P_{AB})^2\over (P_{A_2})(P_{B_2})}={({2x\over 2}\times p)^2\over ({(1-x)\over 2}\times p)({1-x\over 2}\times p)}\)
\(K_p={4x^2\over (1-x)^2}\)
Given that K_p = 1; \({4x^2\over (1-x)^2}=1\)
\(\Rightarrow \) 4x² = (1 - x)²
\(\Rightarrow \) 4x² = 1 + x² - 2x
3x² + 2x - 1 = 0
\(X={-2\pm\sqrt{4-4\times 3\times -1}\over 2(3)}\)
\(X={-2\pm\sqrt{4+12}\over 6}\)
\(={-2\pm\sqrt{16}\over 6}\)
\(={-2+4\over 6};{-2-4\over 6}\)
\(={2\over 6};{-6\over 6}\)
X = 0.33 ; -1 (not possible)
\(\therefore\) [A₂]_eq = 1 - x = 1 - 0.33 = 0.67
[B₂]_eq = 1 - x = 1 - 0.33 = 0.67
[AB]_eq = 2X = 2 x 0.33 = 0.66.

25.

2XY₂(g) ⇌ 2XY(g) + Y₂(g)

	XY₂	XY	Y₂
Initial no.of moles	1	-	-
No.of moles dissociated	x	-	-
No.of moles at equilibrium	(1-x) \(\cong \) 1	x	x/2

Total No. of moles = 1 – x + x + x/2 = 1 + x/2 \(\cong \) 1
[∵ Given that x < < 1 ; 1 – x \(\cong \) 1 and 1 + x/2 \(\cong \) 1]
\(K_p={(P_{XY})^2(P_{Y_2})\over (P_{XY_2})^2}={({x\over 1}\times p)^2({x/2\over 1}\times P)\over ({1\over1}\times P)^2}\)

\(\Rightarrow\) 2K_p = x³P.

26.

Given \(m_{N_2}\) = 28 g \(m_{H_2}\) = 6g
V = 1 L
\((n_{N_2})_{initial}={28\over 28}=1\ mol\)
\((n_{H_2})_{initial}={6\over 2}=3\ mol\)
N₂(g) + 3H₂(g) ⇌ 2 NH₃(g)

	N₂(g)	H₂(g)	NH₃(g)
Initial concentration	1	3	-
Reacted	0.5	1.5	-
Equilibrium concentration	0.5	1.5	1

\([NH_3]=({17\over 17})=1\ mol=1\ mol\)
Weight of N₂ = (no. of moles of N₂) × molar mass of N₂
= 0.5 x 28 = 14 g
Weight of H₂ = (no. of moles of H₂) × molar mass of H₂
= 1.5 x 2 = 3 g

27.

K_C = 4 x 10^–2 for the reaction,
H₂S(g) ⇌ H₂(g) + 1/2 S₂(g)
\(K_c={[H_2][S_2]^{1/2}\over [H_2S]}\)
\(\Rightarrow 4\times 10^{-2}={[H_2][S_2]^{1/2}\over [H_2S]}\)
For the reaction,
2H₂S(g) ⇌ 2H₂(g) + S₂(g)
\(K_c={[H_2]^2[S_2]\over [H_2S]}=(4\times 10^{-2})^2=16\times 10^{-4}\)
For the reaction,
3H₂S(g) ⇌ 3H₂(g) + 3/2 S₂(g)
\(K_c={[H_2]^3[S_2]^{3/2}\over [H_2S]^3}=(4\times 10^{-2})^3=64\times 10^{-6}\)

28.

Let us consider the general reaction in which all reactants and products are ideal gases
\(xA+yB\rightleftharpoons IC+mD\)
The equilibrium constant, K_c is
\({ K }_{ c }=\cfrac { \left[ C \right] ^{ I }\left[ D \right] ^{ m } }{ \left[ A \right] ^{ x }\left[ B \right] ^{ y } } \) .......(1)
and K_pis
\({ K }_{ p }=\cfrac { { p }_{ c }^{ I }\times { p }_{ D }^{ m } }{ { p }_{ a }^{ x }\times { p }_{ B }^{ y } } \) ..........(2)
The ideal gas equation is
PY = nRT
or
\(P=\cfrac { n }{ V } RT\)
Since Active mass = molar concentration = n/V
p = active mass\(\times\)RT
Based on the above expression the partial pressure of the reactants and products can be expressed as,
\({ p }_{ A }^{ x }=\left[ A \right] ^{ x }\left[ RT \right] ^{ x }\)
\({ p }_{ B }^{ y }=\left[ B \right] ^{ y }\left[ RT \right] ^{ y }\)
\({ p }_{ C }^{ 1 }=\left[ C \right] ^{ I }\left[ RT \right] ^{ 1 }\)
\({ p }_{ D }^{ m }=\left[ D \right] ^{ m }\left[ RT \right] ^{ m }\)
On substitution in eqn. 2,
\({ K }_{ p }=\cfrac { { \left[ C \right] }^{ 1 }{ \left[ RT \right] }^{ 1 }{ \left[ D \right] }^{ m }{ \left[ RT \right] }^{ m } }{ { \left[ A \right] }^{ x }{ \left[ RT \right] }^{ x }{ \left[ B \right] }^{ y }{ \left[ RT \right] }^{ y } } \) .........(3)
\({ K }_{ p }=\cfrac { { \left[ C \right] }^{ 1 }{ \left[ D \right] }^{ m }{ \left[ RT \right] }^{ I+m } }{ { \left[ A\quad \right] }^{ x }{ \left[ B \right] }^{ y }{ \left[ RT \right] }^{ x+y } } \)
\({ K }_{ p }=\cfrac { { \left[ C \right] }^{ I }{ \left[ D \right] }^{ m } }{ { \left[ A \right] }^{ x }{ \left[ B \right] }^{ y } } \left[ RT \right] ^{ \left( 1+m \right) -\left( x+y \right) }\) ...........(4)
Sub (1) in (4)
\({ K }_{ p }={ K }_{ C }\left( RT \right) ^{ \left( \Delta { n }_{ g } \right) }\)
where,
\({ \Delta n }_{ g }\) 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.

29.

\(K_c={[NH_3]^4[O_2]^5\over [NO]^4[H_2O]^6}\)
\(4 \mathrm{NO}_{(\mathrm{g})}+6 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})} \rightleftharpoons 4 \mathrm{NH}_{3(\mathrm{~g})}+5 \mathrm{O}_{2(\mathrm{~g})}\)

30.

Let us consider the formation of ammonia in which, 'a' moles nitrogen and 'b' moles hydrogen gas are allowed to react in a container of volume V. Let 'x' moles of nitrogen react with 3x moles of hydrogen to give 2x moles of ammonia.
\({ N }_{ 2 }\left( g \right) +3{ H }_{ 2 }\left( g \right) \rightleftharpoons { 2NH }_{ 3 }\left( g \right) \)

	N₂	H₂	NH₃
Initial number of moles	a	b	0
number of moles reacted	x	3x	0
Number of moles at equilibrium	a - x	b - 3x	2x
Active mass or molaroncentration at equilibrium	\(\cfrac { a-x }{ V } \)	\(\cfrac { b-3x }{ V } \)	\(\cfrac { 2x }{ V } \)

Applying law of mass action
\({ K }_{ C }=\cfrac { \left[ { NH }_{ 3 } \right] ^{ 2 } }{ \left[ { N }_{ 2 } \right] \left[ { H }_{ 2 } \right] ^{ 3 } } \)
= \(\cfrac { \left( \cfrac { 2x }{ V } \right) ^{ 2 } }{ \left( \cfrac { a-x }{ V } \right) \left( \cfrac { b-3x }{ V } \right) ^{ 3 } } \)
= \(\cfrac { \left( \cfrac { 4x }{ V } \right) ^{ 2 } }{ \left( \cfrac { a-x }{ V } \right) \left( \cfrac { b-3x }{ V } \right) ^{ 3 } } \)
\({ K }_{ C }=\cfrac { 4{ x }^{ 2 }{ V }^{ 2 } }{ \left( a-x \right) \left( b-3x \right) ^{ 2 } } \)
The equilibrium constant K_p can also be calculated as follows:
\({ K }_{ p }={ K }_{ C }\left( RT \right) ^{ \left( \Delta { n }_{ g } \right) }\)
\(\Delta \)n_g =n_p - n_r = 2 - 4 = -2
\({ K }_{ p }=\cfrac { { 4x }^{ 2 }{ V }^{ 2 } }{ \left( a-x \right) \left( b-3x \right) ^{ 3 } } \left( RT \right) ^{ -2 }\)
Total number of moles at equilibrium,
n = a - x + b - 3x + 2x = a + b - 2x
\({ K }_{ p }=\cfrac { { 4x }^{ 2 }{ V }^{ 2 } }{ \left( a-x \right) \left( b-3x \right) ^{ 3 } } \times \left[ \cfrac { PV }{ n } \right] ^{ -2 }\)
\({ K }_{ p }=\cfrac { { 4x }^{ 2 }{ V }^{ 2 } }{ \left( a-x \right) \left( b-3x \right) ^{ 3 } } \times \left[ \cfrac { n }{ PV } \right] ^{ 2 }\)
\({ K }_{ p }=\cfrac { { 4x }^{ 2 }{ V }^{ 2 } }{ \left( a-x \right) \left( b-3x \right) ^{ 3 } } \times \left[ \cfrac { a+b-2x }{ PV } \right] ^{ 2 }\)
\({ K }_{ p }=\cfrac { 4{ x }^{ 2 }\left( a+b\quad -2x \right) ^{ 2 } }{ { P }^{ 2 }\left( a-x \right) \left( b-3x \right) ^{ 3 } } \)

31.

Given data:
[A] = [B] = [C]= [D] =1 M
K_c = 100
[D]_eq = ?
Solution:
Let x be the no moles of reactants reacted

	A	B	C	D
Initial concentration	1	1	1	1
At equilibrium (as per reaction stoichiometry)	1-x	1-x	1-x	1-x

\(K_c={[C][D]\over [A][B]}\)
\(100={(1+x)(1+x)\over (1-x)(1-x)}\)
\(\sqrt{100}=\sqrt{{(1+x)(1+x)\over (1-x)(1-x)}}\)
\(10={1+x\over 1-x}\)
10(1 - x) = 1 + x
10 - 10x - 1 - x = 0
9 - 11x = 0
11x = 9
\(x={9\over 11}=0.818\)
[D]eq = 1+x = 1 + 0.818 = 1.818M.

Generic 2

6 x 2 = 12

32.

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 of the reaction \(\alpha \) [Reactant]^x

33.

\( \mathrm{H}_{2(\mathrm{~g})}+\mathrm{I}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{HI}_{(\mathrm{g})} \)
\(\mathrm{K}_{\mathrm{c}}=\frac{4 x^{2}}{(a-x)(b-x)}\)
This expression doesn't involve, V. So, increase or decrease of volume will not affect the equilibrium and hence the yield of the product.

34.

The equilibrium constant is a constant and it is for equilibrium condition. But 'Q', the reaction quotient is not a constant as it is for non - equilibrium condition. 'Q' is the ratio of the product of active masses of a reaction products raised to the respective stoichiometric coefficients in the balanced chemical equation to that of the reactants, under non - equilibrium conditions.
i) If Q = K_c; it is equilibrium
ii) If Q > K_c.; the reaction will proceed in reverse direction
iii) If Q < K_c ; the reaction will proceed in forward direction

35.

This condition is not static and is dynamic, because both the forward and reverse reactions are still occurring with the same rate. No macroscopic change is observed.

36.

2 NO(g) + O₂ (g) ⇌ 2NO₂(g)

	NO₂	O₂	NO₂
Initila Partial Pressure	1	1	-
Reacted	0.96	0.96	-
Equilibrium Partial Pressure	0.04	0.52	0.96

\(K_p={P^2_{NO_2}\over P^2_{NO_2}.Po_2}\)
\(={0.96\times 0.96\over 0.04\times 0.04\times 0.52}\)
= 11.07 x 10² (atm)^-1
K_eq = 41.6 x 10² M^-1.

37.

	Fe³⁺	SCN^-	[Fe(SCN)²⁺
Initial concentration (M)	1x 10^-3 (10 x 10^-4)	8 x 10^-4	-
Reacted	2 x 10^-4	2 x 10^-4	-
Equilibrium concentration	8 x 10^-4	6 x 10^-4	2 x 10^-4

\(K_{eq}={[Fe(SCN)]^{2+}\over [Fe^{3+}][SCN^-]}\)
\(={2\times 10^{-4}M\over 8\times 10^{-4}M\times 6\times 10^{-4}M}\)
= 0.0416 x 10⁴
K_eq = 41.6 x 10² M^-1

Multiple Choice Question

25 x 1 = 25

38.