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Chemical Equilibrium

NEET Chemistry · 93 questions · Page 2 of 10 · Click an option or "Show Solution" to reveal answer

Q11
A 20 litre container at 400 K contains CO 2(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 CO 2 attains its maximum value, will be (Given that : SrCO 3(s) ⇌\rightleftharpoons SrO (s) + CO 2(g) , K p = 1.6 atm)
A 10 litre
B 4 litre
C 2 litre
D 5 litre
Correct Answer
Option D
Solution

Initially, P 1 = 0.4 atm V 1 = 20 L T 1 = 400 K Max. pressure of CO 2 = Pressure of CO 2 at equilibrium For reaction, SrCO 3 (s) ⇌ SrO(s) + CO 2 K p = P CO 2 = 1.6 atm = maximum pressure of CO 2 volume of container at this stage.

V 2 =

nRTP2{{nRT} \over {{P_2}}}

......(i) Since, reaction was not at equilibrium earlier and also container is sealed. ∴\therefore n = constant n =

P1V1RT{{{P_1}{V_1}} \over {RT}}

.....(ii) From (i) and (ii) V 2 =

P1V1RT×RTP2{{{P_1}{V_1}} \over {RT}} \times {{RT} \over {{P_2}}}

=

0.4×201.6{{0.4 \times 20} \over {1.6}}

= 5 L

Q12
Consider the following liquid-vapour equilibrium. Liquid ⇌\rightleftharpoons Vapour Which of the following relations is correct ?
A dln⁡PdT2=−ΔHvT2{{d\ln P} \over {d{T^2}}} = {{ - \Delta {H_v}} \over {{T^2}}}
B dln⁡PdT=ΔHvRT2{{d\ln P} \over {dT}} = {{\Delta {H_v}} \over {R{T^2}}}
C dln⁡GdT2=ΔHvRT2{{d\ln G} \over {d{T^2}}} = {{\Delta {H_v}} \over {R{T^2}}}
D dln⁡PdT=−ΔHvRT{{d\ln P} \over {dT}} = {{ - \Delta {H_v}} \over {RT}}
Correct Answer
Option B
Solution

This is Clausius-Clapeyron equation.

dln⁡PdT=ΔHvRT2{{d\ln P} \over {dT}} = {{\Delta {H_v}} \over {R{T^2}}}
Q13
If the equilibrium constant for N 2(g) + O 2(g) ⇌\rightleftharpoons 2NO (g) is K, the equilibrium constant for 12{1 \over 2} N 2(g) + 12{1 \over 2}O 2(g) ⇌\rightleftharpoons NO (g) will be
A 12{1 \over 2} K
B K
C K 2
D K 1/2
Correct Answer
Option D
Solution

When a reaction is divided by factor n then the new equilibrium constant of the reaction will be

Kn\sqrt[n]K

. Thus, here the given reaction is multiplied by factor

12{1 \over 2}

then the new equilibrium constant will be

K12{K^{{1 \over 2}}}

.

Q14
If the value of equilibrium constant for a particular reaction is 1.6 × \times 10 12 , then at equilibrium the system will contain
A mostly products
B similar amounts of reactants and products
C all reactants
D mostly reactants.
Correct Answer
Option A
Solution

As, 1.6 × 10 12 is very high value of K.

Thus, the reaction proceeds almost to completion and mixture must contain mostly products.

Q15
Which of the following statements is correct for a reversible process in a state of equilibrium?
A Δ\Delta G o = −-2.30 RT log K
B Δ\Delta G o = 2.30 RT log K
C Δ\Delta G = −-2.30 RT log K
D Δ\Delta G = 2.30 RT log K
Correct Answer
Option A
Solution
Δ\Delta

G =

Δ\Delta

G o + 2.303 RT log Q At equilibrium, when

Δ\Delta

G = 0 and Q = K then 0 =

Δ\Delta

G o + 2.303 RT log K

Δ\Delta

G o = – 2.303 RT log K

Q16
For a given exothermic reaction, K p and K' p are the equilibrium constants at temperatures T 1 and T 2 , respectively. Assuming that heat of reaction is constant in temperature range between T 1 and T 2 , it is readily observed that
A K p > K' p
B K p < K' p
C K p = K' p
D K p = 1kp′{1 \over {k{'_p}}}
Correct Answer
Option A
Solution

log K' p K p =

−ΔH2.303R[1T2−1T1]- {{\Delta H} \over {2.303R}}\left[ {{1 \over {{T_2}}} - {1 \over {{T_1}}}} \right]

For exothermic reaction,

Δ\Delta

H = -ve means the temperature T 2 is higher than T 1 . ∴\therefore

[1T2−1T1]\left[ {{1 \over {{T_2}}} - {1 \over {{T_1}}}} \right]

is negative. So log K' p - log K p = -ve ⇒\Rightarrow log K p > log K' p ⇒\Rightarrow K p > K' p

Q17
For the reversible reaction, N 2(g) + 3H 2(g) ⇌\rightleftharpoons 2NH 3(g) + heat The equilibrium shifts in forward direction
A by increasing the concentration of NH 3(g)
B by decreasing the pressure
C by decreasing the concentrations of N 2(g) and H 2(g)
D by increasing pressure and decreasiing temperature.
Correct Answer
Option D
Solution

According to Le-Chatelier’s principle, the equilibrium shifts in that direction so as to oppose the applied change.

Given reaction is exothermic reaction.

Hence according to Le-Chatelier's principle low temperature favours the forward reaction and on increasing pressure equilibrium will shift, towards lesser number of moles i.e. forward direction.

Q18
Using the Gibb's energy change, Δ\Delta G o = +63.3 kJ, for the following reaction, Ag 2 CO 3(s) ⇌\rightleftharpoons 2 Ag + (aq) + CO 3 2−- (aq) the K sp of Ag 2 CO 3(s) in water at 25 o C is (R = 8.314 J K −-1 mol −-1 )
A 3.2 × \times 10 −-26
B 8.0 × \times 10 −-12
C 2.9 × \times 10 −-3
D 7.9 × \times 10 −-2
Correct Answer
Option B
Solution

We know,

Δ\Delta

G o = – 2.303 RT log K sp ∴\therefore 63300 = – 2.303 × 8.314 × 298 log K sp ⇒\Rightarrow log K sp = -11.09 ⇒\Rightarrow K sp = 10 -11.09 = 8.0 × 10 –12

Q19
Given that the equilibrium constant for the reaction, 2SO 2(g) + O 2(g) ⇌\rightleftharpoons 2SO 3(g) has a value of 278 at a particular temperature. What is the value of the equilibrium constant for the following reaction at the same temperature ? SO 3(g) ⇌\rightleftharpoons SO 2(g) + 12{1 \over 2} O 2(g)
A 1.8 × \times 10 −-3
B 3.6 × \times 10 −-3
C 6.0 × \times 10 −-2
D 1.3 × \times 10 −-5
Correct Answer
Option C
Solution

2SO 2(g) + O 2(g) ⇌\rightleftharpoons 2SO 3(g) , K SO 3(g) ⇌\rightleftharpoons SO 2(g) +

12{1 \over 2}

O 2(g) , K' =

1K\sqrt {{1 \over K}}

∴\therefore K' =

1278=\sqrt {{1 \over {278}}} =

6.0 ×\times 10 −-2

Q20
Given the reaction between 2 gases represented by A 2 and B 2 to give the compound AB (g) , A 2 (g) + B 2(g) ⇌\rightleftharpoons 2AB (g) At equilibrium, the concentration of A 2 = 3.0 × \times 10 −-3 M, of B 2 = 4.2 × \times 10 −-3 M, of AB = 2.8 × \times 10 −-3 M If the reaction takes place in a sealed vessel at 527 o C, then the value of K c will be
A 2.0
B 1.9
C 0.62
D 4.5
Correct Answer
Option C
Solution

A 2 (g) + B 2(g) ⇌\rightleftharpoons 2AB (g) ; K c =

[AB]2[A2][B2]{{{{\left[ {AB} \right]}^2}} \over {\left[ {{A_2}} \right]\left[ {{B_2}} \right]}}

=

(2.8×10−3)23×10−3×4.2×10−3{{{{\left( {2.8 \times {{10}^{ - 3}}} \right)}^2}} \over {3 \times {{10}^{ - 3}} \times 4.2 \times {{10}^{ - 3}}}}

= 0.62

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