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Published on: 13/05/2022

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Questions:

Generic 3

5 x 3 = 15

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

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]

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.

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.

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.

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Answers Key:

Generic 3

5 x 3 = 15

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.

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

\(K_c={[AB]^2\over [A_2][B_2]}\)
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.

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