Equilibrium Question 566

Question: $ 2NO_2 $ ⇌ $ 2NO+O_2;K=1.6\times {10^{-12}} $

$ NO+\frac{1}{2}O_2 $ ⇌ $ NO_2{K}’=- $ [CPMT 1996]

Options:

A) $ {K}’=\frac{1}{K^{2}} $

B) $ {K}’=\frac{1}{K} $

C) $ {K}’=\frac{1}{\sqrt{K}} $

D) None of these

Show Answer

Answer:

Correct Answer: C

Solution:

$ 2NO_2 $ ⇌ $ 2NO+O_2 $ –(i) $ K=1.6\times {10^{-12}} $

$ NO+\frac{1}{2}O_2 $ ⇌ $ NO_2 $ –(ii) Reaction (ii) is half of reaction (i) $ K=\frac{{{[NO]}^{2}}[O_2]}{{{[NO_2]}^{2}}} $ –(i) $ {K^{’}}=\frac{[NO_2]}{[NO]{{[O_2]}^{{1}/{2};}}} $ –(ii) On multiplying (i) and (ii) $ K\times {K^{’}}=\frac{{{[NO]}^{2}}[O_2]}{{{[NO_2]}^{2}}}\times \frac{[NO_2]}{[NO]{{[O_2]}^{{1}/{2};}}} $

$ =\frac{[NO]{{[O_2]}^{{1}/{2};}}}{[NO_2]}=\frac{1}{{K^{’}}} $

$ K\times {K^{’}}=\frac{1}{{K^{’}}} $ ; $ K=\frac{1}{{K^{‘2}}} $ ; $ {K^{’}}=\frac{1}{\sqrt{K}} $ .



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