SATHEE: Equilibrium Question 566

Equilibrium Question 566

Question: $2 N O_{2}$ ⇌ $2 N O + O_{2}; K = 1.6 \times 10^{- 12}$

$N O + \frac{1}{2} O_{2}$ ⇌ $N O_{2} K^{'} = -$ [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:

$2 N O_{2}$ ⇌ $2 N O + O_{2}$ –(i) $K = 1.6 \times 10^{- 12}$

$N O + \frac{1}{2} O_{2}$ ⇌ $N O_{2}$ –(ii) Reaction (ii) is half of reaction (i) $K = \frac{{[N O]}^{2} [O_{2}]}{{[N O_{2}]}^{2}}$ –(i) $K^{^{'}} = \frac{[N O_{2}]}{[N O] {[O_{2}]}^{1 / 2;}}$ –(ii) On multiplying (i) and (ii) $K \times K^{^{'}} = \frac{{[N O]}^{2} [O_{2}]}{{[N O_{2}]}^{2}} \times \frac{[N O_{2}]}{[N O] {[O_{2}]}^{1 / 2;}}$

$= \frac{[N O] {[O_{2}]}^{1 / 2;}}{[N O_{2}]} = \frac{1}{K^{^{'}}}$

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

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Equilibrium Question 566

Question: 2NO2 ⇌ 2NO+O2;K=1.6×10−12

Options:

Answer:

Solution:

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Question: $2 N O_{2}$ ⇌ $2 N O + O_{2}; K = 1.6 \times 10^{- 12}$