SATHEE: Equilibrium Question 596

Equilibrium Question 596

Question: If equilibrium constants of reaction, $N_{2} + O_{2}$ ⇌ $2 N O$ is $K_{1}$ and $\frac{1}{2} N_{2} + \frac{1}{2} O_{2}$ ⇌ $N O$ is $K_{2}$ , then [BHU 2004]

Options:

A) $K_{1} = K_{2}$

B) $K_{2} = \sqrt{K_{1}}$

C) $K_{1} = 2 K_{2}$

D) $K_{1} = \frac{1}{2} K_{2}$

Show Answer

Answer:

Correct Answer: B

Solution:

$N_{2} + O_{2}$ ⇌ $2 N O$ -..(i) $\frac{1}{2} N_{2} + \frac{1}{2} O_{2}$ ⇌ $N O$ –(ii)

For equation number (i) $K_{1} = \frac{{[N O]}^{2}}{[N_{2}], [O_{2}]}$ -.. (iii)

For equation number (ii) $K_{2} = \frac{[N O]}{{[N_{2}]}^{1 / 2} {[O_{2}]}^{1 / 2}}$ -… (iv)

From equation (iii) & (iv) it is clear that

$K_{2} = {(K_{1})}^{1 / 2} = \sqrt{K_{1}}$ ;

Hence, $K_{2} = \sqrt{K_{1}}$

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Equilibrium Question 596

Question: If equilibrium constants of reaction, N2+O2 ⇌ 2NO is K1 and 12N2+12O2 ⇌ NO is K2 , then [BHU 2004]

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Question: If equilibrium constants of reaction, $N_{2} + O_{2}$ ⇌ $2 N O$ is $K_{1}$ and $\frac{1}{2} N_{2} + \frac{1}{2} O_{2}$ ⇌ $N O$ is $K_{2}$ , then [BHU 2004]