Equilibrium Question 596

Question: If equilibrium constants of reaction, $ N_2+O_2 $ ⇌ $ 2NO $ is $ K_1 $ and $ \frac{1}{2}N_2+\frac{1}{2}O_2 $ ⇌ $ NO $ is $ K_2 $ , then [BHU 2004]

Options:

A) $ K_1=K_2 $

B) $ K_2=\sqrt{K_1} $

C) $ K_1=2K_2 $

D) $ K_1=\frac{1}{2}K_2 $

Show Answer

Answer:

Correct Answer: B

Solution:

$ N_2+O_2 $ ⇌ $ 2NO $ -..(i) $ \frac{1}{2}N_2+\frac{1}{2}O_2 $ ⇌ $ NO $ –(ii)

For equation number (i) $ K_1=\frac{{{[NO]}^{2}}}{[N_2],[O_2]} $ -.. (iii)

For equation number (ii) $ K_2=\frac{[NO]}{{{[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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