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} $
