Equilibrium Question 480
Question: If $ K_{c} $ is the equilibrium constant for the formation of $ NH_3 $ , the dissociation constant of ammonia under the same temperature will be [DPMT 2001]
Options:
A) $ K_{c} $
B) $ \sqrt{K_{c}} $
C) $ K_c^{2} $
D) $ 1/K_{c} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ NH_3 $ ⇌ $ \frac{1}{2}N_2+\frac{3}{2}H_2 $
$ K_{c}=\frac{{{[N_2]}^{{1}/{2};}}{{[H_2]}^{{3}/{2};}}}{NH_3} $ and $ \frac{1}{2}N_2+\frac{3}{2}H_2 $ ⇌ $ NH_3 $
$ K_{c}=\frac{[NH_3]}{{{[N_2]}^{{1}/{2};}}{{[H_2]}^{{3}/{2};}}} $ So for dissociation $ =\frac{1}{K_{c}} $
BETA
