Equilibrium Question 638
Question: For the equilibrium $ N_2+3H_2 $ ⇌ $ 2NH_3,K_{c} $ at 1000K is $ 2.37\times {10^{-3}} $ . If at equilibrium $ [N_2]=2M,,[H_2]=3M $ , the concentration of $ NH_3 $ is [JIPMER 2000]
Options:
A) 0.00358 M
B) 0.0358 M
C) 0.358 M
D) 3.58 M
Show Answer
Answer:
Correct Answer: C
Solution:
$ K_{c}=\frac{{{[NH_3]}^{2}}}{[N_2]{{[H_2]}^{3}}} $
$ 2.37\times {10^{-3}}=\frac{x^{2}}{[2]{{[3]}^{3}}}=x^{2}=0.12798 $ x = 0.358 M.
