Equilibrium Question 650
Question: For the reaction equilibrium $ N_2O_4 $ ⇌ $ 2N{O_{2(g)}} $ , the concentrations of $ N_2O_4 $ and $ NO_2 $ at equilibrium are $ 4.8\times {10^{-2}} $ and $ 1.2\times {10^{-2}},mol,litr{e^{-1}} $ respectively. The value of $ K_{c} $ for the reaction is [AIEEE 2003]
Options:
A) $ 3.3\times 10^{2} $ $ mollitr{e^{-1}} $
B) $ 3\times {10^{-1}} $ $ mollitr{e^{-1}} $
C) $ 3\times {10^{-3}} $ $ mollitr{e^{-1}} $
D) $ 3\times 10^{3} $ $ mollitr{e^{-1}} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ K=\frac{{{[NO_2]}^{2}}}{[N_2O_4]} $
$ =\frac{{{[1.2\times {10^{-2}}]}^{2}}}{[4.8\times {10^{-2}}]} $
$ =0.3\times {10^{-2}} $
$ =3\times {10^{-3}} $
