Equilibrium Question 527
Question: The decomposition of $ N_2O_4 $ to $ NO_2 $ is carried out at $ 280K $ in chloroform. When equilibrium has been established, 0.2 mol of $ N_2O_4 $ and $ 2\times {10^{-3}} $ mol of $ NO_2 $ are present in 2 litre solution. The equilibrium constant for reaction $ N_2O_4 $ ⇌ $ 2NO_2 $ is [AIIMS 1984]
Options:
A) $ 1\times {10^{-2}} $
B) $ 2\times {10^{-3}} $
C) $ 1\times {10^{-5}} $
D) $ 2\times {10^{-5}} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ K=\frac{{{[NO_2]}^{2}}}{[N_2O_4]}=\frac{{{[ 2\times {{\frac{10}{2}}^{-3}} ]}^{2}}}{[ \frac{0.2}{2} ]}=\frac{{10^{-6}}}{{10^{-1}}}={10^{-5}} $ .
BETA
