Equilibrium Question 665
Question: $ N_2O_4 $ is 10% dissociated at a total pressure $ P_1 $ and 20% dissociated at a total pressure $ P_2 $ . Then ration $ P_1/P_2 $ is
Options:
A) $ \frac{1}{2} $
B) $ \frac{2}{1} $
C) $ \frac{1}{4} $
D) $ \frac{4}{1} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ N_2O_4\rightarrow 2NO_2 $ If $ \alpha $ is the degree of dissociation and P is the total pressure, then $ K_{p}=\frac{4{{\alpha }^{2}}}{1-{{\alpha }^{2}}}P $
when $ \alpha $ =0.10, $ K_{p}=\frac{4{{(0.10)}^{2}}}{1-{{(0.10)}^{2}}}\times P_1=\frac{0.04}{0.99}P_1\simeq 0.04P_1 $
When $ \alpha $ =0.20, $ K_{p}=\frac{4{{(0.20)}^{2}}}{1-{{(0.2)}^{2}}}\times P_2=\frac{0.16}{0.96}P_2\simeq 0.16P_2 $
Thus, $ 0.04P_1=0.16P_2 $ or $ \frac{P_1}{P_2}=\frac{0.16}{0.96}=\frac{4}{1} $
BETA
