Equilibrium Question 765
Question: The partial pressure of $ CH_3OH(g),CO(g) $ and $ H_2(g) $ in equilibrium mixture for the reaction, $ CO(g)+2H_2(g)\rightarrow CH_3OH(g) $ are 2.0, 1.0 and 0.1 atm respectively at $ 427{}^\circ C $ . The value of $ K_{p} $ for the decomposition of $ CH_3OH $ to CO and $ H_2 $ is
Options:
A) $ 10^{2}atm $
B) $ 2\times 10^{2}at{m^{-1}} $
C) $ 50atm^{2} $
D) $ 5\times {10^{-3}}atm^{2} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ K_{p}=\frac{{p_{CH_3OH}}}{p_{CO}\times {p_{H_2}}}=\frac{2}{1\times {{(0.1)}^{2}}}=200; $ For reverse reaction $ \frac{1}{K_{p}}=\frac{1}{200}=5\times {10^{-3}}atm^{2} $
