Question

A vessel at 1000 K contains ${CO}_2$ with a pressure of $0.5$ atm. Some of the ${CO}_2$ are converted into $CO$ on the addition of graphite. If the total pressure at equilibrium is $0.8$ atm the value of $K$ is:

• A. $0.3$ atm
• B. $0.18$ atm
• C. $1.8$ atm
• D. $3$ atm

Answer 1

Answer: (C)

$1.8$ atm

${CO}_2\left(s\right)+C\left(s\right)\rightleftharpoons 2CO\left(g\right)$

Initial : $0.5$atm $0$ $0$

At equ : $0.5-p$ $0$ $2p$

$\therefore$ At equilibrium the total pressure will $=0.5-p+2p=0.5+p$

now $0.5+p=0.8$

$\Rightarrow p=0.3$ atm

now, $K=\frac{p_{CO}^2}{p_{{CO}_2}}=\frac{{\left(2p\right)}^2}{0.5-p}=\frac{{(2\times 0.3)}^2}{0.2}=1.8$ atm