Question

A vessel contains ${CO}_2$ and $CO$ with pressures $2$ atm and $3$ atm respectively at ${27}^{o}C$.

At a temperature of ${27}^{o}C$, the reaction, $2CO\left(g\right)\rightleftharpoons {CO}_2\left(g\right)+C\left(s\right)$, occurs and equilibrium is attained. If the equilibrium pressure is $4.5$ atm, then:

• A. $K_p=2/9{atm}^{-1}$
• B. $p_{CO}:p_{{CO}_2}=4:5$
• C. $K_p=16{atm}^{-1}$
• D. % dissociation of $CO=0.333$

Answer 1

The correct option is B $p_{CO}:p_{{CO}_2}=4:5$

Initial partial pressures of $CO$ and $C{O}_2$ are 3 atm and 2 atm respectively.

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

To reach equilibrium, $2x$ atm of $CO$ reacts to form $x$ atm of $C{O}_2$

The equilibrium partial pressures of $CO$ and $C{O}_2$ are $3-2x$ atm and $2+x$ atm respectively.

Total equilibrium pressure is $(3-2x)+(2+x)=5-x$ atm. But it is equal to 4.5 atm.

$4.5=5-x$
$x=5-4.5$
$x=0.5$
$p_{CO}:p_{{CO}_2}=(3-2x):(2+x)=(3-2(0.5\left)\right):(2+1(0.5\left)\right)=2:2.5=4:5$
$K_p=\frac{P_{C{O}_2}}{P_{CO}^2}$

$K_p=\frac{2.5}{2^2}$

$K_p=0.625at{m}^{-1}$
The dissociation of CO $=\frac{2x}{3}=\frac{2\left(0.5\right)}{3}=0.333$

The % dissociation of CO $=100\times 0.333=33.3$