Question

The value of $K_p$ for the reaction, $C{O}_2\left(g\right)+C\left(s\right)\rightleftharpoons 2CO\left(g\right)$ is $3.0$ at $1000$K. If initially $P_{C{O}_2}=0.48$ bar and $P_{CO}=0$ bar and pure graphite is present, calculate the equilibrium partial pressure of CO and $C{O}_2$.

Answer 1

The given reaction is :-

${CO}_2\left(g\right)+C\left(s\right)\rightleftharpoons 2CO\left(g\right);K_P=3$

Initial pressure : $0.48$ $0$ $0$

At eqm : $(0.48-p)$ $2p$ (let)

Now, $K_P=\frac{{\left[CO\right]}^2}{\left[{CO}_2\right]}{K}_P=\frac{{\left(pCO\right)}^2}{\left({pCO}_2\right)}$

$\Rightarrow K_P=\frac{{\left(2p\right)}^2}{(0.48-p)}$

$\Rightarrow 3(0.48-p)={4p}^2$

$\Rightarrow {4p}^2+3p-1.44=0$

$\Rightarrow p=\frac{-3\pm \sqrt{9+23.04}}{8}=\frac{-3\pm 5.66}{8}$

$\Rightarrow p=\frac{2.66}{8}=0.3325bar$

So, Partial pressure of $CO$ at eqm $=2\times 0.3325=0.665bar$

Partial pressure of ${CO}_2$ at eqm $=0.48-0.3325=0.1475bar$.