Question

At ${817}^{o}C$, $K_p$ for the reaction between ${CO}_2\left(g\right)$ and excess hot graphite (s) is $10atm$.
(a) What are the equilibrium concentrations of gases at ${817}^{o}C$ and a total pressure of $5atm$?
(b) At what total pressure the gas contains $5$% ${CO}_2$ by volume?

Answer 1

(a) $C{O}_2\left(g\right)+C\left(s\right)\rightleftharpoons 2CO\left(g\right)$
At equi, $(5-x)$ atm $x$ atm
$K_p=\frac{{\left[CO\right]}^2}{\left[C{O}_2\right]}$
$10=\frac{x^2}{5-x}$
$x=\frac{-10\pm \sqrt{100+200}}{2}=3.66$
i.e., $p_{CO}=3.66atm;p_{C{O}_2}=1.34atm$
Concentration of $CO=\frac{p}{RT}=\frac{3.66}{0.0821\times 1090}=0.041mol{litre}^{-1}$
Concentration of $C{O}_2=\frac{p}{RT}=\frac{1.34}{0.0821\times 1090}=0.015mol{litre}^{-1}$
(b) $C{O}_2\left(g\right)+C\left(s\right)\rightleftharpoons 2CO\left(g\right)$
$t=0$ $1$ $0$ $0$
$t_{eq.}$ $1-\alpha$ $2\alpha$
$p_i$ $\left(\frac{1-\alpha }{1+\alpha }\right)p$ $\left(\frac{2\alpha }{(1+\alpha )}\right)p$
$K_p=\frac{{\left[\left(\frac{2\alpha }{(1+\alpha )}\right)p\right]}^2}{\left(\frac{1-\alpha }{1+\alpha }\right)p}=\frac{4{\alpha }^{2}p}{1-{\alpha }^2}...\left(i\right)$
$\frac{5}{100}=\frac{1-\alpha }{1+\alpha }$
$\alpha =0.95$
From eq. (i), $10=\frac{4\times {0.95}^2}{1-{0.95}^2}\times p$
$p=0.27atm$