Question

At ${817}^{o}C$ $K_p$ for the reaction between ${CO}_2$ and excess hot graphite to form $2CO\left(g\right)$ is $10$.
(i) What is the analysis (mole fraction) of the gases at equilibrium at ${817}^{o}C$ and a total pressure of $4atm$? What is the partial pressure of ${CO}_2$ at equilirbrium?
(ii) At what total pressure will the gas mixture have $6$% ${CO}_2$ by volume?

Answer 1

$C{O}_2\left(g\right)+C\left(s\right)\rightleftharpoons 2CO\left(g\right)$
$t=0$ $1$ $0$
$t_{eq.}$ $1-\alpha$ $2\alpha$
$p_{eq.}$ $\left(\frac{1-\alpha }{1+\alpha }\right)p$ $\left(\frac{2\alpha }{(1+\alpha )}\right)p$
$K_p=\frac{4{\alpha }^{2}p}{1-{\alpha }^2};10=\frac{4{\alpha }^{2}4}{1-{\alpha }^2}$
On solving, we get $alpha=0.62$
$x_{C{O}_2}=\frac{1-\alpha }{1+\alpha }=\frac{1-0.62}{1+0.62}=0.2345=23.45$% (by volume)
$x_{CO}=0.7655=76.55$% (by volume)
$p_{C{O}_2}=(0.2345\times 4.0)atm=0.938atm$
(ii) Let the total pressure be $P$ atm
$K_p=10=\frac{{\left(p_{CO}\right)}^2}{p_{C{O}_2}}$
$P=0.68atm$