Question

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

Answer 1

(a) ${CO}_2\left(g\right)+C\left(s\right)\rightleftharpoons 2CO\left(g\right);K_p=10atm$
given, $p_{CO\left(g\right)}+p_{{CO}_2\left(g\right)}=5atm$
Let $p_{CO\left(g\right)}=xatm$
So, $p_{{CO}_2\left(g\right)}=(5-x)atm$
$K_p=\frac{{\left[p_{CO}\right]}^2}{p_{{CO}_2}}=\frac{x^2}{(5-x)}$
$10=\frac{x^2}{(5-x)}$
or $x^2+10x-50=0$
On solving
$x=3.66$
$p_{CO\left(g\right)}=3.66atm$
or Mole fraction $CO$ at equilibrium$=73.2$% (by volume)
$p_{{CO}_2\left(g\right)}=1.34atm$
or Mole fraction ${CO}_2$ at equilibrium$=26.8$% (by volume)
(b) Let the total pressure be $Patm$.
Given % ${CO}_2$ (by volume) $=5$ and % $CO$ (by volume)$=95$
$p_{{CO}_2}=\frac{5}{100}\times P=0.05P;p_{CO}=\frac{95}{100}\times P=0.95P$
$K_p=\frac{{\left[p_{CO}\right]}^2}{p_{{CO}_2}}=\frac{0.95P\times 0.95P}{0.05P}=18.05P$
or $P=0.554atm$