Question

At ${817}^{0}C,K_p$ for the reaction between pure $C{O}_2$ and excess hot graphite to form $2CO\left(g\right)$ is 10 atm.
What is the analysis of the gases at equilibrium at ${817}^{0}C$ & a total pressure of 4.0 atm? What is the partial pressure of $C{O}_2$ at equilibrium?

Answer 1

$C{O}_2\left(g\right)+C$ (hot graphite) $\to 2CO$

$P_0$ $0$

$P_0-P$ $2P$

total pressure $=4atm$

$P_0-P+2P=4$

$P_0+P=4\Rightarrow P_0=4-P$

$K_P=\frac{(P_{CO}{)}^2}{P_{C{O}_2}}=\frac{(2P{)}^2}{P_0-P}\frac{4P^2}{P_0-P}$

$\frac{4P^2}{P_0-P}=10$

$4P^2=10P_0-10P$

$4P^2=10)4-P)-10P$

$4P^2+20P-40=0$

$P^2+5P-10=0$

$P=\frac{-5+\sqrt{25+41}}{2}=1.531$

$P_0=4-1.53=2.469$

$P_{C{O}_2}=P_0-P=2.468-1.531=0.937$