A vessel at 1000 K contains $\mathrm{CO}_2$ with a pressure of 0.5 atm. Some of $\mathrm{CO}_2$ is converted into CO on addition of graphite. If total pressure at equilibrium is 0.8 atm , then Kp is :

Question

Aicharya · Accepted Answer

$\begin{array}{lc}\mathrm{CO}_2(\mathrm{~g})+\mathrm{C}(\mathrm{s}) \rightleftharpoons & 2 \mathrm{CO}(\mathrm{g}) \\ 0.5 & - \\ 0.5-\mathrm{x} & 2 \mathrm{x}\end{array}$
$\begin{aligned}\mathrm{P}_{\text {total }}=0.5+\mathrm{x}=0.8 \\ \mathrm{x}=0.3 \\ \mathrm{~K}_{\mathrm{P}}=\frac{(0.6)^2}{0.2}=1.8\end{aligned}$

A vessel at 1000 K contains $\mathrm{CO}_2$ with a pressure of 0.5 atm. Some of $\mathrm{CO}_2$ is converted…

Solution