Question

The equilibrium constant for the reaction, $CO\left(g\right)+H_{2}O\left(g\right)\rightleftharpoons C{O}_2\left(g\right)+H_2\left(g\right)$ at ${986}^{o}C$ is $0.63$. A mixture of $1.0$ mole of water vapor and $3.0$ moles of $CO$ is allowed to come to equilibrium. The equilibrium pressure is $2.0atm$.

(i) How many moles of $H_2$ are present at equilibrium?
(ii) Calculate the partial pressure of gases in the equilibrium mixture.

Answer 1

In the given reaction, $\Delta n=0$, hence, $K_c=K_p$

$CO\left(g\right)+H_{2}O\left(g\right)\rightleftharpoons C{O}_2\left(g\right)+H_2\left(g\right)$

At eq. $(3-x)$ $(1-x)$ $x$ $x$

Applying law of mass action,
$\frac{x\times x}{(3-x)(1-x)}=0.63$

$x^2+6.81x-5.1=0$

or $x=\frac{-6.81\pm {[6.81\times 6.81-4\times (-5.1\left)\right]}^{1/2}}{2}=0.68moleof{H}_2$

(ii) Partial pressure of ${CO}_2=$ Partial pressure of $H_2$

$=$ Mole fraction $\times$Total pressure

$=\frac{0.68}{4}\times 2=0.34atm$

Partial pressure of $CO\left(g\right)=\frac{3-0.68}{4}\times 2=1.16atm$

Partial pressure of $H_{2}O\left(g\right)=\frac{1-0.68}{4}\times 2=0.16atm$