Question

For the reaction, $CO\left(g\right)+2H_2\left(g\right)\leftrightharpoons {CH}_{3}OH\left(g\right)$, hydrogen gas is introduced into a $5L$ flask at ${327}^{o}C$, containing $0.2$ mole of $CO\left(g\right)$ and a catalyst (solid), until the pressure is $4.92$ atmosphere. At this point, $0.1$ mole of ${CH}_{3}OH\left(g\right)$ is formed. Calculate the equilibrium constants $K_p$ and $K_c$.

Answer 1

Let the number of moles of hydrogen introduced be $m$ moles.

Total moles of $CO,H_2$ and $C{H}_{3}OH$ $=m$

Applying, $PV=nRT$

$P=4.92atm,V=5litre,R=0.082,T=(273+327)=600K$

$4.92\times 5=0.082\times 600\times \left(m\right)$

or $m=0.5$ mole

$CO\left(g\right)+2H_2\left(g\right)\leftrightharpoons {CH}_{3}OH\left(g\right)$
No. of moles: $0.2-x$ $y-2x$ $x$ $m=0.2+y,x=0.1$
or $0.2-0.1$ $0.3-0.2$ $0.1$
Active mass: $\frac{0.1}{5}$ $\frac{0.1}{5}$ $\frac{0.1}{5}$

Applying law of mass action,

$K_c=\frac{\left[{CH}_{3}OH\right]}{\left[CO\right]{\left[H_2\right]}^2}=2500$

We know that,

$K_p=K_c{\left(RT\right)}^{\Delta n},\Delta n=-2$

or $K_p=2500{(0.082\times 600)}^{-2}$

$K_p=\frac{2500}{49.2\times 49.2}=1.0327$