Question

$0.15mole$ of $CO$ taken in a $2.5L$ flask is maintained at $705K$ along with a catalyst so that the following reaction can take place: $CO\left(g\right)+2H_2\left(g\right)\leftrightharpoons {CH}_{3}OH\left(g\right)$
Hydrogen is introduced until the total pressure of the system is $8.5atm$ at equilibrium and $0.08$ mole of methanol is formed. Calculate (i) $K_p$ and $K_c$ and (ii) the final pressure if the same amount of $CO$ and $H_2$ as before are used, but with no catalyst so that the reaction does not take place.

Answer 1

Let the total number of moles of gases be $n$ at equilibrium

Applying equation, $PV=nRT$

given, $P=8.5atm;V=2.5L;R=0.0821atmL{mol}^{-1}{K}^{-1};T=705K$

$n=\frac{8.5\times 2.5}{0.0821\times 705}=0.367mol$

$CO\left(g\right)+2H_2\left(g\right)\leftrightharpoons {CH}_{3}OH\left(g\right)$
Initial mole: $0.15$ ${\left(n_{H_2}\right)}_0$ $0$
At equilibrium: $0.15-x$,$[{\left(n_{H_2}\right)}_0-2x]$ $x=0.08$

Number of moles of $CO$ at equilibrium
$=(0.15-0.08)=0.07$ mole

Number of moles of $H_2$ at equilibrium

$=$ Total moles$-$ Moles of $CO-$ Moles of ${CH}_{3}OH$
$=(0.367-0.07-0.08)$

$=0.217$ mole

Applying law of mass action,

$K_c=\frac{\left[{CH}_{3}OH\right]}{\left[CO\right]{\left[H_2\right]}^2}=\frac{\frac{0.08}{2.5}}{\frac{0.07}{2.5}\times {\left(\frac{0.217}{2.5}\right)}^2}=151.6{mol}^{-2}{L}^2$

Now, $K_p=K_c{\left(RT\right)}^{\Delta n}=151.6\times {(0.0821\times 705)}^{-2}=0.045{atm}^{-2}$
(ii) Since $[{\left(n_{H_2}\right)}_0-2x]=0.217$

${\left(n_{H_2}\right)}_0=0.217+2\times 0.08=0.377mole$

Total moles $n_0=0.377+0.15=0.527$

Hence, $p_0=\frac{n_{0}RT}{V}=12.20atm$