Question

At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO₂ in equilibrium with solid carbon has 90.55% CO by mass
$C\left(s\right)+C{O}_2\left(g\right)\rightleftharpoons 2CO\left(g\right)$

Calculate K_c for this reaction at the above temperature.

Answer 1

Let the total mass of the gaseous mixture = 100 g.

Mass of CO = 90.55 g

And, mass of CO₂= (100 – 90.55) = 9.45 g

Now, the number of moles of CO, $n_{CO}=\frac{90.55}{28}=3.234mol$

Number of moles of CO­₂,$n_{C{O}_2}=frac9.4544=0.215mol$

The partial pressure of CO,
$p_{CO}=\frac{n_{CO}}{n_{CO}+n_{C{O}_2}}\times p_{total}$
= $\frac{3.234}{3.234+0.215}\times 1$ = 0.938 atm

Partial pressure of CO₂,
$p_{C{O}_2}$ = $\frac{n_{C{O}_2}}{n_{CO}+n_{C{O}_2}}\times p_{total}$
= $\frac{0.215}{3.234+0.215}\times 1=0.062atm$
$K_p=\frac{P_{CO}^2}{P_{C{O}_2}}$ = $\frac{(0.938{)}^2}{0.062}=14.19$
For the given reaction,

${\Delta }_n{}_g$= 2 –1 = 1

We know that,
$K_P=K_C(RT{)}^{\Delta n}$
14.19= $K_C(0.082\times 1127{)}^1$
$K_C=\frac{14.19}{0.082\times 1127}$
= 0.153