# At 1127 K and 1 atm pressure, a gaseous mixture

Question:

At $1127 \mathrm{~K}$ and $1 \mathrm{~atm}$ pressure, a gaseous mixture of $\mathrm{CO}$ and $\mathrm{CO}_{2}$ in equilibrium with solid carbon has $90.55 \% \mathrm{CO}$ by mass

$\mathrm{C}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g}) \longleftrightarrow 2 \mathrm{CO}(\mathrm{g})$

Calculate $K_{c}$ for this reaction at the above temperature.

Solution:

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

Mass of $\mathrm{CO}=90.55 \mathrm{~g}$

And, mass of $\mathrm{CO}_{2}=(100-90.55)=9.45 \mathrm{~g}$

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

Number of moles of $\mathrm{CO}_{2}, n_{\mathrm{CO}_{2}}=\frac{9.45}{44}=0.215 \mathrm{~mol}$

Partial pressure of CO,

$p_{\mathrm{CO}}=\frac{n_{\mathrm{CO}}}{n_{\mathrm{CO}}+n_{\mathrm{CO}_{2}}} \times p_{\text {total }}$

$=\frac{3.234}{3.234+0.215} \times 1$

$=0.938 \mathrm{~atm}$

Partial pressure of $\mathrm{CO}_{2}$

$p_{\mathrm{CO}_{2}}=\frac{n_{\mathrm{CO}_{2}}}{n_{\mathrm{CO}}+n_{\mathrm{CO}_{2}}} \times p_{\mathrm{total}}$

$=\frac{0.215}{3.234+0.215} \times 1$

$=0.062 \mathrm{~atm}$

Therefore, $K_{\mathrm{p}}=\frac{[\mathrm{CO}]^{2}}{\left[\mathrm{CO}_{2}\right]}$

$=\frac{(0.938)^{2}}{0.062}$

$=14.19$

For the given reaction,

$\Delta n=2-1=1$

We know that,

$K_{\mathrm{P}}=K_{\mathrm{C}}(R T)^{\Delta t}$

$\Rightarrow 14.19=K_{\mathrm{C}}(0.082 \times 1127)^{1}$

$\Rightarrow K_{\mathrm{C}}=\frac{14.19}{0.082 \times 1127}$

$=0.154$ (approximately)