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 $\mathbf{9 0 . 5 5 \%} \mathrm{CO}$ by mass, $\mathrm{C}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g})$. Calculate $\mathrm{K}_{\mathrm{c}}$ for this reaction at the above temperature.

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 $\mathbf{9 0 . 5 5 \%} \mathrm{CO}$ by mass, $\mathrm{C}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g})$. Calculate $\mathrm{K}_{\mathrm{c}}$ for this reaction at the above temperature.,

MARKS App · Accepted Answer

If total mass of mixture of $\mathrm{CO}$ and $\mathrm{CO}_2$ is $100 \mathrm{~g}$, then $\mathrm{CO}$ $=90.55 \mathrm{~g}$ and $\mathrm{CO}_2=100-90.55=9.45 \mathrm{~g}$. Number of moles of $\mathrm{CO}=90 \times 55 / 28=3.234$ Number of moles of $\mathrm{CO}_2=9 \times 45 / 44=0.215$ $\mathrm{p}_{\mathrm{CO}}=3.234 /(3.234+0.215) \times 1 \mathrm{~atm}$ $=0.938 \mathrm{~atm}$ $\mathrm{p}_{\mathrm{CO}_2}=0.215 /(3.234+0.215) \times 1 \mathrm{~atm}=0.062 \mathrm{~atm}$ $\mathrm{K}_{\mathrm{p}}=\frac{\mathrm{p}^2 \mathrm{CO}}{\mathrm{p}_{\mathrm{CO}_2}}=\frac{(0.938)^2}{0.062}=14.19$ $\Delta \mathrm{n}_{\mathrm{g}}=2-1=1$ $\therefore \quad K_p=K_c(R T)^{\Delta n_g}$ or $\mathrm{K}_{\mathrm{c}}=\mathrm{K}_{\mathrm{p}} / \mathrm{RT}=14.19 /(0.0821 \times 1127)$ $=0.153$

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