Let the total mass of the mixture of CO and $\mathrm{CO}_2$ is 100 g, then CO = 90.55 g and $\mathrm{CO}_2$ = 100 – 90.55 = 9.45 g Moles of CO = $\frac{90.55}{28}$ = 3.234 ; Moles of $\mathrm{CO}_2$ = $\frac{9.45}{44}$ = 0.215 Mole of fraction of CO = $\frac{3.234}{3.234+0.215}$ = 0.938 Mole fraction of $\mathrm{CO}_2$ = $\frac{0.215}{3.234+0.215}$ = 0.062 ∴ $p_{\mathrm{CO}}$ = mole fraction × total pressure = 0.938 × 1 atm = 0.938 atm $p_{\mathrm{CO}_2}$ = 0.062 × 1 atm = 0.062 atm $K_p$ for the reaction $\mathrm{C}_{(s)} + \mathrm{CO}_{2(g)}$ ⇌ $2\mathrm{CO}_{(g)}$ $K_p$ = $\frac{p_{\mathrm{CO}}^2}{P_{\mathrm{CO}_2}}$ = $\frac{(0.938)^2}{0.062}$ = 14.19 Now $\Delta n_g$ = 2 - 1 = 1 , $K_p$ = $K_c(RT)^{\Delta n_g}$ or $K_c$ = $\frac{K_p}{RT}$ = $\frac{14.19}{0.0821 \times 1127}$ = 0.153