When \(2.00 \mathrm{~mol}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a \(2.00-\mathrm{L}\) flask at \(303 \mathrm{~K}\), \(56 \%\) of the \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decomposes to \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\) : $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)$$ (a) Calculate \(K_{c}\) for this reaction at this temperature. (b) Calculate \(K_{p}\) for this reaction at \(303 \mathrm{~K}\). (c) Repeat these calculations for \(2.00 \mathrm{~mol}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) in a \(15.00-\mathrm{L}\) vessel at \(303 \mathrm{~K}\).

For given $2.00\,\text{mol}$ of $\text{SO}_{2}\text{Cl}_{2}$ in a $2.00\,\text{L}$ flask, at equilibrium: $[\text{SO}_{2}] = [\text{Cl}_{2}] = (0.56 \times 2.00)\frac{\text{mol}}{2.00\,\text{L}}$ $[\text{SO}_{2}\text{Cl}_{2}] = (2.00 - 0.56 \times 2.00)\frac{\text{mol}}{2.00\,\text{L}}$ Kc can be found using the expression: $K_c = \frac{[\text{SO}_{2}][\text{Cl}_{2}]}{[\text{SO}_{2}\text{Cl}_{2}]}\approx 1.42$ Kp can be found using the expression: $K_p = K_c(RT)^n \approx 3.32 \,\text{atm}$ For the given $2.00\,\text{mol}$ of $\text{SO}_{2}\text{Cl}_{2}$ in a $15.00\,\text{L}$ vessel, at equilibrium: $[\text{SO}_{2}] = [\text{Cl}_{2}] = (0.56 \times 2.00)\frac{\text{mol}}{15.00\,\text{L}}$ $[\text{SO}_{2}\text{Cl}_{2}] = (2.00 - 0.56 \times 2.00)\frac{\text{mol}}{15.00\,\text{L}}$ Kc can be found using the expression: $K_c = \frac{[\text{SO}_{2}][\text{Cl}_{2}]}{[\text{SO}_{2}\text{Cl}_{2}]}\approx 0.602$ Kp can be found using the expression: $K_p = K_c(RT)^n \approx 1.40 \,\text{atm}$