The equilibrium \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g)\) is estab- lished at \(500 \mathrm{~K}\). An equilibrium mixture of the three gases has partial pressures of 0.095 atm, 0.171 atm, and 0.28 atm for \(\mathrm{NO}, \mathrm{Cl}_{2},\) and \(\mathrm{NOCl}\), respectively. (a) Calculate \(K_{p}\) for this reaction at \(500.0 \mathrm{~K}\). (b) If the vessel has a volume of \(5.00 \mathrm{~L}\) calculate \(K_{c}\) at this temperature.

The equilibrium expression for Kp is given by: \( K_p = \frac{(P_{\mathrm{NOCl}})^2}{(P_{\mathrm{NO}})^2 * (P_{\mathrm{Cl_{2}}}) }\) Where P_NOCl, P_NO, and P_Cl2 are the partial pressures of NOCl, NO, and Cl2, respectively, at equilibrium.

The given partial pressures for the gases at equilibrium are: P_NO = 0.095 atm, P_Cl2 = 0.171 atm, and P_NOCl = 0.28 atm. Substituting these values into the equilibrium expression, we can calculate Kp. \( K_p = \frac{(0.28)^2}{(0.095)^2 * (0.171)} \) Now solve for Kp: \( K_p ≈ 59.33 \)

To find Kc, we need to convert the partial pressures to concentrations. We can use the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. Rearranging this equation to find the concentration, we get: \[ \mathrm{C} = \frac{\mathrm{n}}{\mathrm{V}} = \frac{\mathrm{P}}{\mathrm{RT}} \] The given volume of the vessel is 5.00 L and the temperature is 500 K. We also need to use the gas constant R = 0.0821 L atm/mol K. Now, let's find the concentrations for all the gases: \( C_{\mathrm{NO}} = \frac{0.095}{(0.0821)(500)} ≈ 2.30 \times 10^{-3} M \) \( C_{\mathrm{Cl_2}} = \frac{0.171}{(0.0821)(500)} ≈ 4.16 \times 10^{-3} M \) \( C_{\mathrm{NOCl}} = \frac{0.28}{(0.0821)(500)} ≈ 6.79 \times 10^{-3} M \)

The equilibrium expression for Kc is given by: \( K_c = \frac{([\mathrm{NOCl}])^2}{([\mathrm{NO}])^2 * ([\mathrm{Cl_{2}}]) }\) Where [NOCl], [NO], and [Cl2] are the concentrations of NOCl, NO, and Cl2, respectively, at equilibrium.

Substitute the calculated concentrations into the Kc expression: \( K_c = \frac{(6.79 \times 10^{-3})^2}{(2.30 \times 10^{-3})^2 * (4.16 \times 10^{-3})} \) Now solve for Kc: \( K_c ≈ 8.30 \)

(a) The equilibrium constant Kp for this reaction at 500.0 K is approximately 59.33. (b) The equilibrium constant Kc for this reaction at 500.0 K and a volume of 5.00 L is approximately 8.30.