The equilibrium $2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g)\( is established at \)550 \mathrm{~K}$. An equilibrium mixture of the three gases has partial pressures of $10.13 \mathrm{kPa}, 20.27 \mathrm{kPa}\(, and \)35.46 \mathrm{kPa}\( for \)\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.

We are given the partial pressures of the reactants and products at equilibrium. Using the equilibrium expression for Kp, we can calculate its value as follows: The equilibrium expression is \( K_p = \frac{P_{NOCl}^2}{P_{NO}^2 \times P_{Cl_{2}}} \) Now, substitute the given partial pressures into the expression: \( P_{NO} = 10.13 kPa \) \( P_{Cl_{2}} = 20.27 kPa \) \( P_{NOCl} = 35.46 kPa \) \( K_p = \frac{(35.46)^2}{(10.13)^2 \times 20.27} \) Calculate the value of Kp: \( K_p = 3.059 \)

Given that the volume of the vessel is 5.00 L and the temperature is 500.0 K, we can use the relationship between Kp and Kc to find Kc. For a reaction with a change in the number of moles of gas (Δn), the relationship between Kp and Kc is given by: \( K_p = K_c(RT)^{Δn} \) Where R is the gas constant (0.0821 L atm/mol K), T is the temperature in Kelvin, and Δn is the change in moles of gas between reactants and products. For the given reaction: \( Δn = (2 - 2 - 1) = -1 \) Now, solve for Kc: \( K_c = K_p / (RT)^{Δn} \) Since we are given the pressures in kPa, we will use the value of R in L kPa / mol K, which is R = 8.314 L kPa / mol K, and the given temperature to find Kc: \( K_c = \frac{3.059}{(8.314 \times 500.0)^{-1}} \) Calculate the value of Kc: \( K_c = 0.376 \) So the values of Kp and Kc for the given reaction are as follows: (a) \( K_p = 3.059 \) (b) \( K_c = 0.376 \)