Consider the equilibrium $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) $$ Calculate the equilibrium constant \(K_{p}\) for this reaction, given the following information (at \(298 \mathrm{~K}\) ): $$ \begin{array}{l} 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) \quad K_{c}=2.0 \\ 2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \quad K_{c}=2.1 \times 10^{30} \end{array} $$

We are given the following reactions with their respective equilibrium constants: Reaction 1: \(2 NO(g) + Br_2(g) \rightleftharpoons 2 NOBr(g) \) with corresponding \(K_{c1} = 2.0\) Reaction 2: \(2 NO(g) \rightleftharpoons N_2(g) + O_2(g) \) with corresponding \(K_{c2} = 2.1 \times 10^{30}\) We want to find the \(K_{p}\) for the following reaction: Desired Reaction: \(N_2(g) + O_2(g) + Br_2(g) \rightleftharpoons 2 NOBr(g)\)

To form the desired reaction, first reverse Reaction 2, and then add it to Reaction 1: Reverse Reaction 2: \(N_2(g) + O_2(g) \rightleftharpoons 2 NO(g) \) Since Reaction 2 is reversed, the new equilibrium constant becomes \(K'_c = \frac{1}{K_{c2}} = \frac{1}{2.1 \times 10^{30}}\) Add the reversed Reaction 2 and Reaction 1: Reaction 1: \(2 NO(g) + Br_2(g) \rightleftharpoons 2 NOBr(g) \) Reverse Reaction 2: \(N_2(g) + O_2(g) \rightleftharpoons 2 NO(g) \) Desired Reaction: \(N_2(g) + O_2(g) + Br_2(g) \rightleftharpoons 2 NOBr(g)\) Now, multiply the equilibrium constants of the reversed Reaction 2 and Reaction 1.

To calculate the equilibrium \(K'_c\) for the desired reaction, multiply the equilibrium constants of the reversed Reaction 2 and Reaction 1: \(K'_c = K'_c \times K_{c1} = \left(\frac{1}{2.1 \times 10^{30}}\right) \times 2.0\) \(K'_c = \frac{2.0}{2.1 \times 10^{30}}\)

We know that for a reaction between ideal gases, the relationship between \(K_{p}\) and \(K_{c}\) is given by: \(K_{p} = K_{c} \times (RT)^{(\Delta n)}\) Where R is the universal gas constant (\(R = 0.0821 L \cdot atm/(mol \cdot K)\)), T is the temperature in Kelvin, and \(\Delta n\) is the change in the number of moles of gas from reactants to products. In the desired reaction, analyze the moles of gases as follows: Reactants: 1 mole of \(N_2\), 1 mole of \(O_2\), 1 mole of \(Br_2\) (total 3 moles) Products: 2 moles of \(NOBr\) (total 2 moles) \(\Delta n = 2 - 3 = -1\) Now, we can find the \(K_p\): At T = 298 K: \(K_{p} = K'_c \times (RT)^{(\Delta n)} = \frac{2.0}{2.1 \times 10^{30}} \times \left(0.0821 \times 298\right)^{-1}\) Calculate the value of \(K_p\): \(K_{p} \approx 7.22 \times 10^{-31}\) Therefore, the equilibrium constant \(K_{p}\) for the desired reaction \(N_2(g) + O_2(g) + Br_2(g) \rightleftharpoons 2 NOBr(g)\) at 298 K is approximately \(7.22 \times 10^{-31}\).