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 \(298 \mathrm{~K}\) ): $$\begin{aligned} 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) & \rightleftharpoons 2 \mathrm{NOBr}(g) & K_{c} &=2.0 \\ 2 \mathrm{NO}(g) & \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) & K_{c} &=2.1 \times 10^{30} \end{aligned}$$

We are given two reactions with their respective equilibrium constants: 1) \(2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)\), \(K_c = 2.0\). 2) \(2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)\), \(K_c = 2.1 \times 10^{30}\). We need to add and/or manipulate them in such a way that when they're combined, they result in the desired reaction.

First, let's reverse the second reaction because we need \(\mathrm{N}_{2}(g)\) and \(\mathrm{O}_{2}(g)\) in the reactants. The reversed reaction is: \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)\), and its \(K'_c\) is the reciprocal of the given \(K_c\) for the second reaction: \(K'_c = \frac{1}{K_c} = \frac{1}{2.1 \times 10^{30}}\). Now, let's add the reversed second reaction to the first reaction: \(\begin{aligned} (2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)) \qquad & K_c=2.0\\ (+\!\!\!&\!\!\!\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g), \qquad K'_c=\frac{1}{2.1 \times 10^{30}}) \end{aligned}\) This results in: \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)\), which is the desired reaction.

Since we have added the reactions, we multiply their equilibrium constants: $$K_c^{\text{desired}} = K_c \cdot K'_c = 2.0 \cdot \frac{1}{2.1 \times 10^{30}} = \frac{2.0}{2.1 \times 10^{30}}$$ Now, we need to find the equilibrium constant in terms of pressure, \(K_p\). The relationship between \(K_c\) and \(K_p\) is given by: $$K_p = K_c(RT)^{\Delta n}$$ where \(R\) is the gas constant, \(T\) is the temperature in Kelvin, and \(\Delta n\) is the change in moles of gas (the sum of the coefficients of gaseous products minus the sum of the coefficients of gaseous reactants). In our case, the change in moles of gas is: $$\Delta n = (2 - (1+1+1)) = -1$$ Given \(R = 0.0821 \frac{\text{atm} \cdot \text{L}}{\text{mol} \cdot \text{K}}\) and \(T = 298 \, \text{K}\), we can calculate \(K_p\) as follows: $$K_p = \frac{2.0}{2.1 \times 10^{30}}(0.0821 \cdot 298)^{-1} = \frac{2.0}{2.1 \times 10^{30} \cdot 0.0821 \cdot 298}$$

The equilibrium constant \(K_p\) for the desired reaction is: $$K_p = \frac{2.0}{2.1 \times 10^{30} \cdot 0.0821 \cdot 298}$$