Question

Given the following data at \(25^{\circ} \mathrm{C}\) $$ \begin{array}{ll} 2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) & K=1 \times 10^{-30} \\ 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) & K=8 \times 10^{1} \end{array} $$ Calculate \(K\) for the formation of one mole of NOBr from its elements in the gaseous state.

Short answer

Question: Calculate the equilibrium constant for the formation of one mole of NOBr from its elements (N2, O2, and Br2) in the gaseous state, given the following equilibria and their respective constants: 1. 2 NO(g) <=> N2(g) + O2(g), \(K_1 = 1 \times 10^{-30}\) 2. 2 NO(g) + Br2(g) <=> 2 NOBr(g), \(K_2 = 8 \times 10^1\) Answer: The equilibrium constant for the formation of one mole of NOBr from its elements in the gaseous state is \(K = 8 \times 10^{15}\).

Step-by-step solution

Write down the desired reaction and initial reactions

The desired reaction is: $$ \frac{1}{2}\mathrm{N}_{2}(g)+\frac{1}{2}\mathrm{O}_{2}(g)+\frac{1}{2}\mathrm{Br}_{2}(g) \rightleftharpoons \mathrm{NOBr}(g) $$ The initial reactions are: $$ \begin{array}{ll} \text { (1) } 2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) & K_{1} =1 \times 10^{-30} \\\ \text { (2) } 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) & K_{2}=8 \times 10^{1} \end{array} $$

Find the relationship between the desired reaction and given reactions

To get the desired reaction, we can do the following steps: 1. Divide reaction (1) by 2 2. Multiply the new reaction (1) by -1 to reverse its direction 3. Add the modified new reaction (1) to reaction (2)

Modify the given reactions to match the desired reaction

Reaction (1) divided by 2: $$ \text { (1') } \mathrm{NO}(g) \rightleftharpoons \frac{1}{2}\mathrm{N}_{2}(g)+\frac{1}{2}\mathrm{O}_{2}(g), \quad K_{1'} = {K_1}^{\frac{1}{2}} =\sqrt{1 \times 10^{-30}} $$ Now, reverse reaction (1'): $$ \text {(1'')} \frac{1}{2}\mathrm{N}_{2}(g)+\frac{1}{2}\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{NO}(g), \quad K_{1''} =\frac{1}{K_{1'}} = \frac{1}{\sqrt{1 \times 10^{-30}}} $$

Add the modified reaction (1'') and reaction (2) to find the desired reaction

$$ \begin{array}{ll} \text {(1'')} \frac{1}{2}\mathrm{N}_{2}(g)+\frac{1}{2}\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{NO}(g) & K_{1''} =\frac{1}{\sqrt{1 \times 10^{-30}}} \\\ \text {(2)} \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons \mathrm{NOBr}(g) & K_{2}=8 \times 10^{1} \end{array} $$ Adding both reactions yields: $$ \frac{1}{2}\mathrm{N}_{2}(g)+\frac{1}{2}\mathrm{O}_{2}(g)+\frac{1}{2}\mathrm{Br}_{2}(g) \rightleftharpoons \mathrm{NOBr}(g), \quad K = K_{1''} \times K_2 $$

Calculate the desired equilibrium constant (K) for the formation of NOBr

$$ K = K_{1''} \times K_2 = \frac{1}{\sqrt{1 \times 10^{-30}}} \times (8 \times 10^{1}) = 8 \times 10^{15} $$ So the equilibrium constant for the formation of one mole of NOBr from its elements in the gaseous state is \(K = 8 \times 10^{15}\).