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In a hot engine, nitrogen and oxygen from the air form nitric oxide: N₂(g) + O₂(g) ⇌ 2 NO(g)…

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In a hot engine, nitrogen and oxygen from the air form nitric oxide: $\text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2\,\text{NO}(g)$ . Data at $298\ \text{K}$ : $\Delta_f H^\circ(\text{NO}, g) = +90.25\ \text{kJ mol}^{-1}$ ; standard molar entropies $\text{N}_2(g)$ $191.6$ , $\text{O}_2(g)$ $205.1$ and $\text{NO}(g)$ $210.8\ \text{J K}^{-1}\ \text{mol}^{-1}$ . Take $\Delta H^\circ$ and $\Delta S^\circ$ as constant with temperature, $R = 8.314\ \text{J K}^{-1}\ \text{mol}^{-1}$ , and air as $79\%$ nitrogen and $21\%$ oxygen by moles.

What is $\Delta G^\circ$ of the reaction at $2200\ \text{K}$ ?
What is the equilibrium constant $K$ at $2200\ \text{K}$ ?
What is the mole percent of NO when air comes to equilibrium at $2200\ \text{K}$ ?
At what temperature does $K$ reach $1.0 \times 10^{-2}$ ?
What is the mole percent of NO when air comes to equilibrium at that temperature?

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Get $\Delta H^\circ$ and $\Delta S^\circ$ , then $\Delta G^\circ$ and $K$ at each temperature, then the composition.

Data. $\Delta H^\circ = 2 \times 90.25 = 180.5\ \text{kJ mol}^{-1}$ and $\Delta S^\circ = 2(210.8) - 191.6 - 205.1 = 24.9\ \text{J K}^{-1}\ \text{mol}^{-1}$ .

(a) Gibbs energy. $\Delta G^\circ = 180.5 - 2200 \times 24.9 \times 10^{-3} = 180.5 - 54.78 = 125.7\ \text{kJ mol}^{-1}$ .

(b) $K$ . $\ln K = -\dfrac{125.7 \times 1000}{8.314 \times 2200} = -6.873$ , so $K = 1.03 \times 10^{-3}$ .

(c) Composition. For $100\ \text{mol}$ of air, $\dfrac{(2x)^2}{(79 - x)(21 - x)} = 1.03 \times 10^{-3}$ . Solving the quadratic gives $x = 0.642\ \text{mol}$ , so there are $1.28\ \text{mol}$ of NO in $100\ \text{mol}$ of gas: $1.28\%$ . (The quick estimate $4x^2 \approx K \times 1659$ is close, because $x$ is small.)

(d) The temperature. $\Delta H^\circ - T\Delta S^\circ = -RT\ln K$ gives

\displaystyle T = \frac{\Delta H^\circ}{\Delta S^\circ - R\ln K} = \frac{180.5 \times 1000}{24.9 + 38.29}\ \text{K} = 2857\ \text{K}

(e) Composition there. $\dfrac{(2x)^2}{(79 - x)(21 - x)} = 1.0 \times 10^{-2}$ gives $x = 1.918\ \text{mol}$ , so NO is $3.84\%$ . The reaction is endothermic, so more NO forms in a hotter engine.

Check. $K$ at $298\ \text{K}$ carried to $2200\ \text{K}$ with the van 't Hoff equation, $\ln\dfrac{K_2}{K_1} = -\dfrac{\Delta H^\circ}{R}\left(\dfrac{1}{T_2} - \dfrac{1}{T_1}\right)$ , gives the same $K$ , and a search for the root of the equation in (c) gives the same $x$ .

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