AdvancedChemistryHigh School (11–12)Equilibrium › Chemical EquilibriumNumericalJEE Advanced

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

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

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

Have you tried it?

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 $x$ .

← All Problems

← Sulfuryl chloride vapour breaks down on heating: SO₂Cl₂(g) ⇌ SO₂(g) + Cl₂(g)…Next →

More onEquilibrium › Chemical Equilibrium

Discussion (0)

Loading…