Chemistry 9701 · AS & A Level · Nitrogen and sulfur

Nitrogen and sulfur — practice question

Ammonium nitrate fertiliser is obtained from ammonia. In the first stage of its manufacture, ammonia is catalytically oxidised to produce nitrogen monoxide, NO. This takes place at about $1 \times 10^3\,\text{kPa}$ (10 atmospheres) pressure and at $700$ to $850\,^{\circ}\text{C}$. The reaction equation is $4\text{NH}_3(g) + 5\text{O}_2(g) \rightleftharpoons 4\text{NO}(g) + 6\text{H}_2\text{O}(g)$ with $\Delta H^{\circ} = -906\,\text{kJ mol}^{-1}$.
(a)[2]

Write the expression for the equilibrium constant, $K_p$, and state the units.

(b(i))[2]

How will increasing the temperature affect the yield of NO? In each case, explain your answer.

(b(ii))[2]

How will decreasing the applied pressure affect the yield of NO? In each case, explain your answer.

(c)[4]

The standard enthalpy changes of formation of $\text{NH}_3(g)$ and $\text{H}_2\text{O}(g)$ are given below: $\Delta H_f^{\circ}(\text{NH}_3(g)) = -46.0\,\text{kJ mol}^{-1}$ and $\Delta H_f^{\circ}(\text{H}_2\text{O}(g)) = -242\,\text{kJ mol}^{-1}$. Use these data together with the value of $\Delta H^{\circ}_{\text{reaction}}$ below to calculate the standard enthalpy change of formation of $\text{NO}(g)$. Include a sign in your answer. The reaction is $4\text{NH}_3(g) + 5\text{O}_2(g) \rightleftharpoons 4\text{NO}(g) + 6\text{H}_2\text{O}(g)$ with $\Delta H^{\circ} = -906\,\text{kJ mol}^{-1}$.

Worked solution & mark scheme

This 10-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Equilibrium constant form: $K_p = \frac{p(\text{NO})^4 p(\text{H}_2\text{O})^6}{p(\text{NH}_3)^4 p(\text{O}_2)^5}$.” …

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