Chemistry 9701 · AS & A Level · Chemical energetics

Chemical energetics — practice question

Nitrogen monoxide, NO, is produced in a reversible reaction when air is heated to the temperature inside a car engine.
(a(i))

Suggest a ‘dot-and-cross’ electronic arrangement for nitrogen monoxide.

(a(ii))

The enthalpy change of formation of nitrogen monoxide is $+90\ \text{kJ mol}^{-1}$. What is the enthalpy change for the following reaction? $2\text{NO}(g) \rightarrow \text{N}_2(g) + \text{O}_2(g)$

(a(iii))

Explain why nitrogen monoxide is produced in the car engine.

(a(iv))[5]

Using bond enthalpy values from the Data Booklet together with your answer in (ii) above, calculate the bond energy of nitrogen monoxide.

(b(i))

At $800\ \text{K}$, nitrogen monoxide reacts with hydrogen according to the equation: $2\text{H}_2(g) + 2\text{NO}(g) \rightarrow 2\text{H}_2\text{O}(g) + \text{N}_2(g)$ The table below shows how the initial rate of this reaction varies with the partial pressures of the reactants. Find the order of the reaction with respect to each reactant, explaining how you obtained your answer.

(b(ii))[2]

Write down the rate equation and the units of the rate constant.

(b(iii))[1]

The following mechanism has been suggested for this reaction. step 1: $\text{NO} + \text{NO} \rightarrow \text{N}_2\text{O} + \text{O}$ step 2: $\text{H}_2 + \text{O} \rightarrow \text{H}_2\text{O}$ step 3: $\text{H}_2 + \text{N}_2\text{O} \rightarrow \text{N}_2 + \text{H}_2\text{O}$ Show how the overall stoichiometric equation can be obtained from the three separate equations for the steps given above.

(b(iv))[2]

Suggest which one of the three reactions in the mechanism is the rate determining step. Explain your answer.

(c(i))[1]

The following half-reaction data relate to the reaction between $\text{HNO}_3$ and an excess of $\text{FeSO}_4$. $\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \quad E^\circ = +0.77\ \text{V}$ $3\text{H}^+ + \text{NO}_3^- + 2e^- \rightarrow \text{HNO}_2 + \text{H}_2\text{O} \quad E^\circ = +0.94\ \text{V}$ $\text{HNO}_2 + \text{H}^+ + e^- \rightarrow \text{NO} + \text{H}_2\text{O} \quad E^\circ = +0.99\ \text{V}$ Suggest the formula of the nitrogen-containing final product of this reaction.

(c(ii))[1]

Write an equation for how this nitrogen-containing product forms.

(c(iii))[1]

Nitrogen monoxide forms a dark brown complex with an excess of $\text{FeSO}_4(aq)$. What type of bonding is involved in forming the complex?

(c(iv))[1]

Suggest a formula for the complex.

Worked solution & mark scheme

This 14-mark question has a full step-by-step worked solution and mark scheme. One marking point: Accurate dot-and-cross diagram for N$_2$O showing the shared electrons

  • Full mark scheme, point by point
  • Step-by-step worked solution
  • Write your answer & get it marked instantly by AI