Nitrogen oxide is in equilibrium with dinitrogen dioxide.

2NO(g) $\rightleftharpoons$ N₂O₂(g)     ΔH^Θ < 0

Deduce, giving a reason, the effect of increasing the temperature on the concentration of N₂O₂.

A two-step mechanism is proposed for the formation of NO₂(g) from NO(g) that involves an exothermic equilibrium process.

First step: 2NO(g) $\rightleftharpoons$ N₂O₂(g)     fast

Second step: N₂O₂(g) + O₂ (g) → 2NO₂(g)     slow

Deduce the rate expression for the mechanism.

The rate constant for a reaction doubles when the temperature is increased from 25.0 °C to 35 °C.

Calculate the activation energy, E_a, in kJ mol⁻¹ for the reaction using section 1 and 2 of the data booklet.

[N₂O₂] decreases AND exothermic «thus reverse reaction favoured»

Accept “product” for [N₂O₂].

Do not accept just “reverse reaction favoured/shift to left” for “[N₂O₂] decreases”.

[1 mark]

ALTERNATIVE 1:

«from equilibrium, step 1»

$K_c=\frac{\text{[}{\text{N}}_2{\text{O}}_2\text{]}}{{\text{[NO]}}^2}$

OR

[N₂O₂] = K_c[NO]²

«from step 2, rate «= k₁[N₂O₂][O₂] = k₂K[NO]²[O₂]»

rate = k[NO]²[O₂]

ALTERNATIVE 2:

«from step 2» rate = k₂[N₂O₂][O₂]

«from step 1, rate₍₁₎ = k₁[NO]² = k_–1[N₂O₂], [N₂O₂] = $\frac{k_1}{k_{-1}}$ [NO]²»

«rate = $\frac{k_1}{k_{-1}}$ k₂[NO]²[O₂]»

rate = k[NO]²[O₂]

Award [2] for correct rate expression.

[2 marks]

«$\ln \frac{k_1}{k_2}=\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$»

T₂ = «273 + 35 =» 308 K AND T₁ = «273 + 25 =» 298 K

E_a = 52.9 «kJ mol^–1»

Award [2] for correct final answer.

[2 marks]