Home / IB DP Chemistry 16.2 Activation energy HL Paper 2

IB DP Chemistry 16.2 Activation energy HL Paper 2

Question

A mixture of 1.00 mol SO2(g), 2.00 mol O2(g) and 1.00 mol SO3(g) is placed in a 1.00 dm3 container and allowed to reach equilibrium.

2SO2(g) + O2(g) \( \rightleftharpoons \) 2SO3(g)

Nitrogen oxide is in equilibrium with dinitrogen dioxide.

2NO(g) \( \rightleftharpoons \) N2O2(g)     ΔHΘ < 0

Deduce, giving a reason, the effect of increasing the temperature on the concentration of N2O2.[1]

c.i.

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.[2]

c.ii.

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

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

d.
Answer/Explanation

Markscheme

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

Accept “product” for [N2O2].

Do not accept just “reverse reaction favoured/shift to left” for “[N2O2decreases”.

[1 mark]

c.i.

ALTERNATIVE 1:

«from equilibrium, step 1»

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

OR

[N2O2] = Kc[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 = k1[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]

c.ii.

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

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

Ea = 52.9 «kJ mol–1»

Award [2] for correct final answer.

[2 marks]

d.
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