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Date May 2018 Marks available 1 Reference code 18M.2.hl.TZ2.6
Level HL Paper 2 Time zone TZ2
Command term Justify and Deduce Question number 6 Adapted from N/A

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)  2SO3(g)

Nitrogen oxide is in equilibrium with dinitrogen dioxide.

2NO(g)  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)  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.

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 = [ N 2 O 2 ] [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] = k 1 k 1  [NO]2»

«rate = k 1 k 1  k2[NO]2[O2]»

rate = k[NO]2[O2]

 

Award [2] for correct rate expression.

[2 marks]

c.ii.

« ln k 1 k 2 = E a R ( 1 T 2 1 T 1 ) »

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.

Examiners report

[N/A]
c.i.
[N/A]
c.ii.
[N/A]
d.

Syllabus sections

Core » Topic 7: Equilibrium » 7.1 Equilibrium
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