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Date May 2021 Marks available 3 Reference code 21M.2.sl.TZ1.4
Level SL Paper 2 Time zone TZ1
Command term Determine Question number 4 Adapted from N/A

Question

Hydrogen peroxide can react with methane and oxygen to form methanol. This reaction can occur below 50°C if a gold nanoparticle catalyst is used.

Methanol is usually manufactured from methane in a two-stage process.

CH4 (g) + H2O (g) CO (g) + 3H2 (g)
CO (g) + 2H2 (g) CH3OH (l)

Consider the first stage of the reaction.

CH4 (g) + H2O (g) CO (g) + 3H2 (g)

The diagram shows the Maxwell-Boltzmann curve for the uncatalyzed reaction.

Draw a distribution curve at a lower temperature (T2) and show on the diagram how the addition of a catalyst enables the reaction to take place more rapidly than at T1.

[2]
a.

The hydrogen peroxide could cause further oxidation of the methanol. Suggest a possible oxidation product.

[1]
b.

Determine the overall equation for the production of methanol.

[1]
c(i).

8.00 g of methane is completely converted to methanol. Calculate, to three significant figures, the final volume of hydrogen at STP, in dm3. Use sections 2 and 6 of the data booklet.

[3]
c(ii).

Determine the enthalpy change, ΔH, in kJ. Use section 11 of the data booklet.

Bond enthalpy of CO = 1077 kJ mol−1.

[3]
d(i).

State the expression for Kc for this stage of the reaction.

[1]
d(ii).

State and explain the effect of increasing temperature on the value of Kc.

[1]
d(iii).

Markscheme

curve higher AND to left of T1

new/catalysed Ea marked AND to the left of Ea of curve T1

Do not penalize curve missing a label, not passing exactly through the origin, or crossing x-axis after Ea.
Do not award M1 if curve drawn shows significantly more/less molecules/greater/smaller area under curve than curve 1.
Accept Ea drawn to T1 instead of curve drawn as long as to left of marked Ea.

a.

methanoic acid/HCOOH/CHOOH
OR
methanal/HCHO ✔

Accept “carbon dioxide/CO2”.

b.

CH4(g) + H2O(g) CH3OH(l) + H2(g) ✔

Accept arrow instead of equilibrium sign.

c(i).

amount of methane = « 8.00g16.05gmol-1 = » 0.498 «mol» ✔

amount of hydrogen = amount of methane / 0.498 «mol» ✔

volume of hydrogen = «0.498 mol × 22.7 dm3 mol−1 = » 11.3 «dm3» ✔


Award [3] for final correct answer.

Award [2 max] for 11.4 «dm3 due to rounding of mass to 16/moles to 0.5. »

c(ii).

Σbonds broken = 4 × 414 «kJ» + 2 × 463 «kJ» / 2582 «kJ» ✔

Σbonds formed = 1077 «kJ» + 3 × 436 «kJ» / 2385 «kJ» ✔

ΔH «= Σbonds broken − Σbonds formed =( 2582 kJ − 2385 kJ)» = «+»197«kJ» ✔


Award [3] for final correct answer.

Award [2 Max] for final answer of −197 «kJ»

d(i).

Kc=COH23CH4H2O ✔

d(ii).

Kc increases AND «forward» reaction endothermic ✔

d(iii).

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c(i).
[N/A]
c(ii).
[N/A]
d(i).
[N/A]
d(ii).
[N/A]
d(iii).

Syllabus sections

Core » Topic 5: Energetics/thermochemistry » 5.3 Bond enthalpies
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