User interface language: English | Español

Date May 2021 Marks available 1 Reference code 21M.2.hl.TZ1.4
Level HL 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.

Now consider the second stage of the reaction.

CO (g) + 2H2 (g) CH3OH (l)          ΔH = –129 kJ

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 one reason why you would expect the value of ΔH calculated from the Hf values, given in section 12 of data booklet, to differ from your answer to (d)(i).

[1]
d(ii).

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

[1]
d(iii).

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

[1]
d(iv).

The equilibrium constant, Kc, has a value of 1.01 at 298 K.

Calculate ΔG, in kJ mol–1, for this reaction. Use sections 1 and 2 of the data booklet.

[2]
e(i).

Calculate a value for the entropy change, ΔS, in J K–1 mol–1 at 298 K. Use your answers to (e)(i) and section 1 of the data booklet.

If you did not get answers to (e)(i) use –1 kJ, but this is not the correct answer.

[2]
e(ii).

Justify the sign of ΔS with reference to the equation.

[1]
e(iii).

Predict, giving a reason, how a change in temperature from 298 K to 273 K would affect the spontaneity of the reaction.

[1]
e(iv).

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).

bond energies are average values «not specific to the compound» ✔

d(ii).

Kc=COH23CH4H2O ✔

d(iii).

Kc increases AND «forward» reaction endothermic ✔

d(iv).

«ΔG = − RT lnKc»
ΔG = − 8.31 «J K−1 mol−1» × 298 «K» × ln (1.01) / −24.6 «J mol−1» ✔

= −0.0246 «kJ mol–1» ✔


Award [2] for correct final answer.

Award [1 max] for +0.0246 «kJ mol–1».

e(i).

«ΔG = ΔH⦵ TΔS»

ΔG = −129 «kJ mol–1» − (298 «K» × ΔS) = −0.0246 «kJ mol–1» ✔

ΔS = « 129kJmol-1+0.0246kJmol-1×103298K = » −433 «J K–1 mol–1» ✔


Award [2] for correct final answer.

Award [1 max] for “−0.433 «kJ K–1 mol–1»”.

Award [1 max] for “433” or “+433” «J K–1 mol–1».

Award [2] for −430 «J K–1 mol–1» (result from given values).

e(ii).

«negative as» product is liquid and reactants gases
OR
fewer moles of gas in product ✔

e(iii).

reaction «more» spontaneous/ΔG negative/less positive AND effect of negative entropy decreases/TΔS increases/is less negative/more positive
OR
reaction «more» spontaneous/ΔG negative/less positive AND reaction exothermic «so Kc increases » ✔

Award mark if correct calculation shown.

e(iv).

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).
[N/A]
d(iv).
[N/A]
e(i).
[N/A]
e(ii).
[N/A]
e(iii).
[N/A]
e(iv).

Syllabus sections

Core » Topic 1: Stoichiometric relationships » 1.1 Introduction to the particulate nature of matter and chemical change
Show 65 related questions
Core » Topic 1: Stoichiometric relationships
Core

View options