DP Chemistry (last assessment 2024)

Question 21M.2.hl.TZ1.4

Date	May 2021	Marks available	[Maximum mark: 19]	Reference code	21M.2.hl.TZ1.4
Level	hl	Paper	2	Time zone	TZ1
Command term	Calculate, Determine, Draw, Justify, Predict, State, State and explain, Suggest	Question number	4	Adapted from	N/A

[Maximum mark: 19]

21M.2.hl.TZ1.4

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.

(a)

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]

Markscheme

curve higher AND to left of T₁ ✔

new/catalysed E_a marked AND to the left of E_a of curve T₁ ✔

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

(b)

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

[1]

Markscheme

methanoic acid/HCOOH/CHOOH
OR
methanal/HCHO ✔

Accept “carbon dioxide/CO₂”.

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)

(c(i))

Determine the overall equation for the production of methanol.

[1]

Markscheme

CH₄(g) + H₂O(g) $⇌$ CH₃OH(l) + H₂(g) ✔

Accept arrow instead of equilibrium sign.

(c(ii))

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]

Markscheme

amount of methane = « $\frac{8.00 g}{16.05 g {mol}^{- 1}}$ = » 0.498 «mol» ✔

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

volume of hydrogen = «0.498 mol × 22.7 dm³mol⁻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. »

Consider the first stage of the reaction.

CH₄(g) + H₂O (g) $⇌$ CO (g) + 3H₂(g)

(d(i))

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

Bond enthalpy of CO = 1077 kJ mol⁻¹.

[3]

Markscheme

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(ii))

State one reason why you would expect the value of ΔH calculated from the $∆ H_{f}^{⦵}$ values, given in section 12 of data booklet, to differ from your answer to (d)(i).

[1]

Markscheme

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

(d(iii))

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

[1]

Markscheme

$K_{c} = \frac{[CO] {[H_{2}]}^{3}}{[{CH}_{4}] [H_{2} O]}$ ✔

(d(iv))

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

[1]

Markscheme

K_c increases AND «forward» reaction endothermic ✔

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

(e(i))

The equilibrium constant, K_c, 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]

Markscheme

«ΔG^⦵ = − RT lnK_c»
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(ii))

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]

Markscheme

«ΔG^⦵ = ΔH^⦵− TΔS^⦵»

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

ΔS^⦵ = « $\frac{(129 kJ {mol}^{- 1} + 0.0246 kJ {mol}^{- 1}) \times 10^{3}}{298 K}$ = » −433 «J K^–1 mol^–1» ✔

Award [2] for correct final answer.

Award [1 max] for “−0.433 «kJ K^–1mol^–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(iii))

Justify the sign of ΔS with reference to the equation.

[1]

Markscheme

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

(e(iv))

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

[1]

Markscheme

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 K_c increases » ✔

Award mark if correct calculation shown.