Date | May 2021 | Marks available | 2 | Reference code | 21M.2.hl.TZ1.4 |
Level | HL | Paper | 2 | Time zone | TZ1 |
Command term | Calculate | 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.
The hydrogen peroxide could cause further oxidation of the methanol. Suggest a possible oxidation product.
Determine the overall equation for the production of methanol.
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.
Determine the enthalpy change, ΔH, in kJ. Use section 11 of the data booklet.
Bond enthalpy of CO = 1077 kJ mol−1.
State one reason why you would expect the value of ΔH calculated from the values, given in section 12 of data booklet, to differ from your answer to (d)(i).
State the expression for Kc for this stage of the reaction.
State and explain the effect of increasing temperature on the value of Kc.
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.
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.
Justify the sign of ΔS with reference to the equation.
Predict, giving a reason, how a change in temperature from 298 K to 273 K would affect the spontaneity of the reaction.
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.
methanoic acid/HCOOH/CHOOH
OR
methanal/HCHO ✔
Accept “carbon dioxide/CO2”.
CH4(g) + H2O(g) CH3OH(l) + H2(g) ✔
Accept arrow instead of equilibrium sign.
amount of methane = « = » 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. »
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»
bond energies are average values «not specific to the compound» ✔
✔
Kc increases AND «forward» reaction endothermic ✔
«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».
«ΔG⦵ = ΔH⦵ − TΔS⦵»
ΔG⦵ = −129 «kJ mol–1» − (298 «K» × ΔS) = −0.0246 «kJ mol–1» ✔
ΔS⦵ = « = » −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).
«negative as» product is liquid and reactants gases
OR
fewer moles of gas in product ✔
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.