Question 21M.2.sl.TZ1.4
| Date | May 2021 | Marks available | [Maximum mark: 12] | Reference code | 21M.2.sl.TZ1.4 |
| Level | sl | Paper | 2 | Time zone | TZ1 |
| Command term | Calculate, Determine, Draw, State, State and explain, Suggest | Question number | 4 | Adapted from | N/A |
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.
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]
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.
The hydrogen peroxide could cause further oxidation of the methanol. Suggest a possible oxidation product.
[1]
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)
Determine the overall equation for the production of methanol.
[1]
CH4(g) + H2O(g) CH3OH(l) + H2(g) ✔
Accept arrow instead of equilibrium sign.
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]
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. »
Consider the first stage of the reaction.
CH4 (g) + H2O (g) CO (g) + 3H2 (g)
Determine the enthalpy change, ΔH, in kJ. Use section 11 of the data booklet.
Bond enthalpy of CO = 1077 kJ mol−1.
[3]
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»
State the expression for Kc for this stage of the reaction.
[1]
✔
State and explain the effect of increasing temperature on the value of Kc.
[1]
Kc increases AND «forward» reaction endothermic ✔