Date | May 2013 | Marks available | 3 | Reference code | 13M.2.hl.TZ1.8 |
Level | HL | Paper | 2 | Time zone | TZ1 |
Command term | Calculate | Question number | 8 | Adapted from | N/A |
Question
To determine the enthalpy change of combustion of methanol, CH3OH, 0.230 g of methanol was combusted in a spirit burner. The heat released increased the temperature of 50.0 cm3 of water from 24.5 °C to 45.8 °C.
Methanol can be produced according to the following equation.
CO(g)+2H2(g)→CH3OH(l)
The manufacture of gaseous methanol from CO and H2 involves an equilibrium reaction.
CO(g)+2H2(g)⇌CH3OH(g) ΔHΘ<0
Calculate the standard enthalpy change of this reaction, using the values of enthalpy of combustion in Table 12 of the Data Booklet.
Calculate the standard entropy change for this reaction, ΔSΘ, using Table 11 of the Data Booklet and given:
SΘ(CO)=198 JK−1mol−1 and SΘ(H2)=131 JK−1mol−1.
Calculate, stating units, the standard free energy change for this reaction, ΔGΘ, at 298 K.
Predict, with a reason, the effect of an increase in temperature on the spontaneity of this reaction.
1.00 mol of CH3OH is placed in a closed container of volume 1.00 dm3 until equilibrium is reached with CO and H2. At equilibrium 0.492 mol of CH3OH are present. Calculate Kc.
Markscheme
CH3OH+32O2→CO2+2H2O ΔHΘc=−726 (kJmol−1)
CO+12O2→CO2 ΔHΘc=−283 (kJmol−1)
H2+12O2→H2O ΔHΘc=−286 (kJmol−1)
Award [1 max] for three correct values.
Mark can be implicit in calculations.
(ΔHΘR=) 2(−286)+(−283)−(−726);
−129 (kJmol−1);
Award [3] for correct final answer.
Award [2 max] for +129 (kJmol–1).
(ΔSΘ=240−198−2×131=) −220 (JK−1mol−1);
(−129−298(−0.220)=) −63.4 kJmol−1;
Award [1] for correct numerical answer and [1] for correct unit if the conversion has been made from J to kJ for ΔSΘ.
not spontaneous at high temperature;
TΔSΘ<ΔHΘ and ΔGΘ positive;
n(CO)=0.508 (mol);
n(H2)=2×0.508 (mol);
Kc (=0.4920.508×(2×0.508)2)=0.938;
Accept answer in range between 0.930 and 0.940.
Award [3] for correct final answer.
Award [2] for Kc = 1.066 if (c)(ii) is correct.
Examiners report
In (i), the most common error was +129 kJmol−1 but in (ii) the answer was often correct.
In (i), the most common error was +129 kJmol−1 but in (ii) the answer was often correct.
Units tended to get muddled in (iii) and many marks were awarded as “error carried forward”.
Few were able to explain the ΔH and TΔS relationship in detail in (iv).
Equilibrium was well understood in general with many candidates gaining one of the two available marks. “Equal rates” was more often given than the constancy of macroscopic properties for the second mark. The Kc expression was given correctly by the vast majority of candidates (including the correct brackets and indices) but many had difficulty with the equilibrium concentrations in (iii).
The changes in equilibrium position were well understood for the most part although if a mark were to be lost it was for not mentioning the number of moles of gas.