Q23.
Increase in volume
If a volume is quoted it must be less than 300
1
Smaller increase in T above room temperature
Or increased contact between calorimeter and water
Or smaller heat loss by evaporation / from the surface
1
[2]
Q24.
(a)
Penalise C2H6O for ethanol in M1.
M2 and M3
Mark M2 and M3 independently.
Any two conditions in any order for M2 and M3 from
- (enzymes from) yeast or zymase
- 25 °C ≤ T ≤ 42 °C OR 298 K ≤ T ≤ 315 K
- anaerobic / no oxygen / no air OR neutral pH
A lack of oxygen can mean either without oxygen or not having enough oxygen and does not ensure no oxygen, therefore only credit “lack of oxygen” if it is qualified.
Penalise ‘bacteria’, ‘phosphoric acid’, ‘high pressure’ using the list principle.
M4 (fractional) distillation or GLC
Ignore reference to ‘aqueous’ or ‘water’ (ie not part of the list principle).
M5 Carbon-neutral in this context means
There is no net / overall (annual) carbon dioxide / CO2 emission to the atmosphere
OR
There is no change in the total amount / level of carbon dioxide / CO2 present in the atmosphere
For M5 – must be about CO2 and the atmosphere.
The idea that the carbon dioxide / CO2 given out equals the carbon dioxide / CO2 that was taken in from the atmosphere.
5
(b) M1 q = m c ∆T (this mark for correct mathematical formula)
Full marks for M1, M2 and M3 for the correct answer.
In M1, do not penalise incorrect cases in the formula.
M2 = (75 × 4.18 × 5.5)
1724 (J) OR 1.724 (kJ) OR 1.72 (kJ) OR 1.7 (kJ)
(also scores M1)
Ignore incorrect units in M2.
M3 Using 0.0024 mol
therefore ∆H = − 718 (kJ mol−1)
(Accept a range from −708 to −719 but do not penalise more than 3 significant figures)
Penalise M3 ONLY if correct numerical answer but sign is incorrect. Therefore +718 gains two marks.
If units are quoted in M3 they must be correct.
If ∆T = 278.5, CE for the calculation and penalise M2 and M3.
M4 and M5 in any order
Any two from
- incomplete combustion
- heat loss
- heat capacity of Cu not included
- some ethanol lost by evaporation
- not all of the (2.40 × 10−3 mol) ethanol is burned / reaction is incomplete
If c = 4.81 (leads to 1984) penalise M2 ONLY and mark on for M3 = − 827
5
(c) (i) M1 enthalpy / heat / energy change (at constant pressure) or enthalpy / heat / energy needed in breaking / dissociating (a) covalent bond(s)
Ignore bond making.
M2 averaged for that type of bond over different / a range of molecules / compounds
Ignore reference to moles.
2
(ii) M1
∑ B(reactants) − ∑ B(products) = ∆H
OR
Sum of bonds broken − Sum of bonds formed = ∆H
OR
B(C-C) + B(C-O) + B(O-H) + 5B(C-H) + 3B(O=O)
– 4B(C=O) – 6B(O–H) = ∆H = −1279
Correct answer gains full marks.
Credit 1 mark for − 496 (kJ mol−1)
For other incorrect or incomplete answers, proceed as follows
check for an arithmetic error (AE), which is either a transposition error or an incorrect multiplication; this would score 2 marks (M1 and M2).
If no AE, check for a correct method; this requires either a correct cycle with 2CO2 and 3H2O OR a clear statement of M1 which could be in words and scores only M1.
M2 (also scores M1)
348+360+463+5(412)+ 3B(O=O)
(3231) (or 2768 if O–H cancelled)
− 4(805) − 6(463) = ∆H = − 1279
(5998) (or 5535 if O–H cancelled)
3B(O=O) = 1488 (kJ mol−1)
Credit a maximum of one mark if the only scoring point is bonds formed adds up to 5998 (or 5535) OR bonds broken includes the calculated value of 3231 (or 2768).
M3
B(O=O) = 496 (kJ mol−1)
Award 1 mark for −496
Students may use a cycle and gain full marks
3
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