Answers to end-of-chapter questions - Mr. Neibergermrneiberger.weebly.com/uploads/2/4/9/0/24909870/ans_eoc_06.pdf · 2 Answers to end-of-chapter questions: Chapter 6 AS and A Level
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copper(II) nitrate on left and products on right with arrow showing energy going upwards; [1]copper(II) nitrate below products; [1]arrow in upwards direction from copper nitrate to products with ∆H written near the arrow [1]
ii taking time for copper sulfate to dissolve / energy loss to thermometer or air or calorimeter [1]
so temperature recorded lower than expected / energy loss to surroundings and therefore energy released is less [1]
or assumption that the specific thermal capacity
of the solution is the same as that of water [1] the thermal capacity is likely to be slightly
higher so the value calculated for the energy released is too low [1]
Total = 14
2 a CH3COCH3(l) + 4O2(g) → 3CO2(g) + 3H2O(l)2(C C) + 6(C H) + (C O) + 4(O O) → 6(C O) + 6(O H) [1]2(347) + 6(413) + (805) + 4(496) → 6(805) + 6(465) [1]+5961 for bond breaking; −7620 for bond making; realisation that bond breaking is + and bond making is − [1]answer = −1659 kJ [1]
b any two of:the same type of bonds are in different environments;example e.g. C O bonds in carbon dioxide and propanone;average bond energies are generalised / obtained from a number of different bonds of the same type [2]
c bond energies calculated by using enthalpy changes of gaseous compound to gaseous atoms; [1]enthalpy changes of combustion done experimentally using liquid (propanone). [1]
[energy needed to evaporate the propanone for 2 marks] d i Enthalpy change when 1 mol of a
compound [1] is formed from its constituent elements in
their standard states [1] under standard conditions. [1]
ii 3C(graphite) + 3H2(g) + 12 O2(g) → C3H6O(l) [2]
[1 mark for correct equation; 1 mark for correct state symbols]
iii Carbon does not react directly with hydrogen under standard conditions. [1]
e incomplete combustion; [1]heat losses through sides of calorimeter, etc [1]
Total = 11
4 a the energy change when 1 mole [1]is completely combusted in excess oxygen [1]under standard conditions [1]
b i
5O2(g) + P4(white) 5O2(g) + P4(red)
–2984 –2967
P4O10(s)
∆H r
for correct cycle [1] for arrows [1] for correct values on arrows [1] using Hess’s Law, ∆Hr − 2967 = −2984 [1] ∆Hr = −2984 + 2967 = −17 kJ mol−1 [1] ii
P4(red)
Ener
gy
–17 kJ mol–1 P4(white)
P4O10(s)
–2967 kJ mol–1 –2984 kJ mol–1
P4(red) is below P4(white) [1] for arrows from both down to P4O10 [1] for energy label [1] Total = 11
5 a enthalpy change when 1 mol of a compound [1]is formed from its constituent elements in their standard states [1]under standard conditions [1]
b C + 2H2 → CH4 is the equation for ∆Hf [1]∆Hr = sum of ∆Hc of reactants − sum of ∆Hc of products [1]= 2(−286) − 394 − (−891) = −572 − 394 + 891 [1]= −75 kJ mol−1 [1]