Answers to end-of-chapter questions
Chapter 6
1 a
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
2CuO(s) + 4NO2(g) + O2(g)
Energy
∆Hr
2Cu(NO3)2(s)
b
[1]
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]
[1]
Progress of reaction
copper(II) nitrate on left and products
on right with arrow showing energy
going upwards;
copper(II) nitrate below products;
arrow in upwards direction from copper
nitrate to products with ∆H written near
the arrow
2Cu(NO3)2(s)
∆H1
∆Hf [Cu(NO3)2]
∆Hr
[1]
[1]
[1]
2CuO(s) + 4NO2(g) + O2(g)
∆H2
∆Hf [CuO] +
4 × ∆Hf [NO2]
2Cu(s) + 2N2(g) + 6O2(g)
c ∆Hr + ∆H1 = ∆H2
∆Hr + 2(−302.9) = 2(−157.3) + 4(+33.2)
∆Hr + (−605.8) = −181.8,
so ∆Hr = (+)424 kJ mol−1
d i energy released = 100 × 4.18 × 2.9
= 1212.2 J
1212.2 J for 25 g so for 1 mol
249.7
= 1212.2 ×
25.0
= (−)12 107.5 J / 12.1 kJ to 3 significant
figures
[3]
[1]
[1]
[1]
[1]
[energy needed to evaporate the propanone for 2 marks]
d i Enthalpy change when 1 mol of a
compound
is formed from its constituent elements in
their standard states
under standard conditions.
ii 3C(graphite) + 3H2(g) + 12 O2(g)
→ C3H6O(l) [1]
[1]
[1]
[2]
[1 mark for correct equation; 1 mark
for correct state symbols]
iiiCarbon does not react directly with
hydrogen under standard conditions.
[1]
Total = 14
AS and A Level Chemistry © Cambridge University Press
Answers to end-of-chapter questions: Chapter 6
1
240
= 0.01 mol 24000
b heat change = 100 × 4.18 × 33.5
= 14 003 J = 14.0 kJ (to 3 significant figures)
3 a
[1]
[1]
[1]
‒14.0
[1]
0.01
= −1400 kJ mol−1
[1]
d ΔHc = 2(−394) + 3(−286)
[1]
− (−85)
[1]
= −1561
[1]
kJ mol−1
[1]
e incomplete combustion;
[1]
heat losses through sides of calorimeter, etc
[1]
Total = 11
c ∆Hc =
4 a the energy change when 1 mole
is completely combusted in excess oxygen
under standard conditions
b i
5O2(g) + P4(white)
∆H r
–2984
[1]
[1]
[1]
5O2(g) + P4(red)
–2967
P4O10(s)
for correct cycle
for arrows
for correct values on arrows
using Hess’s Law, ΔHr − 2967 = −2984
ΔHr = −2984 + 2967 = −17 kJ mol−1
ii
P4(white)
[1]
[1]
[1]
[1]
[1]
–17 kJ mol–1
Energy
4(C H)
4 × 412
2(O O)
2 × 496
2(C O)
2 × 805
ΔHc = 1648 + 992 − 1610 − 1852
−1
= −822 kJ mol aq + MgCl2(s) + 6H2O(l)
∆H1
2
Answers to end-of-chapter questions: Chapter 6
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
Total = 11
6 a The average energy needed to break
[1]
1 mole of bonds in the gaseous state.
[1]
b bond enthalpies of H2 + I2 = 436 + 151
= +587 kJ mol−1
[1]
bond enthalpies of 2HI = 2 × 299
= +598 kJ mol−1
[1]
enthalpy change = 587 − 598 = −11 kJ mol−1 [1]
c H2 and I2 on left and 2HI on right and
energy label going upwards
[1]
H2 and I2 below 2HI
[1]
arrow going downwards showing ΔHr [1]
Total = 8
–2984 kJ mol–1
P4O10(s)
P4(red) is below P4(white)
[1]
for arrows from both down to P4O10
[1]
for energy label
[1]
Total = 11
[1]
4(O H) [1]
4 × 463 [1]
7 a enthalpy change when 1 mole of solute
is dissolved in a solvent
to form an infinitely dilute solution
b
∆Hr
P4(red)
–2967 kJ mol–1
5 a enthalpy change when 1 mol of a compound
is formed from its constituent elements in
their standard states
under standard conditions
b C + 2H2 → CH4 is the equation for ∆Hf ∆Hr = sum of ΔHc of reactants − sum
of ΔHc of products
= 2(−286) − 394 − (−891) = −572 − 394 + 891
= −75 kJ mol−1
c
CH4 +
2O2 →
CO2 + 2H2O
[1]
[1]
[1]
MgCl2.6H2O(s) + aq
∆H2
MgCl2(aq)
1 mark for each of the three reactions with
the arrows in the correct order/directions
[3]
for ΔH values in correct places
[1]
Total = 7
AS and A Level Chemistry © Cambridge University Press
8 a enthalpy change when reactants converted
to products
in the amounts shown in the equation
under standard conditions
b
∆Hr
2HCl(aq) + MgCO3(s)
MgO(s) + CO2(g) + 2HCl(aq)
9 a 250 × 4.18 × 23.0
= 24 000 J (to 3 significant figures)
b Mr = 32.0
MgCl2(aq) + CO2(g) + H2O(l)
1 mark each for the three reactions
with the arrows in the correct
order/directions
for ΔH values in correct places
[3]
[1]
Total = 7
AS and A Level Chemistry © Cambridge University Press
[1]
[1]
[1]
2.9
= 0.0906 moles
[1]
32.0
24000
c
= 265 000 J mol−1 or (265 kJ mol−1)
[2]
0.0906
d heat loss
[1]
incomplete combustion
[1]
conditions not standard
[1]
Total = 9
∆H2
∆H1
[1]
[1]
[1]
Answers to end-of-chapter questions: Chapter 6
3