Answers to end-of-chapter questions

1Answers to end-of-chapter questions: Chapter 6AS and A Level Chemistry © Cambridge University Press

Chapter 61 a

2CuO(s) + 4NO2(g) + O2(g)

2Cu(NO3)2(s)

Ener

gy

Progress of reaction

∆Hr

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]

b

2Cu(NO3)2(s) 2CuO(s) + 4NO2(g) + O2(g)

2Cu(s) + 2N2(g) + 6O2(g)

∆Hf [Cu(NO3)2]∆Hf [CuO] +4 × ∆Hf [NO2]

∆Hr

∆H1 ∆H2

[3]

c ∆Hr + ∆H1 = ∆H2 [1]∆Hr + 2(−302.9) = 2(−157.3) + 4(+33.2) [1]∆Hr + (−605.8) = −181.8, so ∆Hr = (+)424 kJ mol−1 [1]

d i energy released = 100 × 4.18 × 2.9 = 1212.2 J [1]

1212.2 J for 25 g so for 1 mol

= 1212.2 × 249.725.0

[1]

= (−)12 107.5 J / 12.1 kJ to 3 significant figures [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]

Total = 14

2 Answers to end-of-chapter questions: Chapter 6 AS and A Level Chemistry © Cambridge University Press

3 a 24024000

= 0.01 mol [1]

b heat change = 100 × 4.18 × 33.5 [1]= 14 003 J = 14.0 kJ (to 3 significant figures) [1]

c ∆Hc = 0.01‒14.0 [1]

= −1400 kJ mol−1 [1] d ∆Hc = 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

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]

c CH4 + 2O2 → CO2 + 2H2O

4(C H) 2(O O) 2(C O) 4(O H) [1]4 × 412 2 × 496 2 × 805 4 × 463 [1]

∆Hc = 1648 + 992 − 1610 − 1852 [1]

= −822 kJ mol−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 ∆Hr [1]

Total = 8

7 a enthalpy change when 1 mole of solute [1]is dissolved in a solvent [1]to form an infinitely dilute solution [1]

b

∆H1 ∆H2

∆Hraq + MgCl2(s) + 6H2O(l) MgCl2.6H2O(s) + aq

MgCl2(aq)

1 mark for each of the three reactions with the arrows in the correct order/directions [3]for ∆H values in correct places [1]

Total = 7

3Answers to end-of-chapter questions: Chapter 6AS and A Level Chemistry © Cambridge University Press

8 a enthalpy change when reactants converted to products [1]in the amounts shown in the equation [1]under standard conditions [1]

b

∆H1

∆Hr

∆H2

MgO(s) + CO2(g) + 2HCl(aq)2HCl(aq) + MgCO3(s)

MgCl2(aq) + CO2(g) + H2O(l)

1 mark each for the three reactions with the arrows in the correct order/directions [3]for ∆H values in correct places [1]

Total = 7

9 a 250 × 4.18 × 23.0 [1]= 24 000 J (to 3 significant figures) [1]

b Mr = 32.0 [1]

2 932 0

..

= 0.0906 moles [1]

c 240000 0906.

= 265 000 J mol−1 or (265 kJ mol−1) [2]

d heat loss [1]incomplete combustion [1]conditions not standard [1]

Total = 9