Answers to end-of-chapter questions

Answers to end-of-chapter questions

Chapter 6

1 a

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

Energy

Hr 2Cu(NO3)2(s)

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)

Progress of reaction

2(C C) + 6(C H) + (C O) + 4(O O)

6(C O) + 6(O H) [1]

copper(II) nitrate on left and products

2(347) + 6(413) + (805) + 4(496)

on right with arrow showing energy

6(805) + 6(465)

[1]

going upwards;

[1]

+5961 for bond breaking; -7620 for bond

copper(II) nitrate below products;

[1]

making; realisation that bond breaking is

arrow in upwards direction from copper

+ and bond making is -

[1]

nitrate to products with H written near

answer = -1659kJ

[1]

the arrow

[1]

b any two of: the same type of bonds are in different

b

2Cu(NO3)2(s) Hr

H1

Hf [Cu(NO3)2]

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

H2 Hf [CuO] + 4 ? Hf [NO2]

environments; example e.g. C O bonds in carbon dioxide and propanone;

average bond energies are generalised /

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

[3]

c Hr + H1 = H2

[1]

Hr + 2(-302.9) = 2(-157.3) + 4(+33.2)

[1]

Hr + (-605.8) = -181.8,

so Hr = (+)424kJmol-1

[1]

d i energy released = 100 ? 4.18 ? 2.9

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]

= 1212.2J 1212.2J for 25g so for 1 mol

= 1212.2 ? 249.7 25.0

= (-)12107.5J / 12.1kJ to 3 significant figures

[1]

d i Enthalpy change when 1 mol of a

compound

[1]

is formed from its constituent elements in

[1]

their standard states

[1]

under standard conditions.

[1]

ii 3C(graphite) + 3H2(g) + 12O2(g)

[1]

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

AS and A Level Chemistry ? Cambridge University Press

Answers to end-of-chapter questions: Chapter 6

1

3 a 240 = 0.01mol

[1]

24 000

b heat change = 100 ? 4.18 ? 33.5

[1]

= 14003 J = 14.0kJ (to 3 significant figures) [1]

c

Hc

=

14.0 0.01

[1]

= -1400kJmol-1

[1]

d Hc = 2(-394) + 3(-286)

[1]

- (-85)

[1]

= -1561

[1]

kJmol-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) H r 5O2(g) + P4(red)

?2984

?2967

P4O10(s)

for correct cycle

[1]

for arrows

[1]

for correct values on arrows

[1]

using Hess's Law, Hr - 2967 = -2984 [1]

Hr = -2984 + 2967 = -17kJmol-1

[1]

ii

P4(white)

?17 kJ mol?1

P4(red)

Energy

?2967 kJ mol?1

?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

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]

= -75kJmol-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]

= -822kJmol-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 = +587kJmol-1

of

H2

+

I2

=

436

+

151

[1]

bond enthalpies of 2HI = 2 ? 299

= +598kJmol-1

[1]

enthalpy change = 587 - 598 = -11kJmol-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 aq + MgCl2(s) + 6H2O(l) Hr MgCl2.6H2O(s) + aq

H1

H2

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

2

Answers to end-of-chapter questions: Chapter 6

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

H1

H2

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

1 mark each for the three reactions

with the arrows in the correct

order/directions

for H values in correct places

[3] [1] Total = 7

9 a 250 ? 4.18 ? 23.0

[1]

= 24000J (to 3 significant figures)

[1]

b Mr = 32.0

[1]

2.9 = 0.0906 moles

[1]

32.0

c 24000 = 265000Jmol-1 or (265kJmol-1) [2] 0.0906

d heat loss incomplete combustion conditions not standard

[1] [1] [1] Total = 9

AS and A Level Chemistry ? Cambridge University Press

Answers to end-of-chapter questions: Chapter 6

3

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