AQA A2 Chemistry C17 Thermodynamics Kerboodle Answers

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C17.1 Enthalpy change AQA A2 Chemistry P17 Thermodynamics Kerboodle Answers: Page No.  6

1 a +788 kJ mol^-1

b This is the reverse or the equation for lattice enthalpy formation, so the sign of 6.H is changed.

c The enthalpy change of lattice dissociation.

2 a The electron is auracted by the sodium nucleus, so energy must be put in to remove it.

b The electron is attracted by the chlorine nucleus so energy is given out during this process.

3 a i Al(g) → Al+(g) + e^–

ii

AJ⁺(g) -+ Al²⁺(g) + e^–

 

17.2 Born-Haber cycles AQA A2 Chemistry P17 Thermodynamics Kerboodle Answers: Page No. 12

1 a

b @

-929 kJ mol^-1

 

17.3 More enthalpy changes AQA A2 Chemistry P17 Thermodynamics Kerboodle Answers: Page No. 14

 

1

+22 kJ mol^-1

Small because the energy put in to break the lattice is of similar size to that given out when the ions arc hydrated.

2

They are small and highly charged positive ions so they strongly polarise negative ions. The calculated value would be greater because there is extra covalent bonding.

 

17.4 Why do chemical reactions take place AQA A2 Chemistry P17 Thermodynamics Kerboodle Answers: Page No. 21

 

1 a i Mg(s) + ZnO(s) -» MgO(s) + Zn(s)

Approximately zero, two solids produce two solids.

ii 2Pb(N03)2(s) -»2PbO(s) + 4N02(g) + 0z(g)

Significantly positive, a solid produces several moles of gases.

iii Mg0(s) + C02(g) -+ MgC03(s)

Significantly negative, a gas turns into a solid.

iv H20(l) — H20(g)

Significantly positive, a liquid turns into a gas.

b

i -7.8 J K^-1 mol^-1

ii +876.4 J K^-1 mol^-1

iii -174.8 J K^-1 mol^-1

iv +I 19 J K^-1 mol^-1

2. a i

+493 kJ mol^-1

ii 6000 K

-52 kJ mol^-1

iii At which temperature is the reaction feasible?

It is feasible at 6000 K

b 5523 K

3

-284 J K^-1 mol^-1

 

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Practice questions: Page No. 22-23

 

1 (a)

(i) (At 0 K) particles are stationary / not moving / not vibrating

• No disorder / perfect order / maximum order

(iiAs T increases, particles start to move / vibrate

• Disorder / randomness increases / order decreases

(iii) Mark on temperature axis vertically below second ‘step’.

(iv)

·         L2 corresponds to boiling / evaporating / condensing / l → g / g → l ·         And L1 corresponds to melting / freezing / ·         s →l / l →s ·         Bigger change in disorder for L2 / boiling ·         compared with L1 / melting

(b)

(i

·         ΔG = ΔH −TΔS ·         ΔH= c and (−)ΔS = m / ΔH and ΔS are constants (approx)

(ii)

Because the entropy change / ΔS is positive / TΔS gets bigger

(iii) Not feasible / unfeasible / not spontaneous

(c)

 

(i) + 44.5 J K^-1 mol^-1

(ii) At 5440 ΔH = TΔS

• = 5440 × 44.5 = 242 080
• (OR using given value = 5440 × 98 = 533 120)
• ΔH = 242 kJ mol−1
• (OR using given value ΔH = 533 kJ mol−1)

2

(a) ΔG = ΔH − TΔS

(b) 0.098 or 98

kJ K−1 mol^-1 J K−1 mol^-1

−ΔS/ΔS

(c) ΔG becomes negative

• So reaction becomes spontaneous/feasible

(d) Ammonia liquefies (so entropy data wrong/different)

3

(a) Enthalpy change when 1 mol of an (ionic) compound/lattice (under standard conditions) Is dissociated/broken/separated into its (component) ions

• The ions being in the gaseous state (at infinite separation)

(b) There is an attractive force between the nucleus of an O atom and an external electron.

(c)

• Mg2+ (g) + O(g) + 2e^–
• Mg2+ (g) + O^– (g) + e
• Mg2+ (g) + O2^– (g)
• First new level for Mg2+ and O above last on L
• Next level for Mg2+ and O^– below that
• Next level for Mg2+ and O2^– above that and also above that for Mg2+ and O

(d) LE MgO = 602 + 150 + 736 + 1450 + 248 − 142 + 844

= +3888 kJ mol^-1

(e) Forms a protective layer/barrier of MgO / MgO prevents oxygen attacking Mg.

(f) ΔG = ΔH – TΔS

ΔS = (−602 − (−570)) × 1000 / 298

= −107 J K^-1 mol^-1 / −0.107 kJ K^-1 mol^-1

(g) 1 mol of solid and 0.5 mol of gas reactants form 1 mol solid products

• System becomes more ordered.

 

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