Download - (3) - KS5Chemistry...... Why is the standard entropy of ammonia more positive than the standard entropy of barium chloride? ..... ..... Calculate the standard entropy change of the

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Page 1: (3) - KS5Chemistry...... Why is the standard entropy of ammonia more positive than the standard entropy of barium chloride? ..... ..... Calculate the standard entropy change of the

Maltby Academy 1

1. (a) (i) Define the term standard enthalpy of combustion.

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(ii) The values for the standard enthalpy of combustion of graphite and carbon

monoxide are given below:

Hc /kJ mol–1

C (graphite) –394

CO(g) –283

Use these data to find the standard enthalpy change of formation of carbon

monoxide using a Hess‟s law cycle.

C(graphite) + 2

1O2(g) CO(g)

(3)

(iii) Suggest why it is not possible to find the enthalpy of formation of carbon

monoxide directly.

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Maltby Academy 2

(iv) Draw an enthalpy level diagram below for the formation of carbon monoxide from

graphite.

(1)

(b) Natural gas consists of methane, CH4. When methane burns completely in oxygen the

reaction occurs as shown in the equation

CH4(g) + 2O2(g) CO2(g) + 2H2O(l) Hc = –890 kJ mol–1

Methane does not burn unless lit.

Use this information to explain the difference between thermodynamic and kinetic

stability.

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(Total 12 marks)

2. Thermochemical data, at 298 K, for the equilibrium between zinc carbonate, zinc oxide and

carbon dioxide is shown below.

ZnCO3(s) ZnO(s) + CO2(g) ∆Hο = +71.0 kJ mol

–1

Sο[ZnO(s)] = +43.6 J mol

–1 K

–1

Sο[ZnCO3(s)] = +82.4 J mol

–1 K

–1

Sο[CO2(g)] = +213.6 J mol

–1 K

–1

Page 3: (3) - KS5Chemistry...... Why is the standard entropy of ammonia more positive than the standard entropy of barium chloride? ..... ..... Calculate the standard entropy change of the

Maltby Academy 3

(a) (i) Suggest reasons for the differences between the three standard entropies.

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(ii) Calculate the entropy change for the system, ∆Sο

system, for this reaction. Include the

sign and units in your answer.

(2)

(b) Calculate the entropy change for the surroundings, ∆Sοsurroundings, at 298 K, showing your

method clearly.

(2)

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Maltby Academy 4

(c) (i) Calculate the total entropy change for this reaction, ∆Sοtotal, at 298 K.

(1)

(ii) What does the result of your calculation in (c)(i) indicate about the natural

direction of this reaction at 298 K?

Justify your answer.

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(d) (i) Write an expression for the equilibrium constant, Kp, for this reaction.

(1)

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Maltby Academy 5

(ii) State how you would alter ONE condition to increase the yield of carbon dioxide

from this equilibrium reaction.

Justify your answer.

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(Total 11 marks)

3. The reaction between solid barium hydroxide and solid ammonium chloride can be

represented by the equation below.

Ba(OH)2(s) + 2NH4Cl(s) BaCl2(s) + 2NH3(g) + 2H2O(l) ΔHο = +51.1 kJ mol

–1

The standard entropies, at 298 K, for the reactants and products are:

Sο[Ba(OH)2(s)] = + 99.7 J mol

–1K

–1

Sο[NH4Cl(s)] = + 94.6 J mol

–1K

–1

Sο[BaCl2(s)] = + 123.7 J mol

–1K

–1

Sο[NH3(g)] = + 192.3 J mol

–1K

–1

Sο[H2O(l)] = + 69.9 J mol

–1K

–1

(a) Why is the standard entropy of ammonia more positive than the standard entropy of

barium chloride?

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Maltby Academy 6

(b) Use the values given to calculate the standard entropy change, ΔSοsystem, for this reaction.

Include the sign and units in your answer.

(2)

(c) Calculate the standard entropy change of the surroundings, ΔSοsurroundings, at 298 K for

this reaction.

(2)

(d) Use your answers to (b) and (c) to show that this reaction is feasible at 298 K.

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Page 7: (3) - KS5Chemistry...... Why is the standard entropy of ammonia more positive than the standard entropy of barium chloride? ..... ..... Calculate the standard entropy change of the

Maltby Academy 7

(e) Calculate the minimum temperature, in kelvin, at which the reaction is spontaneous.

(2)

(Total 8 marks)

4. The equation below shows a possible reaction for producing methanol.

CO(g) + 2H2(g) CH3OH(l) ΔHο = 129 kJ mol

–1

(a) The entropy of one mole of each substance in the equation, measured at 298 K, is shown

below.

Substance

Sο

/J mol1

K1

CO(g) 197.6

H2(g) 130.6

CH3OH(l) 239.7

(i) Suggest why methanol has the highest entropy value of the three substances.

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Maltby Academy 8

(ii) Calculate the entropy change of the system, ΔSοsystem, for this reaction.

(2)

(iii) Is the sign of ΔSοsystem as expected? Give a reason for your answer.

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(iv) Calculate the entropy change of the surroundings ΔSοsurroundings, at 298 K.

(2)

(v) Show, by calculation, whether it is possible for this reaction to occur spontaneously

at 298 K.

(2)

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Maltby Academy 9

(b) When methanol is produced in industry, this reaction is carried out at 400 ºC and 200

atmospheres pressure, in the presence of a catalyst of chromium oxide mixed with zinc

oxide. Under these conditions methanol vapour forms and the reaction reaches

equilibrium. Assume that the reaction is still exothermic under these conditions.

CO(g) + 2H2(g) CH3OH(g)

(i) Suggest reasons for the choice of temperature and pressure.

Temperature ........................................................................................................

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Pressure ...............................................................................................................

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(ii) The catalyst used in this reaction is heterogeneous. Explain this term.

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(iii) Write an expression for the equilibrium constant in terms of pressure, Kp, for this

reaction.

CO(g) + 2H2(g) CH3OH(g)

(1)

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Maltby Academy 10

(iv) In the equilibrium mixture at 200 atmospheres pressure, the partial pressure of

carbon monoxide is 55 atmospheres and the partial pressure of hydrogen is 20

atmospheres.

Calculate the partial pressure of methanol in the mixture and hence the value of the

equilibrium constant, Kp. Include a unit in your answer.

(2)

(c) The diagram below shows the distribution of energy in a sample of gas molecules in a

reaction when no catalyst is present. The activation energy for the reaction is EA.

(i) What does the shaded area on the graph represent?

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(ii) Draw a line on the graph, labelled EC, to show the activation energy of the

catalysed reaction. (1)

(Total 17 marks)

Page 11: (3) - KS5Chemistry...... Why is the standard entropy of ammonia more positive than the standard entropy of barium chloride? ..... ..... Calculate the standard entropy change of the

Maltby Academy 11

5. The reaction between nitrogen and hydrogen can be used to produce ammonia.

N2(g) + 3H2(g) 2NH3(g) ΔH

ο = – 92.2 kJ mol

–1

Standard entropies are given below

Sο [N2(g)] = +191.6 J mol

–1 K

–1

Sο [H2(g)] = +130.6 J mol

–1 K

–1

Sο [NH3(g)] = +192.3 J mol

–1 K

–1

(a) Calculate the entropy change of the system, ΔSοsystem, for this reaction. Include a sign and

units in your answer.

(2)

(b) Calculate the entropy change of the surroundings, ΔSοsurroundings, at 298 K. Include a sign

and units in your answer.

(2)

Page 12: (3) - KS5Chemistry...... Why is the standard entropy of ammonia more positive than the standard entropy of barium chloride? ..... ..... Calculate the standard entropy change of the

Maltby Academy 12

(c) (i) Calculate the total entropy change, ΔSοtotal, at 298 K. Include a sign and units in

your answer.

(1)

(ii) Is this reaction feasible at 298 K? Justify your answer.

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(d) In industry the reaction is carried out at about 700 K using an iron catalyst and high

pressures.

(i) The yield of ammonia produced at equilibrium is less at 700 K than at 298 K, if the

pressure remains constant. In terms of entropy, explain why this happens.

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(ii) Higher pressures increase the yield of ammonia at equilibrium. Suggest a reason

why pressures greater than 300 atmospheres are not routinely used.

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Page 13: (3) - KS5Chemistry...... Why is the standard entropy of ammonia more positive than the standard entropy of barium chloride? ..... ..... Calculate the standard entropy change of the

Maltby Academy 13

(iii) Iron is a heterogeneous catalyst. Explain what is meant by heterogeneous.

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(Total 9 marks)

6. (a) The distribution of the energy of particles in a gas at temperature T1 is shown below.

(i) On the diagram above, draw the distribution of energy of particles at a lower

temperature, T2. (2)

Page 14: (3) - KS5Chemistry...... Why is the standard entropy of ammonia more positive than the standard entropy of barium chloride? ..... ..... Calculate the standard entropy change of the

Maltby Academy 14

(ii) Use the diagram to explain why the rate of a reaction increases with an increase in

temperature.

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(iii) Explain fully why a catalyst increases the rate of a reaction.

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Page 15: (3) - KS5Chemistry...... Why is the standard entropy of ammonia more positive than the standard entropy of barium chloride? ..... ..... Calculate the standard entropy change of the

Maltby Academy 15

(b) The fermentation of glucose is an exothermic reaction and is catalysed by enzymes in

yeast.

C6H12O6(aq) → 2C2Η5ΟΗ(aq) + 2CO2(g)

The reaction is slow at room temperature.

(i) Describe, with the aid of a diagram, an experiment you could do to follow the

progress of this reaction at different temperatures.

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Page 16: (3) - KS5Chemistry...... Why is the standard entropy of ammonia more positive than the standard entropy of barium chloride? ..... ..... Calculate the standard entropy change of the

Maltby Academy 16

(ii) Would you expect ∆Ssystem to be positive or negative for this reaction? Justify your

answer with TWO pieces of evidence.

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(iii) Deduce the sign of ∆Ssurroundings. Show your reasoning.

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(Total 15 marks)

7. (a) State Hess‟s Law.

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Page 17: (3) - KS5Chemistry...... Why is the standard entropy of ammonia more positive than the standard entropy of barium chloride? ..... ..... Calculate the standard entropy change of the

Maltby Academy 17

(b) Methane burns in oxygen.

CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

(i) Calculate the enthalpy change for this reaction, using the bond enthalpies

given below.

Bond enthalpy

/ kJ mol–1

C – H +435

O = O +498

C = O +805

H – O +464

(3)

(ii) State the name of this enthalpy change.

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(iii) The value of this enthalpy change, under standard conditions, is –890 kJ mol–1

.

State the meaning of standard conditions.

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(iv) Suggest, with a reason, why the enthalpy change calculated in (i) is different from

the standard value quoted in (iii).

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Page 18: (3) - KS5Chemistry...... Why is the standard entropy of ammonia more positive than the standard entropy of barium chloride? ..... ..... Calculate the standard entropy change of the

Maltby Academy 18

(c) Although the reaction between methane and oxygen is exothermic, it does not occur

unless the mixture is ignited.

Use these facts to explain the difference between thermodynamic and kinetic stability.

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(Total 14 marks)

ô °8. When barium nitrate is heated it decomposes as follows:

Ba(NO3)2(s) → BaO(s) + 2NO2(g) + ½O2(g) ΔH = +505.0 kJ mol–1

(a) Use the following data when answering this part of the question.

Substance Standard entropy,

Sο / J mol

–1 K

–1

Ba(NO3)2(s) + 213.8

BaO(s) + 70.4

NO2(g) + 240.0

O2(g) + 205.0

Page 19: (3) - KS5Chemistry...... Why is the standard entropy of ammonia more positive than the standard entropy of barium chloride? ..... ..... Calculate the standard entropy change of the

Maltby Academy 19

(i) Explain why:

• Sο [NO2(g)] is greater than S

ο [BaO(s)]

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• Sο [Ba(NO3)2(s)] is greater than S

ο [BaO(s)].

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(ii) Calculate the entropy change of the system, ΔSοsystem, for this reaction. Include a

sign and units in your answer.

(2)

(b) Calculate the entropy change of the surroundings, ΔSοsurroundings, for the reaction at 298 K.

Include a sign and units in your answer. (2)

(c) Calculate ΔSοtotal, and explain the significance of the sign for this value.

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Page 20: (3) - KS5Chemistry...... Why is the standard entropy of ammonia more positive than the standard entropy of barium chloride? ..... ..... Calculate the standard entropy change of the

Maltby Academy 20

(d) Calculate the minimum temperature at which the decomposition of barium nitrate should

occur.

You can assume that ΔH and ΔSsystem are not affected by a change in temperature.

(2)

(Total 10 marks)

9. (i) Define the term enthalpy of hydration, ΔHhyd, of an ion.

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Page 21: (3) - KS5Chemistry...... Why is the standard entropy of ammonia more positive than the standard entropy of barium chloride? ..... ..... Calculate the standard entropy change of the

Maltby Academy 21

(ii) The table below gives some information about the sulphates of the Group 2 elements

magnesium and barium.

sulphate lattice energy

/ kJ mol–1

hydration

enthalpy of cation

/ kJ mol–1

solubility

/ mol dm–3

MgSO4 –2874 –1920 1.83

BaSO4 –2374 –1360 9.43 × 10–6

Use the lattice energy and hydration enthalpy values to explain the difference in the

solubility of the two salts.

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(Total 6 marks)

10. (a) (i) Enthalpy or heat change / released when 1 mol of

substance (1)

is burned in excess oxygen / completely (1)

all substances in standard states (at a specified temp)/

at a pressure of 1 atm. (1) 3

Page 22: (3) - KS5Chemistry...... Why is the standard entropy of ammonia more positive than the standard entropy of barium chloride? ..... ..... Calculate the standard entropy change of the

Maltby Academy 22

(ii) Suitable cycle (need not be labelled but if labelled,

these must be correct) (1)

working (1)

answer (1)

e.g. C + O CO ( H CO)

( H Carbon)( H Carbon monoxide)

CO

comb

comb

form2 2

2

1/

(1)

= 394 (283) (1)

= 111 (kJ mol–1

) (1)

Penalise 1 mark if units incorrect 3

(iii) (some) CO2 is always produced in the reaction (1) 1

(iv) 2 2

1/(1)

(Energy) C (+ O )

CO

1

Consequential on (a) (ii)

n.b. if no answer in(a) (ii), correct diagram can still score

(b) Methane (and oxygen) / reactants thermodynamically unstable

w.r.t. products (1)

Must be a comparison

Since reactants are at a higher energy level (than products) (1)

or reverse argument

Reactants / methane, oxygen kinetically stable (1)

Due to high activation energy (1) 4

If no reference to methane in the answer (max 3) [12]

11. (a) (i) Gases have much higher entropies than solids as there are many more

ways of arranging the entities / less ordered / more random(ness)

OR reverse argument (1)

ZnCO3 has more atoms/is more complex than ZnO (1) 2

Page 23: (3) - KS5Chemistry...... Why is the standard entropy of ammonia more positive than the standard entropy of barium chloride? ..... ..... Calculate the standard entropy change of the

Maltby Academy 23

(ii) Sοsystem = (+43.6) + (+213.6) – (+82.4)

=+174.8/175 J mol–1

K–1

method (1)

answer, sign and units (1)

Correct answer, sign and units with no working (2) 2

(b)

As printed

Sοsurroundings =

T

– H

OR = 298

)105.464(– 3 (1)

= – 1560 / 1559 J mol–1

K–1

answer, sign and units (1)

Amended

Sοsurroundings =

T

– H

OR = 298

)100.71(– 3 (1)

= – 238(.3) J mol–1

K–1

answer, sign and units (1)

ONLY accept 3 or 4 SF 2

IF correct answer, sign and units with no working (1)

(c) (i)

Sοtotal = +174.8 – 1558.7

= – 1384 / – 1380 J mol–1

K–1

IF + 174.8 – 1560

= – 1385(.2)

= –1385 / 1390 J mol–1

K–1

IF + 174.8 – 1559

= – 1384 J mol–1

K–1

= – 63.5 / 64 / 63 / 63.2 / 63.4 J mol–1

K–1

ONLY penalise incorrect units OR no units in (a)(ii), (b) and (c)(i) once 1

(ii) Natural direction is right to left /reverse as Sοtotal /total entropy change

is negative / less than zero. 1

MUST be consistent with (i)

Page 24: (3) - KS5Chemistry...... Why is the standard entropy of ammonia more positive than the standard entropy of barium chloride? ..... ..... Calculate the standard entropy change of the

Maltby Academy 24

(d) (i) Kp = p co2 ((g) eqm) 1

(ii) Increase temperature / reduce pressure (1)

Decreases Sοsurroundings (negative) and hence increases S

οtotal / Le Chatelier‟s

principle applied (i.e increasing temperature, reducing pressure) (1) 2 [11]

12. (a) Many more ways of arranging / more disordered gas molecules than solid

(particles) 1

(b) Sοsystem = +123.7 + 2(+192.3) + 2(+69.9) – (+99.7) – 2(+94.6)

= +359(.2) J K–1

mol–1

Method (1) 2

Sign, value, units (1)

(c) Sοsurroundings =

T

H– / =

298

)101.51(– 3 (1)

= – 171.5 / 171 J K–1

mol–1

(1) 2

(d) Sοtotal = S

οsystem + S

οsurroundings So the total entropy change has a positive

value / is greater than zero.

OR

Sοtotal = + 187.7 / +188 J K

–1mol

–1 1

(e) 0 = 359.2 – T

3101.51

Some recognisable correct method (1)

T = 142(.3) / 143 K (1) 2 [8]

Page 25: (3) - KS5Chemistry...... Why is the standard entropy of ammonia more positive than the standard entropy of barium chloride? ..... ..... Calculate the standard entropy change of the

Maltby Academy 25

13. (a) (i) Methanol is the biggest/ most complex molecule / greatest MR /most

atoms/most electrons 1

(ii) Ssystem = 239.7 – 197.6 – 2(130.6)

= –219.1/ –219 J mol–l

K–1

Method (1)

answer + units (1) 2

(iii) yes as 3 molecules 1 OR yes as (2) gases a liquid 1

(iv) Ssurr = –H/T (stated or used) (1)

= –(–129/ 298) = +0.433 kJ mol–1

K–1

/ +433 J mol–1

K–1

/+ 432.9 (1)

–1 for wrong units/ no units / more than 4 SF

–1 for wrong sign/ no sign 2

(v) Stotal = –219.1 + 433 = +213.9 / +213.8 J mol–1

K–1

/ +214 J mol–1

K–1

/

+0.214 kJ mol –1

K–1

(1)

Positive so possible (1) 2

(b) (i) Temperature

Faster at 400°C (1)

even though yield is lower (1)

Pressure

Higher pressure improves yield of methanol (1)

Higher pressure increases rate (1)

Maximum 3 3

(ii) Not in same phase as reactants. ALLOW state instead of phase 1

(iii) Kp = p(CH3OH)/p(CO)×p(H2)2 1

(iv) Partial pressure of methanol = 200 – 55 – 20 = 125 atm (1)

Kp = (125)/55×202

= 5.68 × 10–3

/ 5.7 × 10–3

atm–2

(1) 2

(c) (i) Number of molecules / fraction of molecules with energy EA /number

of molecules which have enough energy to react. 1

(ii) Vertical line / mark on axis to show value to the left of line EA 1

[17]

14. Penalise units only once in this question

(a) (2×192.3)–[191.6 + (130.6 × 3)] (1)

= –198.8/199 J mol–1

K –1

(1) 2

(b) 298

/2.92––

298

/10002.92–– –H / T (1)

= + 309(.4) J mol–1

K–1

/ + 0.309(4) kJ mol–1

K–1

(1) 2

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(c) (i) –198.8 + 309 = + 110 J mol–1

K–1

(3 SF)

OR

– 198.8 + 309.4 = + 111 J mol–1

K–1

(3 SF)

[Do not penalise missing + sign if penalised already in (b)]

NOT 4SF. Penatise SF only once on paper 1

(ii) Yes, as Stotal is positive / total entropy change 1

(d) (i) Higher T makes Ssurroundings decrease (so Stotal is less positive) 1

(ii) Cost (of energy) to provide compression/ cost of equipment

to withstand high P/ maintenance costs.

NOT safety considerations alone 1

(iii) Different phase/state (to the reactants) 1 [9]

15. (a) (i)

Starts at zero and approaching x-axis (1)

Maximum greater and at lower energy(1) – T2 needs only to be just

higher than T1

T2 curve must go below T1 curve approaching the x-axis 2

(ii) As the temperature increases the energy of the particles increases (1)

Use the diagram shading areas

OR more particles to the right hand side of EA line (1)

and so more (successful) collisions/particles have energy greater /

equal or greater than the activation energy (1)

NOT “equal” on its own

NOT mention of “frequency of collisions” on its own 3

Page 27: (3) - KS5Chemistry...... Why is the standard entropy of ammonia more positive than the standard entropy of barium chloride? ..... ..... Calculate the standard entropy change of the

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(iii) A catalyst provides an alternative route with a lower activation energy/

which requires less energy (1)

so more collisions / particles have energy greater than the activation

energy (1) 2

(b) (i) e.g.

Measure the volume of gas given off in a given time / count bubbles /

obscuring cross using limewater (1)

and then repeat over a range of temperatures (1)

No diagram max 3

If method shown cannot possibly work max 1 ie waterbath or

sensible range of temperatures BUT NOT different temperatures

Penalty

–1 for poor diagram

4

(ii) Positive

1 mol goes to 4 moles/particles (so more disorder) /increase in number

of moles/particles (1)

products include a gas (and so more disorder) (1)

NOT 1 mole of compound/element goes to 4 moles of

compound/element

If “negative” 0 (out of 2) 2

(iii) Positive with some explanation e.g. ∆Ssurroundings = – ∆H/T OR

because reaction is exothermic (1)

∆H is therefore negative and so ∆Ssurroundings must be positive (1)

If negative given in (ii) allow TE here 2 [15]

Page 28: (3) - KS5Chemistry...... Why is the standard entropy of ammonia more positive than the standard entropy of barium chloride? ..... ..... Calculate the standard entropy change of the

Maltby Academy 28

16. (a) Heat / enthalpy / energy change (for a reaction) / ∆H (1)

is independent of the pathway / route (between reactants and products)

OR depends only on its initial and final state (1)

Both marks can score from a diagram and equation 2

(b) (i) ∆H = {(4x + 435) + (2x + 498)} (1)

+ {(2x – 805) + (4x – 464)} (1)

IGNORE signs for first two marks, ie marks for total

enthalpies of bonds broken and made.

= – 730 (kJmol–1

) (1)

3rd

mark is consequential on their values for first two marks

+ 730 (kJmol–1

) (max 2) 3

(ii) (Enthalpy of) combustion

DO NOT penalise “standard” 1

(iii) At 1 atm pressure OR 101 / 100 kPa OR 1 bar (1)

stated temperature (1)

ACCEPT 298 K / 25 C 2

(iv) Reaction has H2O(g) (rather than H2O(l)) (1)

So not standard conditions (1) – 2nd

mark is conditional on the 1st

Average bond enthalpies used (so not specific) (1 max) 2

QWC (c) (Exothermic so) products are at lower energy than reactants (1)

Reactants are therefore thermodynamically unstable

(with respect to products) (1) Consequential on 1st mark

NOT „reaction‟ or „system‟ is thermodynamically unstable

Can argue from point of view of products.

Ea is high (for noticeable reaction at room temperature) (1)

NOT „Ea high‟ on its own

So reactants are kinetically stable (with respect to products) (1)

Consequential on 3rd

mark

If “reaction” instead of reactants is used (3 max) 4 [14]

Page 29: (3) - KS5Chemistry...... Why is the standard entropy of ammonia more positive than the standard entropy of barium chloride? ..... ..... Calculate the standard entropy change of the

Maltby Academy 29

17. (a) (i) NO2 is a gas (whereas BaO is a solid) (1)

Ba(NO3)2 has a more complicated structure than BaO (1)

Allow 2nd

mark if a correct statement is combined with a

“neutral” wrong statement 2

Accept Ba(NO3)2 “molecule” has more electrons / is larger

than BaO “molecule” (1)

Accept more atoms/ions/particles

Accept more complicated/complex compound

Reject Ba(NO3)2 has a larger molar mass than BaO

Reject more molecules/elements

(ii) ∆Sοsystem = 70.4 + (2 × 240.0) + (½ × 205.0) – 213.8 (1)

= +439.1 J mol–1

K–1

(1)

–1 per error 2

Accept +439 J mol–1

K–1

Accept J/ mol /K

(b) ∆Sοsurroundings =

T

H (1) =

298

1000505

= – 1700 J mol–1

K–1

(3 s.f.) (1)

Penalise wrong units in (a)(ii) and (b) once only 2

Accept –1690 J mol–1

K–

Accept –1695 J mol–1

K–1

Answers in kJ mol–1

K–1

Reject –1694 J mol–1

K–1

Reject –1694.6 J mol–1

K–1

Reject –1694.63 J mol–1

K–1

(c) ∆Sοtotal = +439.1 – 1695 = – 1260 (J mol

–1 K

–1) (1)

Allow TE [follow through working from (a)(ii) and (b)]

Mark consistently with (a)(ii) and (b)

The reaction isn‟t spontaneous / doesn‟t “go” (at 298K) (1)

Must be consistent with sign in calculation 2

Accept –1256 J mol–1

K–1

Accept –1261 J mol–1

K–1

Accept –1255.5 J mol–1

K–1

Page 30: (3) - KS5Chemistry...... Why is the standard entropy of ammonia more positive than the standard entropy of barium chloride? ..... ..... Calculate the standard entropy change of the

Maltby Academy 30

(d) When just spontaneous, ∆Sοtotal = 0

or implied by calculation i.e a(ii)

505000 OR 505 (1)

∆Sοsurroundings = –439.1 J mol

–1 K

–1

T = 1.439

1000505 = 1150 (K) (1)

Accept 1150.1 K

Accept 877 °C

Accept 1151K with no working (1 max)

Reject 1151K for 2nd

mark

Reject any negative value for T (in K): no 2nd

mark

Reject 1150 °C

ignore 0K

Allow full marks for an answer without working 2 [10]

18. (i) Enthalpy change when 1 mol of gaseous ions (1)

Accept energy or heat

Reject any implication of an endothermic process

e.g. energy required

Reject “….1 mol of gaseous atoms”

is dissolved such that further dilution causes no

further heat change (1)

Accept “added to water”/”reacts with water” instead of

“dissolved”

Reject just “hydrated”

Reject just “completely hydrated”

IGNORE “standard conditions”

Accept is dissolved to form an infinitely dilute solution

OR

Is dissolved in a large/excess/infinite amount of water

Mark each aspect independently 2

Page 31: (3) - KS5Chemistry...... Why is the standard entropy of ammonia more positive than the standard entropy of barium chloride? ..... ..... Calculate the standard entropy change of the

Maltby Academy 31

(ii) EITHER

∆HSOLN = (–∆HLE + ∆HHYD) (1)

Expression quoted or correctly used in at least one of

the calculations below

Accept answer only with no working (1)

∆HSOLN MgSO4 = –(–2874) + (–1920)

= +954 (kJ mol–1

) (1)

∆HSOLN = BaSO4 = –(–2374) + (–1360)

= +1014 (kJ mol–1

) (1)

Accept answer only with no working (1)

Enthalpy of solution of MgSO4 less endothermic/more

exothermic/more negative than for BaSO4, so MgSO4

more soluble than BaSO4 (or reverse argument) (1)

Reject just “solubility/∆Hsoln depends on a balance between

lattice and hydration energies”

OR (both) lattice energies and hydration enthalpies

decrease from MgSO4 to BaSO4 (or down group) (1)

Accept “The hydration energies decrease faster…..”

Reject (–)500 and (–)560 stated without further explanation

(but) lattice energies change less (1)

∆HSOLN = (–∆HLE + ∆HHYD) (1) stated in words or symbols

Reject just “solubility/∆Hsolution depends on a balance between

lattice and hydration energies”

so ∆Hsoln less exothermic/more endothermic/more

positive for BaSO4 so less soluble

OR so ∆Hsoln more exothermic/more negative/less

endothermic for MgSO4 so MgSO4 more soluble (1) 4

[6]