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R1.4 Entropy and spontaneity

Practice exam-style IB Chemistry questions for Entropy and spontaneity, aligned with the syllabus and grouped by topic.

Verified by Dennis M.
Verified by Dennis M.
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Question 1
HL • Paper 1A
Easy
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HL • Paper 1A
Easy
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The reaction CaCO3(s)CaO(s)+CO2(g)CaCO_3(s) \to CaO(s) + CO_2(g) is carried out at constant temperature.

What is the expected sign of the entropy change of the system?

A.

Negative, because a compound is decomposed

B.

Positive, because a gas is produced from solids

C.

Positive, because the reaction is endothermic

D.

Negative, because one solid reactant forms two products

Question 2
HL • Paper 1A
Easy
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HL • Paper 1A
Easy
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What accounts for the predicted entropy of a perfect crystal at 0 K0\ \text{K}?

A.

There is only one possible arrangement in the lowest-energy state

B.

There are no particles present in the crystal lattice

C.

There are no chemical bonds present between particles

D.

There is no enthalpy change associated with the crystal

Question 3
HL • Paper 1A
Easy
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HL • Paper 1A
Easy
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A reaction has ΔG<0\Delta G < 0 at room temperature but occurs immeasurably slowly without a catalyst.

What statement best accounts for this observation?

A.

The reaction is thermodynamically favourable but has a high activation energy

B.

The reaction must have ΔH>0\Delta H > 0 because it is slow

C.

The reaction is non-spontaneous because it has a high activation energy

D.

The reaction must have ΔS<0\Delta S < 0 because it is slow

Question 4
HL • Paper 1A
Easy
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HL • Paper 1A
Easy
Calculator Permitted

Use the standard molar entropies to calculate ΔS\Delta S^\circ for the reaction.

N2O4(g)2NO2(g)N_2O_4(g) \to 2NO_2(g)

SubstanceS/J K1 mol1S^\circ / \text{J K}^{-1}\text{ mol}^{-1}
N2O4(g)N_2O_4(g)304304
NO2(g)NO_2(g)240240
A.

+176 J K1 mol1+176\ \text{J K}^{-1}\text{ mol}^{-1}

B.

+64 J K1 mol1+64\ \text{J K}^{-1}\text{ mol}^{-1}

C.

176 J K1 mol1-176\ \text{J K}^{-1}\text{ mol}^{-1}

D.

64 J K1 mol1-64\ \text{J K}^{-1}\text{ mol}^{-1}

Question 5
HL • Paper 1A
Easy
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HL • Paper 1A
Easy
Calculator Permitted

For a reaction at 298 K298\ \text{K}, ΔH=+45.0 kJ mol1\Delta H^\circ = +45.0\ \text{kJ mol}^{-1} and ΔS=+120 J K1 mol1\Delta S^\circ = +120\ \text{J K}^{-1}\text{ mol}^{-1}.

What are ΔG\Delta G^\circ and the thermodynamic conclusion under standard conditions?

A.

80.8 kJ mol1-80.8\ \text{kJ mol}^{-1}; spontaneous

B.

9.2 kJ mol1-9.2\ \text{kJ mol}^{-1}; spontaneous

C.

+9.2 kJ mol1+9.2\ \text{kJ mol}^{-1}; non-spontaneous

D.

+80.8 kJ mol1+80.8\ \text{kJ mol}^{-1}; non-spontaneous

Question 6
HL • Paper 1A
Easy
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HL • Paper 1A
Easy
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The graph shows ΔG\Delta G^\circ against absolute temperature, TT, for a reaction.

What signs of ΔH\Delta H^\circ and ΔS\Delta S^\circ are indicated?

ΔG° plotted against temperature for a reaction.
A.

ΔH<0\Delta H^\circ < 0 and ΔS>0\Delta S^\circ > 0

B.

ΔH>0\Delta H^\circ > 0 and ΔS>0\Delta S^\circ > 0

C.

ΔH>0\Delta H^\circ > 0 and ΔS<0\Delta S^\circ < 0

D.

ΔH<0\Delta H^\circ < 0 and ΔS<0\Delta S^\circ < 0

Question 7
HL • Paper 1A
Easy
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HL • Paper 1A
Easy
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A reaction is spontaneous only at low temperatures.

What signs of ΔH\Delta H^\circ and ΔS\Delta S^\circ are consistent with this behaviour?

A.

ΔH<0\Delta H^\circ < 0 and ΔS>0\Delta S^\circ > 0

B.

ΔH>0\Delta H^\circ > 0 and ΔS>0\Delta S^\circ > 0

C.

ΔH<0\Delta H^\circ < 0 and ΔS<0\Delta S^\circ < 0

D.

ΔH>0\Delta H^\circ > 0 and ΔS<0\Delta S^\circ < 0

Question 8
HL • Paper 1A
Easy
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HL • Paper 1A
Easy
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For a reversible reaction at a fixed temperature, ΔG\Delta G^\circ is positive.

What does this imply about the equilibrium constant and the likely equilibrium composition?

A.

K>1K>1; products are favoured

B.

K>1K>1; reactants are favoured

C.

K<1K<1; reactants are favoured

D.

K<1K<1; products are favoured

Question 9
HL • Paper 1A
Easy
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HL • Paper 1A
Easy
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For an electrochemical cell reaction under standard conditions, n=2n=2 and Ecell=+0.34 VE^\circ_{\text{cell}}=+0.34\ \text{V}.

What conclusion follows from ΔG=nFEcell\Delta G^\circ=-nFE^\circ_{\text{cell}}?

A.

ΔG>0\Delta G^\circ>0 and the reverse reaction is non-spontaneous

B.

ΔG<0\Delta G^\circ<0 and the reverse reaction is spontaneous

C.

ΔG>0\Delta G^\circ>0 and the cell reaction is spontaneous as written

D.

ΔG<0\Delta G^\circ<0 and the cell reaction is spontaneous as written

Question 10
HL • Paper 2
Easy
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HL • Paper 2
Easy
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Solid ammonium dichromate decomposes on heating according to the equation:

(NH4)2Cr2O7(s)Cr2O3(s)+N2(g)+4H2O(g)(\text{NH}_4)_2\text{Cr}_2O_7(s) \to \text{Cr}_2O_3(s) + N_2(g) + 4H_2O(g)

A

Predict the sign of the entropy change of the system.

[1]
B

Explain your answer to part (a).

[2]
Question 11
HL • Paper 2
Easy
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HL • Paper 2
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The third law of thermodynamics is often described using a perfect crystal at 0 K0\ \text{K}.

A

Outline why a perfect crystal is predicted to have zero entropy at 0 K0\ \text{K}.

[2]
Question 12
HL • Paper 1A
Medium
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HL • Paper 1A
Medium
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A reaction has ΔH=+84.0 kJ mol1\Delta H^\circ = +84.0\ \text{kJ mol}^{-1} and ΔS=+220 J K1 mol1\Delta S^\circ = +220\ \text{J K}^{-1}\text{ mol}^{-1}.

At what temperature condition does the reaction become spontaneous under standard conditions?

A.

Above 382 K382\ \text{K}

B.

Below 382 K382\ \text{K}

C.

Below 2620 K2620\ \text{K}

D.

Above 2620 K2620\ \text{K}

Question 13
HL • Paper 1A
Medium
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HL • Paper 1A
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For a reaction at 298 K298\ \text{K}, ΔG=10.0 kJ mol1\Delta G^\circ = -10.0\ \text{kJ mol}^{-1} and Q=100Q = 100.

Using ΔG=ΔG+RTlnQ\Delta G = \Delta G^\circ + RT\ln Q, what is ΔG\Delta G and the spontaneous direction under these conditions?

A.

1.4 kJ mol1-1.4\ \text{kJ mol}^{-1}; forward reaction

B.

+21.4 kJ mol1+21.4\ \text{kJ mol}^{-1}; reverse reaction

C.

+1.4 kJ mol1+1.4\ \text{kJ mol}^{-1}; reverse reaction

D.

21.4 kJ mol1-21.4\ \text{kJ mol}^{-1}; forward reaction

Question 14
HL • Paper 1A
Medium
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HL • Paper 1A
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At 298 K298\ \text{K}, a reaction has ΔG=17.1 kJ mol1\Delta G^\circ = -17.1\ \text{kJ mol}^{-1}.

What is the approximate value of KK?

A.

1.0×1011.0\times10^{-1}

B.

1.0×1031.0\times10^3

C.

1.0×1031.0\times10^{-3}

D.

1.0×1011.0\times10^1

Question 15
HL • Paper 2
Medium
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HL • Paper 2
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The Haber process is represented by the equation:

N2(g)+3H2(g)2NH3(g)N_2(g) + 3H_2(g) \to 2NH_3(g)

The standard molar entropies are: S[N2(g)]=192 J K1 mol1S^\circ[N_2(g)] = 192\ \text{J K}^{-1}\text{ mol}^{-1}, S[H2(g)]=131 J K1 mol1S^\circ[H_2(g)] = 131\ \text{J K}^{-1}\text{ mol}^{-1} and S[NH3(g)]=193 J K1 mol1S^\circ[NH_3(g)] = 193\ \text{J K}^{-1}\text{ mol}^{-1}.

A

Calculate the standard entropy change for the reaction.

[2]
Question 16
HL • Paper 2
Medium
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HL • Paper 2
Medium
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For a reaction at 298 K298\ \text{K}, ΔH=+82.4 kJ mol1\Delta H^\circ = +82.4\ \text{kJ mol}^{-1} and ΔS=+172 J K1 mol1\Delta S^\circ = +172\ \text{J K}^{-1}\text{ mol}^{-1}.

A

Calculate ΔG\Delta G^\circ for the reaction at 298 K298\ \text{K}.

[2]
B

State whether the reaction is spontaneous under standard conditions at this temperature.

[1]
Question 17
HL • Paper 2
Medium
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HL • Paper 2
Medium
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Phosgene formation is represented by:

CO(g)+Cl2(g)COCl2(g)CO(g) + Cl_2(g) \rightleftharpoons COCl_2(g)

At a particular temperature, K=4.0×103K = 4.0 \times 10^3. For a reacting mixture at the same temperature, Q=2.0×105Q = 2.0 \times 10^5.

A

Deduce the direction in which the reaction mixture will change to reach equilibrium.

[1]
B

State the sign of ΔG\Delta G for the forward reaction under these conditions.

[1]
Question 18
HL • Paper 2
Medium
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HL • Paper 2
Medium
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A voltaic cell has Ecell=+0.460 VE^\circ_{\text{cell}} = +0.460\ \text{V} for its overall reaction. Two moles of electrons are transferred per mole of reaction.

A

Calculate ΔG\Delta G^\circ for the cell reaction using F=9.65×104 C mol1F = 9.65 \times 10^4\ \text{C mol}^{-1}.

[2]
B

Deduce whether the cell reaction is spontaneous as written under standard conditions.

[1]
Question 19
HL • Paper 2
Medium
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HL • Paper 2
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Ammonium nitrate dissolves readily in water, although the process is endothermic.

A

Explain how this process can be spontaneous even though heat is absorbed from the surroundings.

[3]
Question 20
HL • Paper 1B
Medium
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HL • Paper 1B
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The table shows standard molar entropy values, SS^\circ, for selected substances at 298 K298\ \text{K}.

Substance

S^\circ / J K1\text{K}^{-1} mol1\text{mol}^{-1}

H2OH_2O(l)

69.9

Br2Br_2(l)

152.2

H2OH_2O(g)

188.8

Br2Br_2(g)

245.5

A

Identify the substance in the table with the greatest standard molar entropy.

[1]
B

Calculate ΔS\Delta S^\circ for Br2(l)Br2(g)\text{Br}_2(l) \to \text{Br}_2(g).

[1]
C

Explain why the value calculated in (b) is positive.

[2]
Question 21
HL • Paper 1B
Medium
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HL • Paper 1B
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The graph shows the Gibbs energy change, ΔG\Delta G, for the forward reaction as a reversible reaction mixture changes composition at constant temperature.

Gibbs energy change for a forward reaction as composition changes at constant temperature.
A

State the value of ΔG\Delta G at equilibrium.

[1]
B

Identify the region of the graph in which the forward reaction is spontaneous.

[1]
C

Explain why ΔG\Delta G becomes less negative as the reaction proceeds from a reactant-rich mixture toward equilibrium.

[2]
Question 22
HL • Paper 2
Medium
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HL • Paper 2
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The decomposition of calcium carbonate is represented by:

CaCO3(s)CaO(s)+CO2(g)\text{CaCO}_3(s) \to \text{CaO}(s) + CO_2(g)

For this reaction, ΔH=+178.3 kJ mol1\Delta H^\circ = +178.3\ \text{kJ mol}^{-1} and ΔS=+160.5 J K1 mol1\Delta S^\circ = +160.5\ \text{J K}^{-1}\text{ mol}^{-1}.

A

Determine the temperature at which ΔG=0\Delta G^\circ = 0.

[2]
B

State the temperature condition under which the decomposition is spontaneous.

[1]
Question 23
HL • Paper 2
Medium
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HL • Paper 2
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Two reactions have the following thermodynamic signs.

Reaction I: ΔH<0\Delta H^\circ < 0 and ΔS<0\Delta S^\circ < 0

Reaction II: ΔH>0\Delta H^\circ > 0 and ΔS>0\Delta S^\circ > 0

A

Compare how temperature affects the spontaneity of reactions I and II.

[4]
Question 24
HL • Paper 2
Medium
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HL • Paper 2
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A graph of ΔG\Delta G^\circ against temperature is shown for a reaction.

Linear ΔG°-temperature relationship for a reaction.
A

State the thermodynamic quantities represented by the vertical intercept and the gradient of the line.

[2]
B

Deduce the signs of ΔH\Delta H^\circ and ΔS\Delta S^\circ for the reaction and when it is spontaneous.

[2]
Question 25
HL • Paper 2
Medium
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HL • Paper 2
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At 298 K298\ \text{K}, a reaction has ΔG=5.70 kJ mol1\Delta G^\circ = -5.70\ \text{kJ mol}^{-1}.

A

Calculate the equilibrium constant, KK, for the reaction at 298 K298\ \text{K}.

[2]
B

State what the value of KK suggests about the equilibrium mixture.

[1]
Question 26
HL • Paper 1B
Medium
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HL • Paper 1B
Medium
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The table gives standard molar entropy values for the substances in the Haber process.

N2(g)+3H2(g)2NH3(g)\text{N}_2(g) + 3\text{H}_2(g) \to 2\text{NH}_3(g)

Substance

S^\circ / J K1\text{K}^{-1} mol1\text{mol}^{-1}

N2N_2(g)

191.6

H2H_2(g)

130.7

NH3NH_3(g)

192.8

A

Calculate the total standard entropy of the reactants and of the products for the reaction as written.

[2]
B

Determine ΔS\Delta S^\circ for the reaction.

[1]
C

Explain the sign of ΔS\Delta S^\circ in terms of the balanced equation.

[1]
Question 27
HL • Paper 1B
Medium
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HL • Paper 1B
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The thermal decomposition of calcium carbonate is represented by the equation.

CaCO3(s)CaO(s)+CO2(g)\text{CaCO}_3(s) \to \text{CaO}(s) + \text{CO}_2(g)

Thermodynamic data for the substances are shown.

Substance

ΔHf^\circ / kJ mol⁻¹

S^\circ / J K⁻¹ mol⁻¹

CaCO3\text{CaCO}_3(s)

-1207

92.9

CaO\text{CaO}(s)

-635

39.8

CO2\text{CO}_2(g)

-394

213.7

A

Predict whether the entropy of the system increases or decreases during the decomposition.

[1]
B

Calculate ΔS\Delta S^\circ for the decomposition reaction.

[2]
C

Using the data, evaluate whether the decomposition is spontaneous under standard conditions at 298 K298\ \text{K}.

[2]
Question 28
HL • Paper 1B
Medium
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HL • Paper 1B
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The table gives ΔH\Delta H^\circ and ΔS\Delta S^\circ values for four reactions.

Reaction

ΔH^\circ / kJ mol1\text{mol}^{-1}

ΔS^\circ / J K1\text{K}^{-1} mol1\text{mol}^{-1}

A

45

+160

B

-120

+250

C

-80

-120

D

+30

-50

A

Identify the reaction that is non-spontaneous at all temperatures.

[1]
B

Calculate ΔG\Delta G^\circ for reaction A at 298 K298\ \text{K}.

[2]
C

Determine the minimum temperature above which reaction A is spontaneous under standard conditions.

[1]
D

Explain why reaction A can be spontaneous even though it is endothermic.

[1]
Question 29
HL • Paper 1B
Medium
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HL • Paper 1B
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A pure liquid XX is heated at standard pressure. The graph shows ΔG\Delta G for the process X(l)X(g)X(l) \to X(g) as temperature changes. The enthalpy change of vaporization is also shown.

Straight-line plot of ΔG for vaporization of X against temperature, with ΔH vap marked at the y-intercept.
A

State the significance of the temperature at which the graph crosses ΔG=0\Delta G = 0.

[1]
B

Determine the normal boiling point of XX from the graph.

[1]
C

Calculate ΔS\Delta S for vaporization of XX at its boiling point.

[2]
D

Predict whether vaporization of XX is spontaneous at 400 K400\ \text{K}.

[1]
Question 30
HL • Paper 1B
Medium
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HL • Paper 1B
Medium
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Electrochemical data for three standard cells are shown. The relationship between standard Gibbs energy change and standard cell potential is ΔG=nFEcell\Delta G^\circ = -nFE^\circ_{\text{cell}}.

Cell

n

E^\circcell / V

A

2

1.10

B

1

0.76

C

2

-0.34

A

Identify which cells are spontaneous as written under standard conditions.

[1]
B

Calculate ΔG\Delta G^\circ, in kJ mol1\text{kJ mol}^{-1}, for cell A.

[2]
C

Explain why a positive EcellE^\circ_{\text{cell}} indicates a spontaneous cell reaction.

[1]
Question 31
HL • Paper 2
Medium
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HL • Paper 2
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For a reaction at 298 K298\ \text{K}, ΔG=+12.0 kJ mol1\Delta G^\circ = +12.0\ \text{kJ mol}^{-1}. At one instant the reaction quotient is Q=0.010Q = 0.010.

A

Calculate ΔG\Delta G under these non-standard conditions.

[3]
B

Deduce the direction favoured at this instant.

[1]
Question 32
HL • Paper 1B
Hard
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HL • Paper 1B
Hard
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The graph shows how ΔG\Delta G^\circ varies with temperature for a reaction under standard conditions.

Straight-line plot of standard Gibbs energy change against temperature.
A

Determine ΔH\Delta H^\circ for the reaction from the graph.

[1]
B

Calculate ΔS\Delta S^\circ for the reaction from the gradient.

[2]
C

Estimate the temperature at which the reaction becomes spontaneous and state whether it is spontaneous above or below this temperature.

[2]
Question 33
HL • Paper 1B
Hard
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HL • Paper 1B
Hard
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The table shows thermodynamic data for four reactions under standard conditions.

Reaction

ΔH^\circ / kJ mol1\text{mol}^{-1}

ΔS^\circ / J K1\text{K}^{-1} mol1\text{mol}^{-1}

E

-80

+100

F

+60

-90

G

+96

+120

H

-40

-100

A

Identify the reaction that is spontaneous at all temperatures and the reaction that is non-spontaneous at all temperatures.

[2]
B

Calculate the temperature at which reaction G becomes spontaneous under standard conditions.

[2]
C

Use the data to decide whether reaction H is spontaneous at 500 K500\ \text{K}.

[1]
Question 34
HL • Paper 1B
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HL • Paper 1B
Hard
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Ammonium nitrate dissolves endothermically in water. The table gives thermodynamic data for the dissolution process at 298 K298\ \text{K}.

NH4NO3(s)NH4+(aq)+NO3(aq)\text{NH}_4\text{NO}_3(s) \to \text{NH}_4^+(aq) + \text{NO}_3^-(aq)

Process

ΔH^\circ / kJ mol1\text{mol}^{-1}

ΔS^\circ / J K1\text{K}^{-1} mol1\text{mol}^{-1}

NH4NO3\text{NH}_4\text{NO}_3(s) \to NH4+\text{NH}_4^+(aq) + NO3\text{NO}_3^-(aq)

25.7

108.7

A

Calculate ΔSsurroundings\Delta S_{\text{surroundings}} for the dissolution at 298 K298\ \text{K}.

[2]
B

Calculate ΔStotal\Delta S_{\text{total}} and use it to determine whether the dissolution is spontaneous at 298 K298\ \text{K}.

[2]
C

Suggest why the dissolution can be spontaneous even though it is endothermic.

[1]
Question 35
HL • Paper 2
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HL • Paper 2
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Ethanol can exist as a solid, liquid or gas. A perfect crystal of ethanol is considered at very low temperature. The enthalpy of vaporization of ethanol at its normal boiling point, 351 K351\ \text{K}, is 38.6 kJ mol138.6\ \text{kJ mol}^{-1}.

A three-part particle diagram comparing ethanol as a perfect crystal, liquid and gas, showing increasing freedom of movement and dispersal without including numerical entropy values.
A

The entropy of a perfect crystal at 0 K0\ \text{K} is predicted to be zero.

I.

Define a perfect crystal in this context.

[1]
II.

Explain why the entropy is predicted to be zero at 0 K0\ \text{K}.

[1]
B

Calculate ΔS\Delta S for vaporization of ethanol at its normal boiling point, assuming ΔG=0\Delta G=0 at this temperature.

[2]
C

Use your answer to (b) to calculate ΔG\Delta G for vaporization at 298 K298\ \text{K} and explain the result.

[2]
Question 36
HL • Paper 1B
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HL • Paper 1B
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Methanol can be synthesized by the reaction.

CO(g)+2H2(g)CH3OH(g)\text{CO}(g) + 2\text{H}_2(g) \rightleftharpoons \text{CH}_3\text{OH}(g)

The table gives the composition of a reaction mixture at 500 K500\ \text{K} and the value of KK at this temperature.

Species / quantity

Concentration / mol dm3^{-3}

K

CO(g)\text{CO}(g)

0.10

H2(g)\text{H}_2(g)

0.50

CH3OH(g)\text{CH}_3\text{OH}(g)

0.20

Equilibrium constant at 500 K

14

A

Write the expression for the reaction quotient, QQ.

[1]
B

Calculate QQ for the mixture.

[2]
C

Use QQ and KK to calculate ΔG\Delta G for the forward reaction at 500 K500\ \text{K} and state the direction favoured.

[2]
Question 37
HL • Paper 1B
Hard
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HL • Paper 1B
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The reaction N2O4(g)2NO2(g)\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g) has a positive standard Gibbs energy change at 298 K298\ \text{K}. Data for a non-equilibrium mixture are shown.

Property / unit

Value

Temperature / K

298

ΔG^\circ / kJ mol1\text{mol}^{-1}

+4.80

p(N2O4)p(\text{N}_2\text{O}_4) / bar

0.50

p(NO2)p(\text{NO}_2) / bar

0.10

A

Calculate KK for the reaction at 298 K298\ \text{K}.

[2]
B

Predict the likely composition of the equilibrium mixture under standard-state comparison.

[1]
C

Calculate ΔG\Delta G for the non-equilibrium mixture and state the direction favoured.

[2]
Question 38
HL • Paper 2
Hard
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HL • Paper 2
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Calcium carbonate decomposes on heating according to the equation:

CaCO3(s)CaO(s)+CO2(g)\text{CaCO}_3(s) \to \text{CaO}(s) + CO_2(g)

For this reaction, ΔH=+178.3 kJ mol1\Delta H^\circ = +178.3\ \text{kJ mol}^{-1}. The standard entropy values are S(CaCO3(s))=92.9 J K1 mol1S^\circ(\text{CaCO}_3(s)) = 92.9\ \text{J K}^{-1}\text{ mol}^{-1}, S(CaO(s))=39.8 J K1 mol1S^\circ(\text{CaO}(s)) = 39.8\ \text{J K}^{-1}\text{ mol}^{-1} and S(CO2(g))=213.8 J K1 mol1S^\circ(CO_2(g)) = 213.8\ \text{J K}^{-1}\text{ mol}^{-1}.

A

The entropy change of the system is considered first.

I.

Predict, with a reason, the sign of ΔS\Delta S^\circ for the decomposition.

[1]
II.

Calculate ΔS\Delta S^\circ for the reaction.

[2]
B

Calculate the temperature at which the reaction just becomes spontaneous under standard conditions.

[2]
C

Evaluate why a high temperature is used industrially for this decomposition, even though the reaction is endothermic.

[2]
Question 39
HL • Paper 2
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HL • Paper 2
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Ammonia is formed in the Haber process:

N2(g)+3H2(g)2NH3(g)N_2(g)+3H_2(g) \rightleftharpoons 2NH_3(g)

For the forward reaction, ΔH=92.4 kJ mol1\Delta H^\circ = -92.4\ \text{kJ mol}^{-1}. Use S(N2(g))=191.5 J K1 mol1S^\circ(N_2(g))=191.5\ \text{J K}^{-1}\text{ mol}^{-1}, S(H2(g))=130.7 J K1 mol1S^\circ(H_2(g))=130.7\ \text{J K}^{-1}\text{ mol}^{-1} and S(NH3(g))=192.8 J K1 mol1S^\circ(NH_3(g))=192.8\ \text{J K}^{-1}\text{ mol}^{-1}.

A

Consider the entropy change for the forward reaction.

I.

Explain the expected sign of ΔS\Delta S^\circ using the balanced equation.

[1]
II.

Calculate ΔS\Delta S^\circ for the reaction.

[1]
B

Calculate ΔG\Delta G^\circ for the forward reaction at 298 K298\ \text{K} and at 700 K700\ \text{K}.

[3]
C

Discuss the effect of temperature on the thermodynamic favourability of ammonia formation and distinguish this from the rate of reaction.

[3]
Question 40
HL • Paper 2
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HL • Paper 2
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Ammonium nitrate dissolves endothermically in water and is used in some cold packs:

NH4NO3(s)NH4+(aq)+NO3(aq)NH_4NO_3(s) \to NH_4^+(aq)+NO_3^-(aq)

For this process, ΔH=+25.7 kJ mol1\Delta H^\circ = +25.7\ \text{kJ mol}^{-1} and ΔSsystem=+108.7 J K1 mol1\Delta S^\circ_{\text{system}} = +108.7\ \text{J K}^{-1}\text{ mol}^{-1}.

A

The process absorbs heat from the surroundings.

I.

Explain why the entropy change of the system is positive.

[1]
II.

State the sign of the entropy change of the surroundings and give a reason.

[1]
B

Calculate ΔG\Delta G^\circ at 298 K298\ \text{K} and hence determine whether dissolving is spontaneous under standard conditions.

[2]
C

Evaluate the statement: “An endothermic process cannot be spontaneous.”

[3]
Question 41
HL • Paper 2
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HL • Paper 2
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The graph shows how ΔG\Delta G^\circ varies with temperature for two reactions, A and B. The straight-line relationships are:

A: ΔG=84.00.120T\Delta G^\circ = 84.0 - 0.120T

B: ΔG=50.0+0.080T\Delta G^\circ = -50.0 + 0.080T

ΔG\Delta G^\circ is in kJ mol1\text{kJ mol}^{-1} and TT is in K\text{K}.

ΔG° against temperature for reactions A and B, with the zero line marked.
A

Use the form of the Gibbs equation to interpret the graph.

I.

Deduce ΔH\Delta H^\circ and ΔS\Delta S^\circ for reaction A.

[2]
II.

Deduce the signs of ΔH\Delta H^\circ and ΔS\Delta S^\circ for reaction B.

[1]
B

Compare the temperature ranges over which A and B are spontaneous under standard conditions.

[2]
C

Explain why the two reactions show opposite temperature dependences.

[2]
Question 42
HL • Paper 2
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HL • Paper 2
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A standard zinc-copper voltaic cell has the overall reaction:

Zn(s)+Cu2+(aq)Zn2+(aq)+Cu(s)Zn(s)+Cu^{2+}(aq) \to Zn^{2+}(aq)+Cu(s)

For the cell, Ecell=+1.10 VE^\circ_{\text{cell}}=+1.10\ \text{V} and n=2n=2. Use F=96485 C mol1F=96485\ \text{C mol}^{-1} and R=8.31 J K1 mol1R=8.31\ \text{J K}^{-1}\text{ mol}^{-1}.

A labelled galvanic cell diagram with zinc and copper half-cells connected by a salt bridge and voltmeter; labels show $Zn$, $Zn^{2+}$, $Cu$, $Cu^{2+}$ but do not include calculated thermodynamic values.
A

Electrochemical data can be used to predict spontaneity.

I.

Calculate ΔG\Delta G^\circ for the cell reaction.

[2]
II.

State what the sign of ΔG\Delta G^\circ shows about the reaction as written.

[1]
B

Calculate the equilibrium constant at 298 K298\ \text{K} for the cell reaction.

[2]
C

In a non-standard cell, [Zn2+]=1.00 mol dm3[Zn^{2+}]=1.00\ \text{mol dm}^{-3} and [Cu2+]=0.0100 mol dm3[Cu^{2+}]=0.0100\ \text{mol dm}^{-3}. Calculate ΔG\Delta G at 298 K298\ \text{K} and discuss whether the cell can still do electrical work.

[2]
Question 43
HL • Paper 2
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HL • Paper 2
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Two reactions can be coupled so that the product of one is consumed by the other.

Reaction 1: ABA \to B, ΔG=+22.5 kJ mol1\Delta G^\circ = +22.5\ \text{kJ mol}^{-1}

Reaction 2: CDC \to D, ΔG=31.0 kJ mol1\Delta G^\circ = -31.0\ \text{kJ mol}^{-1}

The coupled process is A+CB+DA+C \to B+D.

A

Consider the thermodynamics of the separate reactions.

I.

State whether reaction 1 is spontaneous under standard conditions, with a reason.

[1]
II.

State whether reaction 2 is thermodynamically favourable under standard conditions, with a reason.

[1]
B

Calculate ΔG\Delta G^\circ and KK at 298 K298\ \text{K} for the coupled process.

[3]
C

Evaluate whether a negative ΔG\Delta G^\circ guarantees that the coupled process will occur rapidly.

[1]
Question 44
HL • Paper 2
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HL • Paper 2
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Carbon monoxide is oxidized in catalytic converters:

2CO(g)+O2(g)2CO2(g)2CO(g)+O_2(g) \to 2CO_2(g)

For this reaction, ΔH=566.0 kJ mol1\Delta H^\circ=-566.0\ \text{kJ mol}^{-1}. Use S(CO(g))=197.7 J K1 mol1S^\circ(CO(g))=197.7\ \text{J K}^{-1}\text{ mol}^{-1}, S(O2(g))=205.2 J K1 mol1S^\circ(O_2(g))=205.2\ \text{J K}^{-1}\text{ mol}^{-1} and S(CO2(g))=213.8 J K1 mol1S^\circ(CO_2(g))=213.8\ \text{J K}^{-1}\text{ mol}^{-1}.

A

Consider entropy and the balanced equation.

I.

Predict the sign of ΔS\Delta S^\circ and justify your prediction.

[1]
II.

Calculate ΔS\Delta S^\circ for the reaction.

[2]
B

Calculate ΔG\Delta G^\circ at 298 K298\ \text{K} and interpret its sign.

[2]
C

Discuss how increasing temperature affects the spontaneity of this reaction.

[2]
Question 45
HL • Paper 2
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HL • Paper 2
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Phosphorus pentachloride dissociates in a sealed vessel:

PCl5(g)PCl3(g)+Cl2(g)PCl_5(g) \rightleftharpoons PCl_3(g)+Cl_2(g)

At 500 K500\ \text{K}, ΔG=+13.4 kJ mol1\Delta G^\circ = +13.4\ \text{kJ mol}^{-1} for the forward reaction. At one instant the reaction quotient is Q=0.050Q=0.050.

A

The reaction quotient has the same form as the equilibrium expression.

I.

Write the expression for QQ for this reaction.

[1]
II.

State why pure solids and pure liquids, if present, would be omitted from QQ.

[1]
B

Calculate ΔG\Delta G at this instant and determine the direction in which the mixture will change. Use R=8.31 J K1 mol1R=8.31\ \text{J K}^{-1}\text{ mol}^{-1}.

[3]
C

Calculate KK at 500 K500\ \text{K} and use it to support your conclusion in (b).

[3]
Question 46
HL • Paper 2
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HL • Paper 2
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Dinitrogen tetroxide dissociates according to:

N2O4(g)2NO2(g)N_2O_4(g) \rightleftharpoons 2NO_2(g)

At 298 K298\ \text{K}, ΔG=+4.73 kJ mol1\Delta G^\circ = +4.73\ \text{kJ mol}^{-1} for the reaction as written. In a mixture at 298 K298\ \text{K}, [NO2]=0.0550 mol dm3[NO_2]=0.0550\ \text{mol dm}^{-3} and [N2O4]=0.0250 mol dm3[N_2O_4]=0.0250\ \text{mol dm}^{-3}.

A

Relate the standard Gibbs energy change to the equilibrium constant.

I.

Calculate KK at 298 K298\ \text{K}.

[2]
II.

Interpret the value of KK in terms of the likely equilibrium composition.

[1]
B

For the mixture described, calculate QQ and ΔG\Delta G.

[3]
C

Evaluate whether the sign of ΔG\Delta G^\circ alone is sufficient to predict the direction of change for this mixture.

[2]
Question 47
HL • Paper 2
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HL • Paper 2
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Hydrogen and iodine react reversibly at 700 K700\ \text{K}:

H2(g)+I2(g)2HI(g)H_2(g)+I_2(g) \rightleftharpoons 2HI(g)

At 700 K700\ \text{K}, K=54.0K=54.0. In a mixture at this temperature, [HI]=0.100 mol dm3[HI]=0.100\ \text{mol dm}^{-3}, [H2]=0.300 mol dm3[H_2]=0.300\ \text{mol dm}^{-3} and [I2]=0.200 mol dm3[I_2]=0.200\ \text{mol dm}^{-3}.

ΔG becomes less negative as Q rises toward equilibrium at 700 K.
A

Use the equilibrium constant to determine the standard Gibbs energy change.

I.

Calculate ΔG\Delta G^\circ at 700 K700\ \text{K}.

[2]
II.

State what the sign of ΔG\Delta G^\circ indicates about the standard-state equilibrium comparison.

[1]
B

For the mixture described, calculate QQ and ΔG\Delta G at 700 K700\ \text{K}.

[3]
C

Explain how ΔG\Delta G changes as this mixture approaches equilibrium.

[2]
Question 48
HL • Paper 2
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HL • Paper 2
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The dissolution equilibrium for barium sulfate is:

BaSO4(s)Ba2+(aq)+SO42(aq)BaSO_4(s) \rightleftharpoons Ba^{2+}(aq)+SO_4^{2-}(aq)

At 298 K298\ \text{K}, Ksp=1.1×1010K_{sp}=1.1\times10^{-10}. A mixture contains [Ba2+]=5.0×105 mol dm3[Ba^{2+}]=5.0\times10^{-5}\ \text{mol dm}^{-3} and [SO42]=4.0×106 mol dm3[SO_4^{2-}]=4.0\times10^{-6}\ \text{mol dm}^{-3}.

A

The equilibrium constant can be used to calculate a standard Gibbs energy change.

I.

Calculate ΔG\Delta G^\circ for the dissolution process at 298 K298\ \text{K}.

[2]
II.

Interpret the sign of ΔG\Delta G^\circ for a saturated solution under standard-state comparison.

[1]
B

For the mixture described, calculate QQ and determine whether precipitation or further dissolution is favoured.

[2]
C

Evaluate why forming separated aqueous ions from a solid does not necessarily mean that dissolution is spontaneous.

[2]

R1.3 Energy from fuels

R2.1 How much? The amount of chemical change