Practice exam-style IB Chemistry questions for Entropy and spontaneity (AHL), aligned with the syllabus and grouped by topic.
The standard molar entropy of one mole of a substance is greatest for which state under the same conditions?
Liquid
Aqueous precipitate
Perfect crystalline solid
Gas
At constant temperature and pressure, what does ΔG < 0 indicate for the forward reaction?
It must be rapid
It is thermodynamically spontaneous
It is at equilibrium
It is non-spontaneous
For the reaction
2SO₂
Negative, because all species are gases
Negative, because the number of moles of gas decreases
Positive, because three moles of gas form two moles of gas
Positive, because the reaction forms a compound
For A
−350 J K⁻¹ mol⁻¹
−140 J K⁻¹ mol⁻¹
+850 J K⁻¹ mol⁻¹
+350 J K⁻¹ mol⁻¹
In a graph of ΔG⦵ against temperature for a reaction, what does the gradient represent?
TΔS⦵
ΔH⦵
−TΔH⦵
−ΔS⦵
A reaction has ΔH⦵ < 0 and ΔS⦵ < 0. Under standard conditions, when is it most likely to be spontaneous?
At no temperature
At high temperature only
At low temperature only
At all temperatures
If ΔG⦵ is positive for a reaction at a given temperature, what is true about K?
K = 1; neither side is favoured
K < 0; the reaction cannot occur
K > 1; products are favoured
K < 1; reactants are favoured
State the unit of standard molar entropy.
Explain why Br₂
has a higher entropy than Br₂
under the same conditions.
Define a perfect crystal.
State why the entropy of a perfect crystal at 0 K is predicted to be zero.
Distinguish between Q and K.
State the value of ΔG when Q = K.
For a reaction, ΔH⦵ = +48.0 kJ mol⁻¹ and ΔS⦵ = +120 J K⁻¹ mol⁻¹. What is ΔG⦵ at 298 K?
−12.2 kJ mol⁻¹
+12.2 kJ mol⁻¹
+47.9 kJ mol⁻¹
−35.8 kJ mol⁻¹
For a reaction, ΔH⦵ = +72.0 kJ mol⁻¹ and ΔS⦵ = +240 J K⁻¹ mol⁻¹. Above what temperature is the reaction spontaneous under standard conditions?
1.73 × 10⁴ K
0.00333 K
Above no temperature
300 K
A reaction is thermodynamically spontaneous at 298 K but is not observed to occur over several days. What is the best explanation?
The entropy change must be zero
The activation energy may be large
The reaction must be at equilibrium
ΔG must be positive at 298 K
For an electrochemical cell reaction, n = 2 and E⦵cell = +0.54 V. What is the sign of ΔG⦵ and the spontaneity of the reaction as written?
ΔG⦵ is negative; spontaneous
ΔG⦵ is positive; spontaneous
ΔG⦵ is negative; non-spontaneous
ΔG⦵ is positive; non-spontaneous
For a reversible reaction at a fixed temperature, Q < K. What is the sign of ΔG for the forward reaction?
Equal to ΔG⦵
Negative
Zero
Positive
For N₂O₄
[NO₂]/[N₂O₄]
[NO₂]²/[N₂O₄]
[N₂O₄]²/[NO₂]
[N₂O₄]/[NO₂]²
At 298 K, ΔG⦵ = −12.0 kJ mol⁻¹ and Q = 0.050. What is the sign of ΔG?
Negative
Zero
Positive
Cannot be determined without K
For each change, predict the sign of ΔS for the system and give a reason.
H₂O
→ H₂O
2NH₃
→ N₂
Methanol can be formed by
CO
| Substance | S° / J K⁻¹ mol⁻¹ |
|---|---|
| CO(g) | 198 |
| H₂(g) | 131 |
| CH₃OH(l) | 127 |
→ CH₃OH
The standard molar entropies are: CO
= 198 J K⁻¹ mol⁻¹, H₂
= 131 J K⁻¹ mol⁻¹, CH₃OH
= 127 J K⁻¹ mol⁻¹.
Calculate ΔS⦵ for the reaction.
State why the sign is expected.
For a reaction at 350 K, ΔH⦵ = −25.0 kJ mol⁻¹ and ΔS⦵ = −60.0 J K⁻¹ mol⁻¹.
Calculate ΔG⦵.
State whether the reaction is spontaneous under standard conditions at 350 K.
A reaction has ΔH⦵ = +95 kJ mol⁻¹ and ΔS⦵ = +250 J K⁻¹ mol⁻¹.
Determine the temperature at which ΔG⦵ = 0.
State whether the reaction is spontaneous above or below this temperature.
A dissolving process is endothermic but occurs spontaneously at room temperature.
State the sign of ΔH for the process.
Explain how the process can still be spontaneous.
For the reaction X(aq) ⇌ Y(aq), K = 25.0 at 310 K.
Calculate ΔG⦵ in kJ mol⁻¹. Use R = 8.31 J K⁻¹ mol⁻¹.
State what the value suggests about the equilibrium mixture.
A straight-line plot of ΔG⦵ against temperature for a reaction has a positive vertical intercept and a negative gradient.

State the sign of ΔH⦵.
State the sign of ΔS⦵.
Predict whether increasing temperature makes the reaction more or less spontaneous.
The combustion of a hydrocarbon is exothermic and spontaneous at 298 K.
State the effect of heat released on the entropy of the surroundings.
Explain why ΔG can be negative even if the entropy of the chemical system decreases.
The graph shows ΔG⦵ against temperature for reaction I.

Determine the temperature at which reaction I changes from non-spontaneous to spontaneous under standard conditions.
State the sign of ΔH⦵ for reaction I.
Explain how the sign of ΔS⦵ can be obtained from the graph.
Predict whether reaction I is spontaneous at temperatures above the intercept with the temperature axis.
The table gives standard molar entropies for substances involved in the reaction:
CaCO₃
| Substance | S⦵ / J K⁻¹ mol⁻¹ |
|---|---|
| CaCO₃(s) | 92.9 |
| CaO(s) | 39.8 |
| CO₂(g) | 213.7 |
→ CaO
Calculate ΔS⦵ for the reaction using the table.
Explain why the sign of ΔS⦵ is expected from the equation.
Suggest why the physical states must be included when using entropy data.
The diagram shows a heating curve for a pure substance with labelled solid, liquid and gas regions.

Identify the region with the highest entropy.
State whether entropy increases or decreases during freezing.
Explain the entropy change during vaporization in terms of matter and energy dispersal.
At 298 K, a reaction has ΔG⦵ = +5.71 kJ mol⁻¹. What is K? Use R = 8.31 J K⁻¹ mol⁻¹.
10.0
2.30
0.100
0.0100
For A(aq) ⇌ B(aq), ΔG⦵ = +8.0 kJ mol⁻¹ at 298 K. In a mixture where Q is very small, the forward reaction can be spontaneous because
ΔG⦵ becomes negative at equilibrium
K becomes equal to zero
the entropy term is omitted outside standard conditions
RT ln Q can be sufficiently negative
For a reaction at 298 K, ΔG⦵ = −18.5 kJ mol⁻¹ and ΔS⦵ = +75.0 J K⁻¹ mol⁻¹.
Rearrange the Gibbs equation to make ΔH⦵ the subject.
Calculate ΔH⦵.
At 298 K an electrochemical reaction transfers 3 mol of electrons per mole of reaction and has E⦵cell = +0.42 V. Use F = 9.65 × 10⁴ C mol⁻¹.
Calculate ΔG⦵ in kJ mol⁻¹.
State whether the reaction is spontaneous as written.
For A
⇌ B(g), ΔG⦵ = +4.00 kJ mol⁻¹ at 298 K. In a mixture, Q = 0.100. Use R = 8.31 J K⁻¹ mol⁻¹.
Calculate ΔG in kJ mol⁻¹.
Predict the direction in which the reaction will proceed.
For PCl₅
⇌ PCl₃
Calculate Q.
State the direction in which the reaction will proceed to reach equilibrium.
For a reaction at 298 K, ΔG⦵ = +16.0 kJ mol⁻¹.
Calculate K. Use R = 8.31 J K⁻¹ mol⁻¹.
Comment on the claim: “A positive ΔG⦵ means no products will form.”
A graph shows ΔG⦵ for four reactions at 298 K, labelled A–D, with their ΔH⦵ and ΔS⦵ sign combinations indicated.
| Reaction | ΔH° sign | ΔS° sign | As T increases, ΔG° |
|---|---|---|---|
| A | − | + | Decreases |
| B | + | − | Increases |
| C | + | + | Decreases |
| D | − | − | Increases |
Identify the reaction that is spontaneous at all temperatures.
Identify the reaction that is non-spontaneous at all temperatures.
For the reaction with ΔH⦵ > 0 and ΔS⦵ > 0, state whether it is favoured at high or low temperature.
Explain your answer to
using ΔG⦵ = ΔH⦵ − TΔS⦵.
The graph shows ΔG against ln Q for a reaction at a fixed temperature.

State the relationship between the gradient of the graph and temperature.
Determine ΔG⦵ from the graph.
Identify the point on the graph corresponding to equilibrium.
Explain how the graph shows that ΔG becomes less negative as the reaction approaches equilibrium from a reactant-rich mixture.
The table shows concentrations for three mixtures of the equilibrium:
H₂
| Mixture | [H₂] / mol dm⁻³ | [I₂] / mol dm⁻³ | [HI] / mol dm⁻³ | Kc |
|---|---|---|---|---|
| A | 0.0400 | 0.0200 | 0.200 | 50.0 |
| B | 0.0300 | 0.0300 | 0.150 | 50.0 |
| C | 0.0500 | 0.0200 | 0.300 | 50.0 |
⇌ 2HI
at the same temperature. The equilibrium constant Kc is given.
Write the expression for Qc.
Use the table to determine which mixture is at equilibrium.
For a mixture with Qc < Kc, state the sign of ΔG for the forward reaction.
Explain why pure solids, if present in an equilibrium expression, would be omitted from Q.
The table gives ΔG⦵ values for three reactions at 298 K.
| Reaction | ΔG⦵ / kJ mol⁻¹ |
|---|---|
| 1 | −8.6 |
| 2 | +0.4 |
| 3 | +14.2 |
For reaction 1, calculate K using ΔG⦵ = −RT ln K.
Identify which reaction has the most product-favoured equilibrium.
Explain the relationship between the sign of ΔG⦵ and the magnitude of K.
A student investigates a reaction and plots ΔG⦵ against temperature. The best-fit line and uncertainty range are shown.

Determine whether ΔH⦵ is endothermic or exothermic from the graph.
Determine whether ΔS⦵ is positive or negative from the graph.
Estimate the temperature at which the reaction becomes spontaneous.
Evaluate one limitation of using this straight-line extrapolation to predict spontaneity at much higher temperatures.
A table gives ΔH⦵ and ΔS⦵ for four proposed industrial reactions.
| Reaction | ΔH⦵ / kJ mol⁻¹ | ΔS⦵ / J K⁻¹ mol⁻¹ |
|---|---|---|
| A | −84 | +118 |
| B | −42 | −180 |
| C | +96 | +220 |
| D | +58 | −140 |
Identify the reaction that is spontaneous at 298 K by calculating or comparing ΔG⦵ values.
Identify any reaction whose spontaneity changes with temperature.
For one reaction that changes spontaneity with temperature, determine whether increasing temperature favours products or reactants.
Explain why the conclusion is thermodynamic and does not predict reaction rate.
A reaction has ΔH⦵ = +110 kJ mol⁻¹ and ΔS⦵ = +220 J K⁻¹ mol⁻¹.
Calculate the temperature at which ΔG⦵ = 0.
Discuss the spontaneity of the reaction below and above this temperature, and explain why an endothermic reaction can be spontaneous.
Two processes are considered:
I. NH₄NO₃
→ NH₄⁺(aq) + NO₃⁻(aq)
II. 2NO₂
→ N₂O₄
Predict the sign of ΔS for each process.
Compare and contrast the entropy changes in the two processes in terms of dispersal of matter and energy, and discuss how entropy contributes to spontaneity.
Entropy values for substances in their standard states are absolute values, unlike standard enthalpies of formation.
State the third-law prediction for a perfect crystal at 0 K and define a perfect crystal.
Explain how entropy data can be used to calculate standard entropy changes and why balanced coefficients and physical states are essential.
The graph shows Gibbs energy of a reacting mixture as a function of reaction progress for a reversible reaction at constant temperature and pressure.

Identify the point corresponding to equilibrium.
State the sign of ΔG for the forward reaction at a point to the left of equilibrium.
State the sign of ΔG for the forward reaction at a point to the right of equilibrium.
Explain why ΔG⦵ for the reaction need not be zero even though ΔG is zero at equilibrium.
For the equilibrium
A
⇌ B
ΔG⦵ = +6.50 kJ mol⁻¹ at 298 K.
Calculate K at 298 K. Use R = 8.31 J K⁻¹ mol⁻¹.
Evaluate how the direction of spontaneous change depends on the composition of the mixture, referring to Q, K and ΔG.
The reaction
C₂H₄
| Species | S⦵ / J K⁻¹ mol⁻¹ |
|---|---|
| C₂H₄(g) | 220 |
| H₂(g) | 131 |
| C₂H₆(g) | 230 |
→ C₂H₆
has ΔH⦵ = −137 kJ mol⁻¹. Standard molar entropy data are provided: C₂H₄
= 220 J K⁻¹ mol⁻¹, H₂
= 131 J K⁻¹ mol⁻¹, C₂H₆
= 230 J K⁻¹ mol⁻¹.
Calculate ΔS⦵ for the reaction.
Evaluate the effect of increasing temperature on the spontaneity of the reaction under standard conditions.
A process has ΔH = −40 kJ mol⁻¹ and ΔSsystem = −80 J K⁻¹ mol⁻¹ at 298 K.
Calculate ΔG at 298 K.
Discuss the relationship between ΔG, total entropy change, system entropy and surroundings entropy for this process.
For the reaction
2A
⇌ B
at 350 K, ΔG⦵ = −9.50 kJ mol⁻¹. A mixture contains A and B such that Q = 18.0.
Calculate ΔG for the mixture. Use R = 8.31 J K⁻¹ mol⁻¹.
Evaluate whether the mixture is at equilibrium and predict how its composition will change.
Two reactions have the following thermodynamic data at 298 K.
Reaction X: ΔH⦵ = −60 kJ mol⁻¹, ΔS⦵ = −150 J K⁻¹ mol⁻¹
Reaction Y: ΔH⦵ = +60 kJ mol⁻¹, ΔS⦵ = +150 J K⁻¹ mol⁻¹
Calculate ΔG⦵ for both reactions at 298 K.
Compare and contrast the temperature dependence of spontaneity for X and Y.