Entropy and spontaneity practice questions.

A reaction has ΔH < 0 (exothermic) and ΔS > 0 (entropy increases). At which temperatures is it spontaneous?

• at all temperatures
• at no temperature
• only at high temperature
• only at low temperature

Answer: at all temperatures

ΔG = ΔH − TΔS. With ΔH negative and −TΔS also negative (because ΔS > 0), ΔG is negative at every temperature, so the reaction is always spontaneous.

Define entropy, S, as understood in thermodynamics.

• A measure of the dispersal or distribution of energy and matter in a system; greater disorder means greater entropy
• The total heat content of a system at constant pressure
• The maximum total amount of energy that is available to do useful work in a process carried out at constant temperature and pressure
• The rate at which a reaction reaches equilibrium

Answer: A measure of the dispersal or distribution of energy and matter in a system; greater disorder means greater entropy

Entropy quantifies how energy and matter are dispersed among available microstates; more possible arrangements (more disorder) means higher entropy.

For a reaction ΔH = −92 kJ mol⁻¹ and ΔS = −198 J K⁻¹ mol⁻¹. Determine ΔG at 298 K.

• -33.0 kJ mol⁻¹
• -151 kJ mol⁻¹
• +59 000 kJ mol⁻¹
• +33.0 kJ mol⁻¹

Answer: -33.0 kJ mol⁻¹

ΔG = ΔH − TΔS = −92 − (298 × −0.198) = −92 + 59.0 = −33.0 kJ mol⁻¹ (convert ΔS to kJ: −0.198 kJ K⁻¹ mol⁻¹).

Determine the sign of ΔS° for the reaction 2SO₂(g) + O₂(g) → 2SO₃(g).

• Negative, because the number of moles of gas decreases from 3 to 2
• Positive, because a new bond is formed
• Zero, because it is a reversible equilibrium reaction at constant temperature
• Positive, because energy is released

Answer: Negative, because the number of moles of gas decreases from 3 to 2

Gas moles fall from 3 to 2, reducing the number of ways to disperse matter/energy, so ΔS° is negative.

A reaction has ΔH = +178 kJ mol⁻¹ and ΔS = +161 J K⁻¹ mol⁻¹ (the decomposition of CaCO₃). Calculate the minimum temperature, in K, above which the reaction becomes spontaneous.

• 1106 K
• 884 K
• 1.11 K
• 339 K

Answer: 1106 K

At the crossover ΔG = 0, so T = ΔH/ΔS = 178 000 J / 161 J K⁻¹ = 1106 K; above this ΔG < 0.

A reaction has ΔH > 0 and ΔS < 0. Determine its spontaneity.

• Non-spontaneous at all temperatures
• Spontaneous at all temperatures
• Spontaneous only at high temperature
• Spontaneous only at low temperature

Answer: Non-spontaneous at all temperatures

ΔG = ΔH − TΔS: with ΔH positive and −TΔS positive (since ΔS < 0), ΔG is positive at every temperature, so it is never spontaneous.