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Difficulty: Very hardEntropy, Free Energy and Reaction Spontaneity

Match each set of thermodynamic conditions for a chemical reaction on the left with its corresponding temperature-dependent spontaneity behavior on the right, based on the Gibbs free energy relationship ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S.

  • Endothermic process (ΔH>0\Delta H > 0) accompanied by an increase in entropy (ΔS>0\Delta S > 0)Spontaneous only at high temperatures where T>ΔHΔST > \frac{\Delta H}{\Delta S}
  • Exothermic process (ΔH<0\Delta H < 0) accompanied by a decrease in entropy (ΔS<0\Delta S < 0)Spontaneous only at low temperatures where T<ΔHΔST < \frac{\Delta H}{\Delta S}
  • Endothermic process (ΔH>0\Delta H > 0) accompanied by a decrease in entropy (ΔS<0\Delta S < 0)Non-spontaneous at all temperatures (ΔG>0\Delta G > 0 under all conditions)
  • Exothermic process (ΔH<0\Delta H < 0) accompanied by an increase in entropy (ΔS>0\Delta S > 0)Spontaneous at all temperatures (ΔG<0\Delta G < 0 under all conditions)

Answer

1. Endothermic with ΔS>0\Delta S > 0 matches Spontaneous only at high temperatures (T>ΔHΔST > \frac{\Delta H}{\Delta S}).
2. Exothermic with ΔS<0\Delta S < 0 matches Spontaneous only at low temperatures (T<ΔHΔST < \frac{\Delta H}{\Delta S}).
3. Endothermic with ΔS<0\Delta S < 0 matches Non-spontaneous at all temperatures.
4. Exothermic with ΔS>0\Delta S > 0 matches Spontaneous at all temperatures.
Each pair is matched by evaluating the sign of ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S. An endothermic reaction with positive entropy change requires high temperature to make TΔS>ΔHT\Delta S > \Delta H. An exothermic reaction with negative entropy change requires low temperature to keep ∣ΔH∣>T∣ΔS∣|\Delta H| > T|\Delta S|. An endothermic reaction with negative entropy change yields positive ΔG\Delta G at all temperatures. An exothermic reaction with positive entropy change yields negative ΔG\Delta G at all temperatures.

Step-by-Step Solution

1
Recall the fundamental thermodynamic criterion for spontaneity
A reaction is spontaneous when Gibbs free energy change is negative (ΔG<0\Delta G < 0), given by ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S.
Temperature TT in Kelvin is always positive, so the sign of ΔG\Delta G depends on the combination of signs of ΔH\Delta H and ΔS\Delta S.
2
Analyze Case 1: ΔH>0\Delta H > 0 and ΔS>0\Delta S > 0
ΔG=(+value)−T(+value)\Delta G = (+\text{value}) - T(+\text{value}). For ΔG<0\Delta G < 0, TΔS>ΔH  ⟹  T>ΔHΔST\Delta S > \Delta H \implies T > \frac{\Delta H}{\Delta S}.
The reaction is driven by entropy increase at elevated temperatures.
3
Analyze Case 2: ΔH<0\Delta H < 0 and ΔS<0\Delta S < 0
ΔG=(−value)−T(−value)=−∣ΔH∣+T∣ΔS∣\Delta G = (-\text{value}) - T(-\text{value}) = -|\Delta H| + T|\Delta S|. For ΔG<0\Delta G < 0, ∣ΔH∣>T∣ΔS∣  ⟹  T<ΔHΔS|\Delta H| > T|\Delta S| \implies T < \frac{\Delta H}{\Delta S}.
The reaction is enthalpy-driven and requires low temperatures so that the positive −TΔS-T\Delta S term does not outweigh ΔH\Delta H.
4
Analyze Case 3: ΔH>0\Delta H > 0 and ΔS<0\Delta S < 0
ΔG=(+value)−T(−value)=(+value)+T(+value)>0\Delta G = (+\text{value}) - T(-\text{value}) = (+\text{value}) + T(+\text{value}) > 0 always.
Both enthalpy and entropy factors oppose spontaneity.
5
Analyze Case 4: ΔH<0\Delta H < 0 and ΔS>0\Delta S > 0
ΔG=(−value)−T(+value)<0\Delta G = (-\text{value}) - T(+\text{value}) < 0 always.
Both enthalpy release and entropy increase favor spontaneity under all conditions.

Key Concept

Dependence of Gibbs Free Energy and Reaction Spontaneity on the Signs of Enthalpy and Entropy Changes
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Topics

Entropy, Free Energy and Reaction Spontaneity