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Difficulty: MediumEntropy, Free Energy and Reaction Spontaneity

Match each thermodynamic condition or combination on the left with its corresponding reaction spontaneity description on the right.

  • ΔG<0\Delta G < 0The reaction is spontaneous under the given conditions.
  • ΔG=0\Delta G = 0The system has reached dynamic equilibrium.
  • ΔH<0\Delta H < 0 and ΔS>0\Delta S > 0The reaction is spontaneous at all temperatures.
  • ΔH>0\Delta H > 0 and ΔS<0\Delta S < 0The reaction is non-spontaneous at all temperatures.

Answer

ΔG<0\Delta G < 0 matches with 'The reaction is spontaneous under the given conditions'; ΔG=0\Delta G = 0 matches with 'The system has reached dynamic equilibrium'; ΔH<0\Delta H < 0 and ΔS>0\Delta S > 0 matches with 'The reaction is spontaneous at all temperatures'; ΔH>0\Delta H > 0 and ΔS<0\Delta S < 0 matches with 'The reaction is non-spontaneous at all temperatures'.
Matching ΔG<0\Delta G < 0 to spontaneity under specified conditions, ΔG=0\Delta G = 0 to dynamic equilibrium, ΔH<0\Delta H < 0 and ΔS>0\Delta S > 0 to spontaneity at all temperatures, and ΔH>0\Delta H > 0 and ΔS<0\Delta S < 0 to non-spontaneity at all temperatures follows directly from the Gibbs-Helmholtz relation ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S.

Step-by-Step Solution

1
Recall the fundamental thermodynamic criterion for spontaneity involving Gibbs free energy change (ΔG)(\Delta G).
A reaction is spontaneous when ΔG<0\Delta G < 0, non-spontaneous when ΔG>0\Delta G > 0, and at dynamic equilibrium when ΔG=0\Delta G = 0.
Gibbs free energy combines enthalpy and entropy factors to determine direction of feasible chemical changes.
2
Analyze the Gibbs-Helmholtz equation ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S for sign combinations.
If ΔH<0\Delta H < 0 and ΔS>0\Delta S > 0, ΔG=(−value)−T(+value)\Delta G = (-\text{value}) - T(+\text{value}), which is always negative at any absolute temperature T>0 KT > 0\text{ K}.
Exothermic enthalpy releases energy while positive entropy increases disorder, driving spontaneity unconditionally.
3
Evaluate the opposing sign combination where ΔH>0\Delta H > 0 and ΔS<0\Delta S < 0.
ΔG=(+value)−T(−value)=+value+T(value)\Delta G = (+\text{value}) - T(-\text{value}) = +\text{value} + T(\text{value}), which is always positive.
Endothermic process with decreasing entropy is thermodynamically unfavorable at all temperatures.

Key Concept

Gibbs Free Energy Equation and Reaction Spontaneity
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Topics

Entropy, Free Energy and Reaction Spontaneity