Gibbs Free Energy: Is This Reaction Spontaneous?
For a reaction with delta H = -120 kJ and delta S = -0.200 kJ/K, calculate delta G at T = 298 K and state whether the reaction is spontaneous.
Given
Enthalpy change: delta H = -120 kJ
Entropy change: delta S = -0.200 kJ/K
Temperature: T = 298 K
Find
Delta G at 298 K and whether the reaction is spontaneous.
Principle
Gibbs free energy combines enthalpy and entropy into one verdict on spontaneity: delta G = delta H - T delta S. A negative delta G means the reaction can proceed on its own under constant temperature and pressure. Temperature decides which term dominates — enthalpy at low T, entropy at high T.
Formula
Solution
Compute the entropy term: T x delta S = 298 K x (-0.200 kJ/K) = -59.6 kJ.
Substitute into delta G = delta H - T delta S: delta G = (-120 kJ) - (-59.6 kJ).
Simplify: -120 + 59.6 = -60.4 kJ.
Since delta G = -60.4 kJ < 0, the reaction is spontaneous at 298 K.
Unit check
K x kJ/K gives kJ, so both terms on the right-hand side of delta G = delta H - T delta S are in kJ, and delta G comes out in kJ. Consistent units are what make the subtraction legal.
What this means physically
Even though the entropy change is unfavorable (more order), the strongly exothermic enthalpy wins at room temperature, so delta G is negative and the reaction proceeds spontaneously. Raise the temperature enough and the -T delta S penalty would eventually flip delta G positive — entropy dominates at high T.
Students mix units — plugging delta H in kJ and delta S in J/K into the same subtraction gives a wildly wrong answer. Convert delta S to kJ/K (divide by 1000) before subtracting, and watch the sign of the T delta S term.