MCAT topics → Thermodynamics and Kinetics → Gibbs Free Energy and Spontaneity

Gibbs Free Energy and Spontaneity

MCAT trap: Anchors spontaneity judgment on ΔH sign alone without computing the TΔS term. Spontaneity requires evaluating ΔG = ΔH − TΔS; a positive ΔH can be overcome by a sufficiently large positive ΔS at high temperature, yielding negative ΔG.

Gibbs free energy is the MCAT's core framework for deciding whether a reaction will run on its own. The fundamental equation is ΔG = ΔH − TΔS, and the sign of ΔG at constant temperature and pressure tells you everything: negative means spontaneous (exergonic), positive means nonspontaneous (endergonic), zero means equilibrium. The exam tests this at multiple levels — pure recall of the equation, quadrant-style reasoning about ΔH/ΔS sign combinations, quantitative conversion between ΔG° and K, and passage-based data interpretation where you have to figure out at what temperature a reaction flips spontaneity.

What makes this concept genuinely tricky is that students anchor too hard on one variable. The most common error is treating enthalpy as the whole story — assuming a positive ΔH automatically means nonspontaneous. It doesn't. If ΔS is large and positive, the TΔS term can dominate at high temperature and make ΔG negative. The other major trap is conflating ΔG° with ΔG. Standard free energy only applies at standard conditions (1 M concentrations, 1 atm, 298 K). Under any other conditions you need ΔG = ΔG° + RT ln Q, and the actual Q value shifts ΔG substantially — sometimes even reversing the direction of spontaneity.

The MCAT also loves the equilibrium link: ΔG° = −RT ln K. A large positive K means a large negative ΔG°, and vice versa. You won't need a calculator for this — just understand the qualitative relationship and the algebra well enough to solve for one variable given the other. Reactions where ΔH and ΔS have the same sign are the most interesting cases because spontaneity is temperature-dependent, and there's a specific crossover temperature (T = ΔH/ΔS) the exam can ask you to identify.

Common misconceptions

Common mistake
Wrong: A reaction with positive ΔH is always nonspontaneous because enthalpy drives it uphill.
Right: Spontaneity requires evaluating ΔG = ΔH − TΔS; a positive ΔH can be overcome by a sufficiently large positive ΔS at high temperature, yielding negative ΔG.
Enthalpy alone does not determine spontaneity — ΔG does, and ΔG has two terms. A reaction with positive ΔH can still have a negative ΔG if ΔS is large and positive, because the −TΔS term becomes large and negative at high temperatures. The classic real-world example is dissolving ammonium nitrate in water: endothermic (positive ΔH) but spontaneous because entropy increases dramatically. Always compute or estimate both terms before drawing a conclusion.
Common mistake
Wrong: When product concentrations are low (Q < K), ΔG becomes more positive (less spontaneous in the forward direction).
Right: When Q < K (products low relative to equilibrium), ΔG = ΔG° + RT ln Q is more negative, driving the reaction forward more spontaneously.
When Q < K, the system hasn't reached equilibrium yet and has more capacity to form products — meaning the forward reaction is thermodynamically favored and ΔG is more negative, not more positive. Think of it this way: low product concentration means the system is far from equilibrium in the forward direction, so it gets more 'pull' toward products. The equation ΔG = ΔG° + RT ln Q confirms this: a smaller Q means a more negative ln Q, which pushes ΔG downward.
Common mistake
Wrong: ΔG° < 0 means the reaction is spontaneous under all conditions.
Right: ΔG° < 0 means the reaction is spontaneous only under standard conditions; actual spontaneity depends on ΔG = ΔG° + RT ln Q.
ΔG° is a fixed constant for a given reaction at a given temperature — it only tells you about spontaneity when all species are at standard concentrations (1 M, 1 atm). The actual free energy change ΔG depends on the reaction quotient Q via ΔG = ΔG° + RT ln Q, so even a reaction with ΔG° < 0 can become nonspontaneous if products accumulate enough (high Q). Equilibrium is precisely the point where ΔG = 0, not where ΔG° = 0.
Common mistake
Gap: Unaware that a specific crossover temperature can be calculated to determine when a reaction switches spontaneity
The crossover temperature where spontaneity switches is T = ΔH/ΔS; above or below this temperature the sign of ΔG reverses for reactions where ΔH and ΔS have the same sign.
Set ΔG = 0 and solve: 0 = ΔH − TΔS gives T = ΔH/ΔS. This crossover temperature matters when ΔH and ΔS have the same sign, because those are the cases where temperature determines which term wins. For example, if both ΔH and ΔS are positive, the reaction is only spontaneous above this crossover temperature. On the MCAT, a passage might give you ΔH and ΔS values and ask you to predict spontaneity at a specified temperature — compute T = ΔH/ΔS first, then compare.
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What the exam tests

  1. Know the Gibbs equation ΔG = ΔH − TΔS cold — you need to correctly identify the sign of ΔG given values or signs of ΔH, T, and ΔS.
  2. Classify spontaneity for all four combinations of ΔH and ΔS signs: know which combinations are always spontaneous, always nonspontaneous, and temperature-dependent.
  3. Use ΔG° = −RT ln K to convert qualitatively or quantitatively between the standard free energy change and the equilibrium constant — including identifying which direction corresponds to K > 1 vs K < 1.
  4. From a table or graph of ΔH and ΔS data, calculate or identify the crossover temperature where a reaction switches between spontaneous and nonspontaneous (T = ΔH/ΔS).

Can you avoid these mistakes?

A reaction has ΔH = +50 kJ/mol and ΔS = +200 J/(mol·K). At 200 K and at 300 K, is the reaction spontaneous? Calculate ΔG at both temperatures and identify the crossover temperature.
ΔG° for a reaction is −20 kJ/mol. A student concludes the reaction will always proceed spontaneously in the forward direction. What is wrong with this conclusion, and under what condition would the reaction actually be at equilibrium or run in reverse?
If the equilibrium constant K for a reaction is much greater than 1, what is the sign of ΔG°? What does this tell you about where equilibrium lies — heavily toward products or reactants?
A reaction has Q < K. Without plugging in numbers, explain qualitatively why ΔG is negative (forward reaction spontaneous) rather than positive, using the equation ΔG = ΔG° + RT ln Q.

Related topics

Enthalpy and Hess's Law Entropy and the Second Law First Law of Thermodynamics and Internal Energy Calorimetry and Heat Capacity

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