Gibbs Free Energy & Equilibrium Calculator (ΔG = −RT ln K)

If you're computing Gibbs free energy and equilibrium constants and your K comes out negative, something went wrong—K is always positive. The relationship ΔG° = −RT ln K connects standard free energy change to the equilibrium constant, and it falls out of the definition of ΔG under standard conditions. At equilibrium, ΔG = 0. The general expression ΔG = ΔG° + RT ln Q reduces to 0 = ΔG° + RT ln K (since Q = K at equilibrium). Rearranging gives ΔG° = −RT ln K.

The units demand attention. R = 8.314 J·mol⁻¹·K⁻¹ gives ΔG° in joules. If your thermodynamic tables list ΔG° in kJ/mol, convert before plugging in: either multiply ΔG° by 1000 or use R = 0.008314 kJ·mol⁻¹·K⁻¹. Mixing J and kJ is the most frequent arithmetic error in this calculation and makes K wrong by e¹⁰⁰⁰ instead of e¹—off by hundreds of orders of magnitude.

The exponential relationship means small changes in ΔG° produce enormous changes in K. At 298 K, every 5.7 kJ/mol decrease in ΔG° multiplies K by 10. So ΔG° = −5.7 kJ/mol gives K ≈ 10, ΔG° = −11.4 gives K ≈ 100, and ΔG° = −57 gives K ≈ 10¹⁰. This steep sensitivity is why tabulated ΔG° values need several significant figures to be useful for calculating K.

A negative ΔG means the forward reaction is thermodynamically favorable at those specific conditions. It can proceed without external energy input. But "favorable" is not the same as "fast" or "complete." Diamond combustion to CO₂ has ΔG° = −397 kJ/mol at 298 K—overwhelmingly favorable—yet diamonds persist for billions of years because the activation energy is enormous. Thermodynamics says the reaction can happen; kinetics says it won't happen at any useful rate.

A positive ΔG means the reverse reaction is favored. Students often say "the reaction won't happen," which is wrong. It means the forward reaction is not spontaneous at those conditions—the reverse direction is favored. The system still reaches equilibrium, but with more reactants than products. A positive ΔG° of +10 kJ/mol gives K = exp(−10000/(8.314 × 298)) = exp(−4.04) = 0.018. The reaction does proceed—just not very far.

When ΔG = 0 (not ΔG°, but actual ΔG at current conditions), the system is at equilibrium. No net change occurs. This is the point where Q = K. The sign of ΔG tells you which direction the system will move from its current state, while the sign of ΔG° tells you which side of unity K falls on (products or reactants favored at standard conditions).

ΔG° is a single number for a reaction at a given temperature—all species at standard-state activity (1 M for solutes, 1 bar for gases). Real mixtures almost never match standard state, so you need ΔG = ΔG° + RT ln Q to find the actual driving force. The RT ln Q correction shifts ΔG up or down depending on where Q sits relative to K.

When Q is very small (nearly pure reactants), ln Q is large and negative, making ΔG more negative than ΔG°. The reaction has a stronger driving force forward than the standard value suggests. When Q is very large (nearly pure products), ln Q is large and positive, making ΔG more positive. The reaction pushes backward even if ΔG° is negative. This is how a reaction with a favorable ΔG° can still be unfavorable under specific conditions—you've loaded the system with too many products.

At equilibrium, ΔG = 0, which means RT ln Q = −ΔG°, which means Q = K. This is the connection point between the two equations. ΔG° tells you K. Q tells you where you are. ΔG tells you which way you'll move. Students who substitute K into the Q position of the ΔG equation always get ΔG = 0—that's correct but trivially true and doesn't answer any useful question.

Temperature is the only variable that changes K itself. Concentration changes shift Q, not K. Pressure changes shift Q (for gas-phase reactions), not K. Catalysts change rate, not K. But raise or lower the temperature, and K moves.

The direction depends on ΔH. For an exothermic reaction (ΔH < 0), heating decreases K—products become less favored. For an endothermic reaction (ΔH > 0), heating increases K—products become more favored. This follows from the van't Hoff equation: ln(K₂/K₁) = −(ΔH/R)(1/T₂ − 1/T₁). If ΔH is negative and T₂ > T₁, the right side is negative, so K₂ < K₁.

The van't Hoff equation assumes ΔH is roughly constant over the temperature range. For narrow ranges (say 298 to 350 K), this is usually fine. Over hundreds of degrees, ΔH itself changes with temperature (because heat capacities of products and reactants differ), and you need the Kirchhoff equation to correct ΔH before plugging into van't Hoff. For a general chemistry course, the constant-ΔH approximation is standard. For physical chemistry, you'll integrate Cp.

Why does ΔG = ΔH − TΔS sometimes predict spontaneity wrong? It doesn't—if you use the right temperature. ΔG° values from tables assume 298 K. Using them at 500 K gives the wrong ΔG° because ΔH° and ΔS° themselves change with temperature. Either recalculate ΔH° and ΔS° at the new temperature using heat capacity data, or use ΔG = ΔH − TΔS with the understanding that ΔH and ΔS are approximate at temperatures far from 298 K.

Can ΔG° be zero? Yes. It means K = 1 exactly. At standard conditions, neither products nor reactants are favored. The system reaches equilibrium with roughly equal amounts of each (adjusted for stoichiometry). This is a special case, not an error.

What about biochemistry's ΔG°'? Biochemical standard state uses pH 7 ([H⁺] = 10⁻⁷ M) instead of the thermodynamic standard of [H⁺] = 1 M. Since many biological reactions involve H⁺, this matters enormously. ΔG°' for ATP hydrolysis is about −30 kJ/mol at pH 7, but ΔG° (at [H⁺] = 1 M, which is pH 0) would be quite different. Always check which standard state your data uses.

Does a more negative ΔG° always mean a faster reaction? No. ΔG° relates to equilibrium position (K), not rate. Rate depends on activation energy (Ea), which is entirely separate. A reaction with ΔG° = −200 kJ/mol but Ea = 300 kJ/mol is extremely favorable but painfully slow. Thermodynamics and kinetics answer different questions.

Problem: The decomposition of N₂O₄(g) ⇌ 2 NO₂(g) has ΔH° = +57.2 kJ/mol and ΔS° = +175.8 J/(mol·K). Find K at 298 K and at 400 K.

At 298 K, K = 0.14 (reactants slightly favored—ΔG° is positive). At 400 K, K = 52 (products strongly favored—ΔG° flipped negative). The crossover at 325 K is where K = 1 and ΔG° = 0. This reaction is endothermic with positive ΔS, so heating helps—the TΔS term eventually overwhelms the unfavorable ΔH. Notice all calculations used ΔG° in J/mol with R = 8.314 J·mol⁻¹·K⁻¹. Converting ΔS from J to kJ first would have avoided that issue entirely.