Equilibrium Constant Calculator

Calculate the equilibrium constant K from the standard Gibbs free energy of reaction ΔG° using ΔG° = −R·T·ln(K), or solve the formula for ΔG° or the temperature T – simply leave the desired variable blank. The calculator automatically converts between units (kJ/mol, J/mol, K, °C) and also indicates whether the equilibrium favors the products or the reactants.

Enter Values

Do you have ΔH and ΔS given and need to find ΔG° first? Use our Gibbs Free Energy Calculator.

Standard Gibbs free energy of reaction (always molar, e.g., −57.2 kJ/mol).
Absolute temperature at which the equilibrium is established.
Equilibrium constant (dimensionless). Can span many orders of magnitude – scientific notation like 3.4e-4 is allowed.

Explanation: Equilibrium Constant and ΔG°

How are ΔG° and the Equilibrium Constant Connected?

The standard Gibbs free energy of reaction ΔG° indicates how far a reaction proceeds until it reaches equilibrium under standard conditions. If ΔG° is strongly negative, the equilibrium lies almost completely on the side of the products.

The exact relationship is established via the equilibrium constant K, which describes the ratio of product to reactant activities at equilibrium.

Basic Formula

ΔG° = −R · T · ln(K)

Overview of the Three Variables

Standard Gibbs Free Energy of Reaction (ΔG°)

ΔG° describes the driving force of a reaction under standard conditions (1 bar, usually 298 K, standard concentrations). Negative values favor the product side.

Universal Gas Constant (R)

R = 8.314 J/(mol·K) is a fixed physical constant. Because R itself is defined on a molar basis, ΔG° in this formula is always a molar quantity as well.

Temperature (T)

T is the absolute temperature in Kelvin at which equilibrium is established. The higher T is, the stronger the effect of ΔG° on K becomes.

Equilibrium Constant (K)

K is dimensionless and describes how far a reaction proceeds until it reaches equilibrium at a given temperature. K can be very small (hardly any conversion) or very large (near-complete conversion).

Why ΔG° is Always Molar Here

Unlike in the general Gibbs free energy calculator (ΔG = ΔH − T·ΔS), the "per mole" part cannot be canceled out here: the gas constant R is strictly defined in J/(mol·K). Therefore, ΔG° must always be entered as a molar value (e.g., kJ/mol) – the unit selection reflects this accordingly.

Solving the Formula for Each Variable

Depending on which variable is requested, the formula is rearranged accordingly:

Solving for K

K = e^(−ΔG° / (R·T))
Used when ΔG° and T are known.

Solving for T

T = −ΔG° / (R·ln(K))
Used when ΔG° and K are known. Note: at K = 1, T cannot be determined (ΔG° = 0 for any temperature).

What Does K Tell Us About the Position of Equilibrium?

K > 1 means that there are more products than reactants at equilibrium. K < 1 means the opposite. K = 1 means that products and reactants are present in comparable amounts at equilibrium (more precisely, equal activities).

Example Problems

Two examples demonstrate how to calculate depending on the requested variable.

Example 1: Calculating K

A reaction has ΔG° = −20 kJ/mol at T = 298 K. What is the equilibrium constant K?

Given

ΔG° = −20000 J/mol, T = 298 K, R = 8.314 J/(mol·K)

Solution

K = e^(−ΔG° / (R·T))

K = e^(20000 / (8.314 · 298)) = e^(8.07) ≈ 3200

K ≈ 3200 → The equilibrium heavily favors the products.

Example 2: Calculating ΔG° from a Known K

At T = 310 K, a value of K = 2.5·10⁻³ was measured. What is ΔG°?

Given

T = 310 K, K = 0.0025

Solution

ΔG° = −R·T·ln(K)

ΔG° = −8.314 · 310 · ln(0.0025) = −8.314 · 310 · (−5.99) ≈ 15440 J/mol

ΔG° ≈ +15.4 kJ/mol → positive, the equilibrium favors the reactants.

Tips and Common Mistakes

Common Pitfalls

Using log Instead of ln

The formula uses the natural logarithm (ln), not the common logarithm (log₁₀). Mixing them up introduces an error factor of approximately 2.303.

Using ΔG Instead of ΔG°

This formula applies to the standard Gibbs free energy of reaction ΔG° (under standard conditions), not the instantaneous ΔG under arbitrary concentrations. For non-standard conditions, you additionally need the reaction quotient Q (ΔG = ΔG° + R·T·ln(Q)).

Using Temperature in °C Instead of Kelvin

The formula always requires the absolute temperature in Kelvin, not degrees Celsius. Don't forget to add 273.15.

Where is this Used?

The relationship between ΔG° and K is central to chemical thermodynamics:

  • Predicting how far a reaction will proceed (estimating yield)
  • Calculating solubility products and acid-base equilibria
  • Evaluating enzyme reactions and metabolic equilibria in biochemistry

Frequently Asked Questions about the Equilibrium Constant

Using the formula $ \Delta G^\circ = -R \cdot T \cdot \ln(K) $, the equilibrium constant K can be calculated directly from the standard Gibbs free energy of reaction ΔG° (and vice versa). The more negative ΔG° is, the larger K becomes and the further the equilibrium lies on the product side.

K > 1 means that at equilibrium, there are more products than reactants present (equilibrium lies on the product side). K < 1 means the opposite. At K = 1, products and reactants are present in comparable amounts at equilibrium, and ΔG° = 0.

The thermodynamic derivation of the formula strictly leads to the natural logarithm (ln), not the common logarithm (log₁₀). Confusing the two introduces an error by a factor of ln(10) ≈ 2.303 – which makes a substantial difference for larger values of K.

ΔG° applies only under standard conditions (1 bar, usually 298 K, standard concentrations) and is linked to K through this formula. Under arbitrary, non-standard concentrations, ΔG = ΔG° + R·T·ln(Q) applies instead, where Q is the reaction quotient for the current (non-equilibrium) concentrations.

Unlike in the general Gibbs-Helmholtz equation (ΔG = ΔH − T·ΔS), the "per mole" cannot cancel out here because the gas constant R is fixed in units of J/(mol·K). Therefore, ΔG° must strictly be entered as a molar value (e.g., kJ/mol) for the units to be consistent.

When K = 1, ln(K) = 0, which makes ΔG° = −R·T·0 = 0 – regardless of the temperature used. In this case, any temperature satisfies the equation, so T cannot be uniquely determined.