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Equilibrium Concentration Calculator

Enter the initial concentration of A and the equilibrium constant Keq.

How it is calculated

The tool handles the single reaction A ⇌ B with 1:1 stoichiometry, starting with only A (no B at the start). Let x be the amount of A that turns into B. At equilibrium Keq = x ÷ (A0 − x), which solves to x = Keq × A0 ÷ (1 + Keq). Then [B] = x, [A] = A0 − x and conversion = x ÷ A0 × 100. A0 is in mol/L and Keq is dimensionless; both must be above zero. Concentrations are shown to four decimals. Reactions with other coefficients, such as 2A ⇌ B, or with product already present at the start, need a different equation.

Worked example

Start with [A]0 = 0.50 mol/L and Keq = 4. Then x = 4 × 0.50 ÷ (1 + 4) = 2.0 ÷ 5 = 0.4000 mol/L, which is [B] at equilibrium. The remaining reactant is 0.50 − 0.40 = 0.1000 mol/L. Check: 0.40 ÷ 0.10 = 4, matching Keq. Conversion is 0.40 ÷ 0.50 × 100 = 80.00%.

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Common mistakes

  • Using this for reactions that are not 1:1, such as 2A ⇌ B or A + B ⇌ C. The equation here is only valid for A ⇌ B; other coefficients give a quadratic or higher equation.
  • Entering a nonzero starting amount of B. The tool assumes only A at the start, so an initial [B] must be handled with a full ICE table instead.
  • Typing Keq as a percentage or as Keq of the reverse reaction. For B ⇌ A the constant is 1 ÷ Keq, which flips the answer.
  • Entering moles instead of mol/L. The result is only a concentration if the starting value is one; for a different volume, divide the moles by the volume first.

About this calculator

For a simple reversible reaction A ⇌ B starting with only reactant A present, the equilibrium constant Keq = [B]/[A] tells you the ratio the concentrations settle into once the forward and reverse reaction rates become equal — a large Keq means the reaction favors product B at equilibrium, while a small Keq means most of the material stays as A.

Given the initial concentration of A and Keq, an ICE table (Initial, Change, Equilibrium) approach lets you solve directly for both equilibrium concentrations: [B] = Keq/(1+Keq) times the initial concentration, and [A] is what remains. This is a foundational calculation in general chemistry for reactions simple enough to have a 1:1 stoichiometry and no other competing species.

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Comments (1)

  • Editorial team

    For this 1:1 reaction the fraction converted depends only on Keq, not on the starting amount. Starting from 10 mol/L or from 0.1 mol/L with Keq = 4 gives 80.00% conversion both times, only the concentrations scale. So diluting the mixture will not shift this equilibrium; it would for a reaction with unequal coefficients. If a textbook answer disagrees with yours, post the reaction and constant you used.

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