Equilibrium Constant Calculator

Calculate equilibrium constant Kc or Kp from equilibrium concentrations or partial pressures.

Type

Product concentrations/pressures (multiply together) ([C]^c × [D]^d)

Reactant concentrations/pressures (multiply together) ([A]^a × [B]^b)

Δn gas (mol products − mol reactants) (for Kc↔Kp conversion)

Temperature T (K) (K)

Please check your inputs and try again.

What K Tells You — and What It Does Not

The equilibrium constant expresses the ratio of products to reactants once a reaction has settled, with each concentration raised to its stoichiometric coefficient. Its magnitude answers one question only: where does the equilibrium lie?

K_c = [C]^c[D]^d / [A]^a[B]^b

K value	Position of equilibrium	Interpretation
K > 10³	Far to the right	Essentially complete — little reactant remains
10⁻³ to 10³	Intermediate	Appreciable amounts of both present
K < 10⁻³	Far to the left	Barely proceeds — mostly reactant

What K does not tell you is how fast the reaction gets there. Diamond converting to graphite has a favourable equilibrium constant, yet the process takes geological time. Thermodynamics sets the destination; kinetics sets the speed, and the two are entirely independent.

K_c, K_p and Q

K_p = K_c(RT)^Δn

where Δn is moles of gaseous product minus moles of gaseous reactant. When Δn = 0 the two are numerically equal. Note that R must be 0.0821 L·atm/(mol·K) when pressures are in atmospheres.

The reaction quotient Q uses the identical expression but with whatever concentrations are present at that moment, not necessarily at equilibrium. Comparing Q with K tells you which way the reaction will go:

Comparison	Direction	Why
Q < K	Forward, toward products	Too few products relative to equilibrium
Q = K	No net change	System is at equilibrium
Q > K	Reverse, toward reactants	Too many products relative to equilibrium

Pure solids and pure liquids are omitted from both expressions because their activity is essentially constant. Adding more solid CaCO₃ to a decomposition equilibrium changes nothing.

Worked Examples

Example 1: H2+I2⇌2HI: [HI]²/([H2][I2])=0.04²/(0.01×0.01)

Kc=0.0016/0.0001

Result: Kc=16 — products moderately favored

Equilibrium lies toward HI

Example 2: CO+3H2⇌CH4+H2O, Δn=-2

Convert Kc=10⁴ to Kp at 1000K

Result: Kp=Kc×(RT)^-2=10⁴/(82.06)²

Kp much smaller: fewer gas moles in products

Example 3: Reversing the reaction

H₂ + I₂ ⇄ 2HI has K_c = 16

Result: Reverse reaction: K = 1/16 = 0.0625

Reversing a reaction inverts K. Doubling the coefficients squares it, and adding two reactions multiplies their constants.

Example 4: Using Q to predict direction

K_c = 16, but current [HI] = 0.10, [H₂] = [I₂] = 0.01

Result: Q = 100 > K, so the reaction runs in reverse

Too much product relative to equilibrium. HI decomposes until Q falls back to 16.

Example 5: Heterogeneous equilibrium

CaCO₃(s) ⇄ CaO(s) + CO₂(g)

Result: K = P_CO2 only

Both solids are excluded. The equilibrium pressure of CO₂ depends only on temperature, regardless of how much solid is present.

Common Mistakes

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Including solids and pure liquids in the expression

Their activity is effectively fixed at 1, so they never appear in K. For CaCO₃(s) ⇄ CaO(s) + CO₂(g), the expression is simply K = [CO₂].

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Forgetting the stoichiometric exponents

Each concentration is raised to its coefficient. For 2HI ⇄ H₂ + I₂ the HI term is squared — omitting the exponent changes the answer by orders of magnitude.

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Assuming a large K means a fast reaction

K describes the equilibrium position, not the rate. Many thermodynamically favourable reactions are kinetically inaccessible without a catalyst.

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Using the wrong R when converting to K_p

Use 0.0821 L·atm/(mol·K) for pressures in atmospheres, or 8.314 J/(mol·K) only when working in pascals and cubic metres. Mixing them is a frequent error.

Frequently Asked Questions

K >>1 vs <<1?

K>>1 (>10⁴): reaction essentially complete, products favored. K<<1 (<10⁻⁴): barely proceeds, reactants favored. K~1: significant amounts of both at equilibrium.

Q vs K?

Q (reaction quotient) uses actual concentrations. If QK: proceeds reverse. If Q=K: at equilibrium. Same formula, different context.

What does the magnitude of K mean?

K above about 10³ means the reaction goes essentially to completion; below 10⁻³ it barely proceeds. Between those values, appreciable amounts of both reactants and products coexist.

Does a large K mean a fast reaction?

No. K describes only the equilibrium position. A reaction can have a very favourable K and still be immeasurably slow without a catalyst.

Why are solids and pure liquids excluded?

Their activity is essentially constant regardless of quantity, so they contribute a fixed factor absorbed into K. Adding more solid does not shift the equilibrium.

What is the difference between Q and K?

They use the same expression, but Q uses current concentrations while K uses equilibrium ones. Comparing them predicts which direction the reaction will move.

How do I convert K_c to K_p?

Multiply by (RT)^Δn, where Δn is moles of gaseous products minus gaseous reactants. When Δn is zero the two values are identical.

Formula Explorer connections

Interpretation: This formula describes how reactants, products, ions or phases distribute when opposing processes reach equilibrium. Assumption: Use equilibrium rather than initial concentrations, correct stoichiometric exponents, and the specified temperature; activities may replace concentrations in nonideal systems.

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