Hong Life CalculatorsUnderstanding the equilibrium constant calculator — concept and principle
The equilibrium constant tells you how far a reversible reaction proceeds before forward and reverse rates balance. For gas-phase reactions it can be written with molar concentrations (Kc) or with partial pressures (Kp). This calculator converts Kc to Kp at a given temperature and also turns Kc into the standard Gibbs energy change ΔG°.
How it is calculated
- Kp = Kc (RT)^Δn, with R = 0.0820574 L·atm/(mol·K) and T in kelvin. Kc is in mol/L units, so the resulting Kp is on an atmosphere basis.
- Δn = (moles of gaseous products) − (moles of gaseous reactants), taken from the balanced equation. Solids and liquids are not counted.
- ΔG° = −RT ln K, with R = 8.314462618 J/(mol·K), divided by 1000 to give kJ/mol, shown twice: from Kp in bar, Kc (0.0831446 T)^Δn, for the usual 1 bar standard state, and from Kc for 1 mol/L.
The principle
For an ideal gas, partial pressure and concentration are linked by p = (n/V)RT = cRT. Every pressure term in Kp is therefore a concentration multiplied by RT. The RT factors of products (numerator) and reactants (denominator) cancel except for the surplus Δn, leaving (RT)^Δn. When Δn = 0 the two constants are equal.
The link to ΔG° comes from thermodynamics: at equilibrium ΔG = 0, so ΔG° = −RT ln K. A K above 1 means a negative ΔG° (products favoured); a K below 1 means a positive ΔG°. The value depends on the standard state: Kp in bar gives ΔG° for 1 bar, Kc gives ΔG° for 1 mol/L, and they differ whenever Δn ≠ 0.
Worked example
- Inputs: Kc = 0.5, Δn = −2 (for example a reaction like N₂ + 3H₂ ⇌ 2NH₃), T = 700 K.
- RT = 0.0820574 × 700 = 57.440 L·atm/mol.
- (RT)^−2 = 1 ÷ 57.440² = 3.031 × 10⁻⁴.
- Kp = 0.5 × 3.031 × 10⁻⁴ = 0.0001515.
- ΔG° (1 bar): Kp in bar = 0.5 × (0.0831446 × 700)⁻² = 1.476 × 10⁻⁴, so ΔG° = −8.314462618 × 700 × ln(1.476 × 10⁻⁴) ÷ 1000 = +51.34 kJ/mol.
- ΔG° (1 mol/L): −8.314462618 × 700 × ln 0.5 ÷ 1000 = +4.034 kJ/mol.
| Δn (moles gas products − reactants) | Kp (atm) |
|---|---|
| -0.5 | 0.0659723734 |
| -1 | 0.0087047081 |
| -1.5 | 0.0011485405 |
| -2 | 0.0001515439 |
| -2.5 | 0.0000199954 |
| -3 | 0.0000026383 |
| -4 | 4.59311e-8 |
How to read the result
K much larger than 1 (say above 10³) means the mixture lies far toward products; much smaller than 10⁻³ means it lies toward reactants; values in between mean appreciable amounts of both. A negative Δn makes Kp smaller than Kc at high temperature, because the products occupy fewer gas moles.
Common mistakes
- Counting solids, liquids or solvent in Δn.
- Entering temperature in °C instead of K.
- Using R = 8.314 in the Kp conversion, which gives a kPa basis rather than atm.
Frequently asked questions
When are Kp and Kc equal?
When Δn = 0, meaning the balanced equation has the same number of gas moles on both sides, such as H₂ + I₂ ⇌ 2HI.
Do I use atm or bar for Kp?
The Kp shown uses R in L·atm, so it is on an atm basis; the 1 bar ΔG° row uses R = 0.0831446 L·bar/(mol·K). The two Kp values differ slightly when Δn ≠ 0.
Sources: Atkins, P. and de Paula, J., Atkins' Physical Chemistry (Oxford University Press), chapter on chemical equilibrium · IUPAC Compendium of Chemical Terminology (Gold Book), "equilibrium constant" · CODATA recommended value of the molar gas constant, NIST