Library / Foundations: Chemistry / FCH-27
Foundations: Chemistry
Equilibrium constant Kp from partial pressures
Equilibrium constant of a gas reaction written with partial pressures, in the common school convention.
- Ref
- FCH-27
- Rev
- v1
- Field
- Foundations: Chemistry
- Result
- Kp
For a gas-phase reaction aA + bB ⇌ cC + dD at equilibrium, Kp = (pCc pDd)/(pAa pBb), where each p is the equilibrium partial pressure and each exponent is the coefficient in the balanced equation. Here the pressures are taken in bar relative to a 1 bar standard state, so Kp has no unit; with other pressure units its unit depends on Δn = (c + d) − (a + b). It depends only on temperature. Put 0 for the exponent of a species that is not in the reaction.
Equilibrium constant of a gas reaction written with partial pressures, in the common school convention. Its units follow from the powers; use only gases, since pure solids and liquids are left out.
Calculator
- Change in moles of gas, Δn = (c + d) − (a + b)−2
- ln Kp−5.257
- Standard Gibbs energy change at 298 K, −RT ln Kp13.03 kJ/mol
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- With these pressures Kp = 0.005208. A value well below 1 means the mixture is mostly reactants; well above 1, mostly products. The standard Gibbs energy change at 298 K is 13 kJ/mol.
- The equilibrium constant depends only on temperature: changing the total pressure shifts the pressures, but they shift together so that the same Kp is restored.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing () | Local step | Change in result |
|---|---|---|---|---|---|---|
| pB Partial pressure of B | 0.01 to 100 bar | 1.125e-6 | 1,125,000 | 1,125,000 | +10% | −24.9 % |
| pC Partial pressure of C | 0.01 to 100 bar | 2.315e-7 | 23.15 | 23.15 | +10% | +21 % |
| b Coefficient of B | 0 to 6 | 2.411e-5 | 1.125 | 1.125 | +1 | −83.3 % |
| pA Partial pressure of A | 0.01 to 100 bar | 1.042e-4 | 1.042 | 1.042 | +10% | −9.09 % |
| c Coefficient of C | 0 to 6 | 0.002315 | 0.02637 | 0.02405 | +1 | +50 % |
| a Coefficient of A | 0 to 6 | 1.628e-4 | 0.01042 | 0.01025 | +1 | −50 % |
| pD Partial pressure of D | 0.01 to 100 bar | 0.005208 | 0.005208 | 0 | +10% | 0 % |
| d Coefficient of D (0 if there is no D) | 0 to 6 | 0.005208 | 0.005208 | 0 | +1 | 0 % |
Sweep of Partial pressure of C
| Partial pressure of C (bar) | Equilibrium constant Kp |
|---|---|
| 0.01 | 2.315e-7 |
| 0.03162 | 2.315e-6 |
| 0.1 | 2.315e-5 |
| 0.3162 | 2.315e-4 |
| 1 | 0.002315 |
| 3.162 | 0.02315 |
| 10 | 0.2315 |
| 31.62 | 2.315 |
| 100 | 23.15 |
Sweep one variable
Variables and how each one contributes
Equilibrium constant Kp
What the equation gives you. Everything else feeds it.
Partial pressure of A
Equilibrium partial pressure of reactant A, in bar. Lowers Kp, because pAa is in the denominator; the larger the exponent a, the stronger the effect. Raising it by 0.5 bar (to 2.5 bar) moves the result by −20 %, from 0.005208 to 0.004167.
Partial pressure of B
Equilibrium partial pressure of reactant B, in bar. Lowers Kp, because pBb is in the denominator; with a coefficient of 3 on B the effect is strong. Raising it by 0.5 bar (to 6.5 bar) moves the result by −21.3 %, from 0.005208 to 0.004096.
Partial pressure of C
Equilibrium partial pressure of product C, in bar. Raises Kp, because pCc is in the numerator; the larger the exponent c, the stronger the effect. Raising it by 0.25 bar (to 1.75 bar) moves the result by +36.1 %, from 0.005208 to 0.007089.
Partial pressure of D
Equilibrium partial pressure of product D, in bar. It has no effect while its exponent d is 0. Has no effect while d = 0, and raises Kp with the power d when d is above 0. Raising it by 0.25 bar (to 1.25 bar) moves the result by +0 %, from 0.005208 to 0.005208.
Coefficient of A
Stoichiometric coefficient of A, a whole number (0 if absent). Lowers Kp when pA is above 1 bar and raises it when below, because pAa grows or shrinks with the exponent. Raising it by 1 (to 2) moves the result by −50 %, from 0.005208 to 0.002604.
Coefficient of B
Stoichiometric coefficient of B, a whole number (0 if absent). Lowers Kp when pB is above 1 bar and raises it when below. Raising it by 1 (to 4) moves the result by −83.3 %, from 0.005208 to 8.681×10−4.
Coefficient of C
Stoichiometric coefficient of C, a whole number (0 if absent). Raises Kp when pC is above 1 bar and lowers it when below. Raising it by 1 (to 3) moves the result by +50 %, from 0.005208 to 0.007813.
Coefficient of D (0 if there is no D)
Stoichiometric coefficient of D, a whole number (0 if absent). Has no effect when pD is 1 bar, raises Kp when pD is above 1 bar and lowers it when below. Raising it by 1 (to 1) moves the result by +0 %, from 0.005208 to 0.005208.
Worked example with the values above
- Products: pC2 × pD0 = 2.25 × 1 = 2.25.
- Reactants: pA1 × pB3 = 2 × 216 = 432.
- Kp = 2.25 / 432 = 0.005208.
Assumptions and limits
- Ideal gases at equilibrium, with partial pressures in bar against a 1 bar standard state.
- Exponents equal to the stoichiometric coefficients of the balanced equation, and a single temperature.
Common mistakes
- Including solids or pure liquids in the expression: they are left out.
- Using initial rather than equilibrium pressures.
- Mixing pressure units and then quoting Kp as dimensionless.
Statistics and uncertainty
Equilibrium pressures are measured to a few per cent, and each is raised to its coefficient, so the relative error in Kp is the sum of the coefficient-weighted relative errors: with a coefficient of 3 on B, a 2 % error in pB becomes a 6 % error in Kp.
Where it is used
- Predicting the composition of a gas mixture at equilibrium.
- Finding the equilibrium constant from measured partial pressures, for example in a synthesis or combustion product mix.
Maths you need
- Partial pressure: the share of the total pressure from one gas
- Balanced equation: the coefficients give the exponents