Kp Calculator

Convert Kp and Kc, or compute Kp from the partial pressures of a gaseous reaction.

Thirteen pressure units, six gas-constant scales, and reverse solves for temperature, mole counts, pressures and coefficients.

Updated October 1, 2026

CreatorFrank Zhao

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Introduction

An equilibrium constant written in terms of partial pressures, $K_{\mathrm{p}}$ , is not a separate piece of chemistry from the constant you already know. It is the same balance between products and reactants, measured against a different yardstick. This calculator covers both directions of that conversion and also builds $K_{\mathrm{p}}$ directly from the partial pressures of a reaction.

There are two ways in, chosen from the menu at the top of the form. Pick convert between Kp and Kc when you already know the concentration-based constant and want the pressure one — or the reverse, for example to find what temperature would produce a target $K_{\mathrm{p}}$ . Pick calculate Kp from partial pressures when a lab or a textbook problem hands you the equilibrium pressures themselves. The equilibrium constant calculator covers the concentration side of the same reaction.

The one thing that surprises people

Changing the pressure scale changes the number you get. That is not a rounding artefact — it is arithmetic. The same reaction reported in atm and in kPa gives two different $K_{\mathrm{p}}$ values, and the gap between them is fixed and predictable. Section the two formulas shows the exact relationship, so you can convert any published value without guessing which scale it used.

Quick start

Converting a concentration constant to a pressure constant

Choose "convert between Kp and Kc" from the menu

The form then asks for the concentration constant, the pressure scale, the temperature, and the number of moles of gaseous products and reactants.

Pick the pressure scale your pressures are quoted in

Standard practice in most textbooks is atmospheres. If your problem statement quotes pressures in kPa or in bar, pick that instead — it changes the result by a known factor, not an unknown one.

Enter Kc, the temperature, and the mole counts

Only gases count toward the mole change. For a reaction whose water is a liquid, that water contributes nothing. If you are unsure which numbers belong, balance the equation first with the chemical equation balancer.

Read the result

The bottom box fills in and turns blue, because the engine solved it from your four inputs. Any of the five boxes can be overwritten instead, and the calculator re-solves the remainder.

Worked example — the ammonia synthesis reaction. For $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ at 298 K with $K_{\mathrm{c}} = 0.0227$ , there are two moles of gaseous product and four moles of gaseous reactant, so $\Delta n = 2 - 4 = -2$ .

K_{\mathrm{p}}

=

K_{\mathrm{c}} \times (R \times T)^{\Delta n}

=

0.0227 \times (0.082057366080959684185 \times 298)^{-2}

=

3.796 \times 10^{-5}

Building Kp from equilibrium partial pressures

Choose "calculate Kp from partial pressures"

The form asks for up to four component pressures and the stoichiometric coefficient of each.

Put each species in the box that matches its side of the equation

Reactants go in A and B, products in C and D. Coefficients that your equation does not use stay at $0$ .

Fill every pressure box, then read Kp

A species whose coefficient is zero is raised to the power zero and drops out of the expression entirely, so whatever number sits in its box makes no difference to the answer. The box still needs a value before the engine will commit to a result.

Worked example — for the same ammonia reaction with equilibrium pressures of 1 atm for nitrogen, 3 atm for hydrogen and 2 atm for ammonia:

K_{\mathrm{p}}

=

\frac{P_{NH_3}^{\,2} \times P_{\text{unused}}^{\,0}}{P_{N_2}^{\,1} \times P_{H_2}^{\,3}}

=

\frac{2^2}{1^1 \times 3^3}

=

0.148148

The two formulas

The first formula relates the two constants. $\Delta n$ is the change in the number of gaseous moles: total coefficients on the product side minus the total on the reactant side.

K_{\mathrm{p}} = K_{\mathrm{c}} \times (R \times T)^{\Delta n}

The second builds the pressure constant from the partial pressures themselves. Every species is raised to its own stoichiometric coefficient — this is the single most important thing to get right.

K_{\mathrm{p}} = \frac{P_{\mathrm{C}}^{\,c} \times P_{\mathrm{D}}^{\,d}}{P_{\mathrm{A}}^{\,a} \times P_{\mathrm{B}}^{\,b}}

Why the pressure scale changes the answer

$K_{\mathrm{p}}$ is a ratio built from powers of pressures, so multiplying every pressure by a conversion factor $f$ multiplies the whole constant by $f^{\Delta n}$ . That is the entire rule, and it is exact.

K_{\mathrm{p},\,\text{new}}

=

K_{\mathrm{p},\,\text{old}} \times f^{\Delta n}

, \qquad f = \frac{f_{\text{new}}}{f_{\text{old}}}

Going back to the ammonia example with $\Delta n = -2$ : converting pressures from atm to kPa multiplies them by $f = 101.325$ , so the constant must fall by $101.325^2$ .

\frac{K_{\mathrm{p}}(\mathrm{kPa})}{K_{\mathrm{p}}(\mathrm{atm})}

=

101.325^{-2}

=

0.0000974017534385406

The calculator makes this automatic: the scale you choose sets the gas constant to match, so the reported value is always in the scale you asked for.

The gas constant

The value used here is $8.31446261815324\ \mathrm{J\,mol^{-1}\,K^{-1}}$ . That figure is exact, not rounded: since the 2019 SI redefinition the Boltzmann constant and the Avogadro constant are both defining constants, and their product is therefore fixed without uncertainty (BIPM, SI defining constants). Expressed in litres per atmosphere it becomes a repeating decimal, so the atm and torr entries are carried to twenty significant digits rather than truncated to the eight a printed table would give.

What you can solve backwards

Overwrite any input and the engine re-solves the rest, so from a single $K_{\mathrm{p}}$ and three pressures you can recover the fourth, and from a $K_{\mathrm{p}}$ , a $K_{\mathrm{c}}$ and the temperature you can recover $\Delta n$ or the temperature itself. Taking a root is how that works: isolating $P_{\mathrm{D}}$ with a coefficient of two is a square root.

P_{\mathrm{D}}^{\,d} = \frac{K_{\mathrm{p}} \times P_{\mathrm{A}}^{\,a} \times P_{\mathrm{B}}^{\,b}}{P_{\mathrm{C}}^{\,c}}

, \qquad P_{\mathrm{D}} = \left(\frac{234 \times 1 \times 1}{1}\right)^{1/2}

=

15.29705854077835449

Worked examples

Recovering a missing partial pressure

A reactor analysis gives you the equilibrium constant and three of the four pressures, but not the fourth. Leave that box empty and the engine solves it.

With $K_{\mathrm{p}} = 234$ , the other three pressures all at 1, and coefficients $a = b = c = 1$ and $d = 2$ , the missing pressure comes back as $\sqrt{234} = 15.29705854077835449$ , not 117. Anything dividing 234 by the coefficient instead of raising the pressure to it is wrong the moment the coefficient is not one.

A reaction with a solid product

For $2H_{2(g)} + O_{2(g)} \rightleftharpoons 2H_{2}O_{(s)}$ the water is a solid at ordinary temperatures, so it contributes no pressure term at all. Give it a coefficient of zero.

With hydrogen at 1 and oxygen at 3, the expression collapses to $K_{\mathrm{p}} = \frac{1}{1^{2} \times 3} = 0.333333$ . Type any numbers at all into the two unused pressure boxes — they are raised to the power zero and cannot move the answer.

Matching a value from a textbook

Published constants rarely agree to the last digit, and that is normal. Two things explain almost all of it: which pressure scale was used, and how many digits the source constant was printed with.

A tabulated gas constant of $8.314\ 462\ 1$ means $8.3144621$ — the trailing digits in parentheses are the uncertainty, not part of the value. Using that eight-digit figure instead of the exact one shifts the ammonia answer in the sixth significant digit, and nowhere else. If your two figures differ only that far, neither is wrong.

Going from pressure to concentration

Converting between a partial pressure and a molar concentration needs the molar concentration at equilibrium, so measure that first with the molarity calculator and then feed both constants into the conversion mode here.

Common mistakes

Multiplying by the coefficient instead of raising to it

Writing $P_{\mathrm{D}} \times d$ where the law of mass action calls for $P_{\mathrm{D}}^{\,d}$ happens more often than you would think, and it slips through unnoticed whenever $d = 1$ or $d = 2$ . Any solution that divides by the coefficient rather than taking a root is giving the wrong answer for a coefficient of three or five. The ammonia example above is the quickest way to tell the two apart.

Forgetting that only gases count

Water left as a liquid, or any solid, must carry a coefficient of zero. Counting it changes $\Delta n$ in the conversion mode and adds a spurious factor in the pressure mode.

Mixing pressure scales between the two modes

A partial pressure entered in kPa in one mode and a gas constant chosen in atm in the other will not agree, and the mismatch is silently absorbed into the answer. Use one scale throughout a single problem.

Expecting a temperature back-solve when Δn is zero

If the gaseous mole count does not change, $K_{\mathrm{p}}$ and $K_{\mathrm{c}}$ are the same number regardless of temperature, so the temperature simply is not determined. The calculator leaves that box empty rather than inventing an answer, and shows why.

Limitations

What this calculator does not model

The gas-law expression assumes ideal behaviour. At high pressures, or close to condensation, real gases need a fugacity correction and the result will drift.
The temperature is used exactly as entered. The engine does not correct for a thermometer, and it does not account for the heat of reaction.
In the pressure mode every pressure box needs a value before $K_{\mathrm{p}}$ is produced. For a species that does not take part, any number will do.
Two situations genuinely have no solution and are reported as such rather than estimated: a temperature back-solve when $\Delta n = 0$ , and a mole back-solve when $R \times T = 1$ .
Results are for coursework, design comparison and checking someone else's arithmetic. Do not use them as the sole basis of a safety case or a regulatory submission.

Questions

Why does my Kp change when I switch the pressure scale?

Because $K_{\mathrm{p}}$ is built from powers of pressures, so changing the scale changes the number by exactly $f^{\Delta n}$ . The constant itself has not moved; the unit it is measured against has. See the two formulas for the conversion.

Can Kp be zero or negative?

No. A real equilibrium leaves some of every species present, so the constant is strictly positive. A zero or negative entry is flagged and shown in red rather than quietly folded into a result.

What if my mole counts are equal on both sides?

Then $\Delta n = 0$ , the power of $R \times T$ becomes one, and $K_{\mathrm{p}} = K_{\mathrm{c}}$ exactly. It is a useful shortcut, and it is also why the temperature can no longer be recovered from the two constants.

Should I enter the coefficient of a solid or liquid?

No — leave it at zero. Its activity is fixed at one and it has no partial pressure to enter, so it drops out of the expression entirely.

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