If I have been given an equation that involves a gas-aqueous phase equilibria, how do I write the equilibrium constant?

I have come across several questions that give equations having BOTH gas and the aqueous phase and I couldn't comprehend how to attempt them.

Should we find the concentrations of the gas too and use it for the equilibria?

But in several questions I was given partial pressure of the gas at equilibrium but I was never given the temperature to find concentration.

What is a proper way of representing the equilibrium constant?

Here is an example:

$$\ce{RNH2(g) + H2O(l) <=> RNH3+(aq) + OH-(aq)}$$

How do I write the equilibrium constant for this, if I have been given concentrations of $\ce{RNH3+(aq)}$ and $\ce{OH-(aq)}$ and the partial pressure of $\ce{RNH2}$ at equilibrium?

  • $\begingroup$ It would help if you would give an example of a specific problem. $\endgroup$
    – MaxW
    May 9, 2020 at 1:19
  • 1
    $\begingroup$ Yes I have edited my question now $\endgroup$ May 9, 2020 at 5:57

2 Answers 2


I dare say most equilibrium constant don't cite specific units because the appropriate units are implied implicitly. In the case of the reaction:

$$\ce{R-NH2(g) <=>[aqueous] R-NH3+ + OH-}$$

I'd write the mathematical equilibrium equation as:

$$K_{eq} = \dfrac{\ce{[R-NH3+][OH-]}}{p_\ce{R-NH2}}$$

and give the constant the appropriate units: $\pu{mol^2\cdot L^{-2} \cdot kPa^-1}$. (I just picked kPa as one of the zillion units for pressure. Use whatever units you were given the vapor pressure in: atm, torr, mm Hg, lbs/in^2, slugs/furlong^2 ... whatever)

This is obviously a "overall" equilibrium. According to Henry's law there should be an equilibrium between the vapor pressure of $\ce{R-NH2}$ and the concentration of $\ce{R-NH2}$ in solution. Since you have no way of calculating the concentration of $\ce{R-NH2}$ in solution you can't break the overall reaction into the two different equilibria.

  • 1
    $\begingroup$ There are no units in any equilibrium constant under any circumstances. The reason is that the activity coefficient is what truly defines the equilibrium constant, and activity is unitless. Typically, we just hide the fact that the pressures are measured with respect to a standard pressure of 1 atm and a standard concentration of 1 M. $\endgroup$ Oct 23, 2023 at 14:43

So there are different ways to express equilibrium there is the Kc value which, as the c denotes, is based on the concentration. In a question that was giving you moles or molarity of a substance that is a gas, we can find the concentration and using the coefficients to express a Kc.

Kc = [products]/[reactants] - take note of the bracket which means it is concentration.

However, in questions when you are given partial pressure, you must express the equilibrium expression as a value equating to Kp. The K constant is particular to a given temperature which is why increasing or decreasing the temperature will affect your Keq, however, you do not need the value of temperature to write an equilibrium expression.

Kp = (Pa)^a(Pb)^b - these are pressures and must be written with a parenthesis.

The third kind of equilibrium is dealing with substances in aqueous, for example, ions. In this case, the K value is Ksp or solubility of products and similarly, you are finding the concentration of the ions that dissociate.

In most cases however dealing with Kc, gases, and aqueous solutions are treated the same meaning we can directly put them in the Kc equation for equilibrium. It is only pure solids and liquids that are not expressed as their concentration does not change.

I hope this helps!


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