Suppose we have a heterogenous equilibrium :

$$\ce{A(aq) +B(aq) <=> C(g) +D(aq)}$$

Which equilibrium constant is used here?

Both pressure and concentration terms are there. So, according to me, both $K_\mathrm{p}$ and $K_\mathrm{c}$ can be written separately. If that is the case, can we relate both of them by $K_\mathrm{p} = K_\mathrm{c} (RT)^{\Delta n}$?

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    $\begingroup$ You seem to have a misconception in the actual concept of equilibrium constant. The actual values to be written are the activity. For simplicity, we say if the entire reaction is in water, $K_\mathrm{c}$ is taken and for gaseous reactions $K_\mathrm{p}$ is taken. This should help. The value of activity would be one for the solvent/a solid $\endgroup$ – Safdar Faisal Jul 23 '20 at 12:34
  • $\begingroup$ But there is no pure solid in the equation, aqueous reactants as well as gaseous reactants $\endgroup$ – Aditya Prakash Jul 23 '20 at 12:58
  • $\begingroup$ That was just an added information. Note the fact activity should be taken $\endgroup$ – Safdar Faisal Jul 23 '20 at 13:01
  • $\begingroup$ So gas's partial pressure can be taken as its activity and molar concentrations of the aqueous reactants can be taken as their activity and equilibrium constant's expression will have both pressure and molar concentration terms? $\endgroup$ – Aditya Prakash Jul 23 '20 at 13:05
  • $\begingroup$ This explains the different activity constants $\endgroup$ – Safdar Faisal Jul 23 '20 at 13:07

Your conception of the equilibrium constant is flawed. $K_\mathrm{p}$ is preferably used when the reaction only contains gaseous and solid components whereas $K_\mathrm{c}$ is used primarily for a reaction taking place in water.

In reality, however, both of these equilibrium constants are special cases of equilibrium. This is because in the actual scenario, what we use to calculate the equilibrium constant is a concept known as activity.

$K_\mathrm{p}$ and $K_\mathrm{c}$ are said to be special cases of $K$ since, in $K_\mathrm{p}$, the activity is expressed in terms of the partial pressure of the components and in $K_\mathrm{c}$, the activity is expressed in terms of concentrations of the products and reactants.

In case of reaction as stated above:

$$\ce{A(aq) + B(aq) <=> C(g) + D(aq)}$$

The equilibrium constant would be expressed as follows:

$$K = \frac{a_\mathrm{C}\cdot a_\mathrm{D}}{a_\mathrm{A}\cdot a_\mathrm{B}}$$

Now, for the aqueous components, the activity is expressed in terms of their concentration in the solvent. For the gaseous component, we express it in terms of its partial pressure. Substituting these conditions, we get:

$$K = \frac{P_\mathrm{C}[\mathrm{D}]}{[\mathrm{A}][\mathrm{B}]}$$

An example of where this is used is in electrochemistry.

A galvanic cell may be made which has a heterogenous reaction in which solid, aqueous and gaseous components may be given.

The Nernst equation used to calculate the EMF of such a cell is given as:

$$E_\mathrm{cell} = E^\circ_\mathrm{cell}-\frac{2.303RT}{nF}\log_{10}Q$$

Here, $Q$ is the reaction quotient. Using this, we calculate the EMF taking all the activities into consideration.

Note: The activity of a solid or the solvent is taken to be $1$ in such cases.

  • $\begingroup$ You don't need brackets for activity and concentration. i.e., either $C_{\ce{A}}$ or preferably $[\ce{A}]$. $\endgroup$ – Zhe Jul 23 '20 at 14:56
  • $\begingroup$ Fixed. Thanks @Zhe $\endgroup$ – Safdar Faisal Jul 23 '20 at 15:02

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