If two or more reactions are added to give another, the equilibrium constant for the reaction is the product of the equilibrium constants of the equations added.

How is this true? What is the reason?


This is a direct result from the law of mass action.

When you have an equilibrium

$$\ce{A <=> B}$$

with the equilibrium constant

$$K_1=\frac {[\ce{B}]}{[\ce{A}]}\tag1$$

and a second equilibrium

$$\ce{B <=> C}$$

with the equilibrium constant

$$K_2=\frac {[\ce{C}]}{[\ce{B}]}\tag2$$

you can combine both chemical equations:

$$\begin{align} \ce{A \;&<=> B}\\ \ce{B \;&<=> C}\\ \hline \ce{A \;&<=> C} \end{align}$$

Then you can rearrange Equation $\text(1)$

$$\begin{align} K_1&=\frac {[\ce{B}]}{[\ce{A}]}\tag1\\[6pt] \Leftrightarrow [\ce{B}]&=K_1\cdot[\ce{A}] \end{align}$$

and substitute $[\ce{B}]$ in Equation $\text(2)$

$$\begin{align} K_2&=\frac {[\ce{C}]}{[\ce{B}]}\tag2\\[6pt] K_2&=\frac {[\ce{C}]}{K_1\cdot[\ce{A}]}\\[6pt] \Leftrightarrow K_1\cdot K_2&=\frac {[\ce{C}]}{[\ce{A}]} \end{align}$$

Thus the equilibrium constant for the combined equilibrium is the product $K_2\cdot K_1$.

| improve this answer | |
  • $\begingroup$ Because reactant $\ce{B}$ doesn't show up in the net reaction this doesn't mean that its concentration is zero right? $\endgroup$ – ado sar Oct 26 at 18:19

The equilibrium constant is related to the Gibbs free energy change: $K \propto \exp(-\Delta_0 G / RT)$. When adding equations, you can add the free energies:

$\Delta_0 G = \sum_i \Delta_0 G_i$,

so the equilibrium constant becomes the product:

$K\propto \exp(-\Delta_0 G / RT) = \exp\left(-\sum_i\Delta_0 G_i / RT \right) = \prod_i \exp\left(-\Delta_0 G_i / RT \right)$.

| improve this answer | |
  • $\begingroup$ You're right, my $\Delta G$ was not per mole but per particle which is not very usual. I corrected it, thanks. $\endgroup$ – albapa Nov 23 '15 at 22:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.