4.7 Transforming K

This section is a bit of a mathematical exercise. We can change a reversible chemical reaction and rationalize the change in the equilibrium constant.

Reversing the reaction

For example, the reaction

\[\mathrm{N_2O_4}(g) \rightleftharpoons 2\mathrm{NO}_2 (g)\]

has the following equilibrium expression

\[K = \dfrac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} = 4.46\times 10^{-3}\] indicating that the reaction is reactant, N₂O₄, is favored. If we reverse the reaction such that

\[2\mathrm{NO}_2 (g) \rightleftharpoons \mathrm{N_2O_4}(g) \] the equilibrium expression is

\[K' = \dfrac{[\mathrm{N_2O_4}]}{[\mathrm{NO_2}]^2}\]

and the resulting equilibrium constant is

\[K' = \dfrac{1}{K} = \dfrac{1}{4.46\times 10^{-3}} = 224.22\]

and the product, N₂O₄, is favored.

Multiplying the reaction

Take the following reaction

\[\mathrm{N_2O_4}(g) \rightleftharpoons 2\mathrm{NO}_2 (g)\]

and double everything (multiply by two) to give

\[2\mathrm{N_2O_4}(g) \rightleftharpoons 4\mathrm{NO}_2 (g)\]

Let us compare the original equilibrium expression with the new one.

\[\begin{align*} K &= \dfrac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} \\[1.5ex] K' &= \dfrac{[\mathrm{NO_2}]^4}{[\mathrm{N_2O_4}]^2} \end{align*}\]

By multiplying every species by 2, the exponents in the equilibrium expression were multiplied by two. The new equilibrium constant is then

\[K' = K^2 = (4.46\times 10^{-3})^2 = 1.99\times 10^{-5}\]

We generalize this by stating that multiplying a reaction by n gives rise to a new equilibrium constant such that

\[K' = K^n\]

Adding reactions together

For a multi-step reaction

\[\begin{align*} \mathrm{A} &\rightleftharpoons \mathrm{2B} &&\qquad K_1 = \dfrac{[\mathrm{B}]^2}{[\mathrm{A}]} \\[1.5ex] \mathrm{2B} &\rightleftharpoons \mathrm{3C} &&\qquad K_2 = \dfrac{[\mathrm{C}]^3}{[\mathrm{B}]^2} \end{align*}\]

when added together gives

\[\mathrm{A} \rightleftharpoons \mathrm{3C}\]

giving a new equilibrium constant such that

\[\begin{align*} K' &= K_1 \times K_2 \\[2ex] &= \dfrac{[\mathrm{B}]^2}{[\mathrm{A}]} \times \dfrac{[\mathrm{C}]^3}{[\mathrm{B}]^2} \\[1ex] &= \dfrac{[\mathrm{C}]^3}{[\mathrm{A}]} \end{align*}\]

Therefore, multiply the equilibrium constants of both reactions together to give the final equilibrium constant.