6.7 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.