Transforming K
Below are the ways that a K is transformed when transforming a chemical equation.
Reversing a chemical equation
When reversing a chemical equation, take the inverse of the equilibrium constant.
Consider the following general reaction
\[\mathrm{A} + \mathrm{2B} \rightleftharpoons \mathrm{3C}\]
The corresponding equilibrium expression is
\[K=\frac{[C]^{3}}{[\mathrm{A}][\mathrm{B}]^{2}}\]
Reversing the reaction
\[ \mathrm{3C} \rightleftharpoons\mathrm{A} + \mathrm{2B}\]
Gives the following equilibrium expression
\[K_{\mathrm{reverse}}=\frac{[\mathrm{A}][\mathrm{B}]^{2}}{[C]^{3}} = \dfrac{1}{K}\]
Multiplying a chemical equation
When multiplying a reaction by a number, raise the equlibrium constant to that number
Consider the following general reaction
\[\mathrm{A} + \mathrm{2B} \rightleftharpoons \mathrm{3C}\]
The corresponding equilibrium expression is
\[K=\frac{[C]^{3}}{[\mathrm{A}][\mathrm{B}]^{2}}\]
If you multiply all the stoichiometric coefficients by n
\[n\mathrm{A} + 2n\mathrm{B} \rightleftharpoons 3n\mathrm{C}\]
the equilibrium expression becomes
\[K'=\frac{[C]^{3n}}{[\mathrm{A}]^n[\mathrm{B}]^{2n}} = K^n\]
Add two or more equations together
When adding equations together, multiply each equilibrium constant together to get the new equilibrium constant.
Consider two reactions
\[\mathrm{A} \rightleftharpoons \mathrm{2B} \qquad K_1 = \dfrac{[\mathrm{B}]^2}{[\mathrm{A}]}\]
\[\mathrm{2B} \rightleftharpoons \mathrm{3C} \qquad K_2 = \dfrac{[\mathrm{C}]^3}{[\mathrm{B}]^2}\]
Adding these two reactions together gives
\[\mathrm{A} \rightleftharpoons 3\mathrm{C}\]
The equilibrium expression (Koverall) becomes
\[\begin{align*} K_{\mathrm{overall}} &= K_1 \times K_2 \\[2ex] &= \dfrac{[\mathrm{B}]^2}{[\mathrm{A}]} \times \dfrac{[\mathrm{C}]^3}{[\mathrm{B}]^2} \\[1ex] &= \dfrac{[\mathrm{C}]^3}{[\mathrm{A}]} \end{align*}\]