# 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*}

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