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 expresion 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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