# Equilibrium Expression

An equilibrium expression for a general reaction

$aA + bB \rightleftharpoons cC + dD$

can be written as

$\dfrac{[C]^c[D]^d}{[A]^a[B]^b} = Q~\mathrm{or}~K$ This is a ratio of the product concentrations to the reactant concentrations, each raised to the power of the stoichiometric coefficient in the balanced chemical equation.

Note: Solids (s) and pure liquids (l) are never written into an equilibrium expression.

An equilibrium expression for a reaction involving gases can be written out in terms of their partial pressures (instead of their concentrations) as

$\dfrac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b} = Q_P~\mathrm{or}~K_P$

## Reactant Quotient, Q

Imagine a line that represents a ratio of reactants and products for a given chemical reaction. If you were at the beginning of the line, it would represent having all reactants and no products. If you were at the end of the line, it would represent having all products and no reactants.

Anywhere in between would represent a mixture of both reactants and products. Making a ratio of the concentrations of products to reactants at that position would give the reaction quotient, Q, written as

$Q = \dfrac{[\mathrm{products}]}{[\mathrm{reactants}]}$

The further to the right on the line you were, the larger the product concentration and the smaller the reactant concentration. Here, Q would be greater than 1.

The further to the left on the line you were, the smaller the product concentration and the larger the reactant concentration.

## Equilibrium Constant, K

The equilibrium constant, K, represents the concentrations of products to reactants at equilibrium for a reaction. A reversible reaction will always proceed toward this state of equilibrium, K, no matter where you begin the reaction, Q.

For example, imagine starting a reversible reaction with all reactants and no products. Your starting location is Q. Imagine that the equilibrium location, K, is toward the right on this line. The reaction will “proceed to the right” until it reaches K.

You can imagine, in this situation, that the distance, x, between Q (the starting point) and K (the ending point) is “large” . That is, the reaction must proceed quite a bit before it hits equilibrium.

You can also start on the opposite end with all product. Here, x would be small.

A reaction can also start anywhere on the line. If Q is less than K, the reaction will proceed to the right. If Q is greater than K, the reaction will proceed to the left.

$\mathrm{If}~Q < K~\qquad \mathrm{reaction~proceeds~right}$ $\mathrm{If}~Q > K~\qquad \mathrm{reaction~proceeds~left}$

## Practice Problems

### Problem 1

For a reaction, Q = 0.125 and K = 1.35 × 108. Which way will the reaction proceed?

Solution

Since Q < K, reaction will proceed to the right.

### Problem 2

For the reaction

$\mathrm{A} \rightleftharpoons \mathrm{2B}$

the equilibrium concentrations are

\begin{align*} [\mathrm{A}]_{\mathrm{eq}} &= 0.346~M\\ [\mathrm{B}]_{\mathrm{eq}} &= 1.053~M \end{align*}

Solve for the equilibrium constant, K.

Solution

\begin{align*} K &= \dfrac{[B]^2}{[A]} \\[1ex] &= \dfrac{(1.053)^2}{0.346}\\[1ex] &= 3.20 \end{align*}

### Problem 3

For the reaction

$\mathrm{A} \rightleftharpoons \mathrm{2B} \qquad K =3.20$

the initial concentrations are

\begin{align*} [\mathrm{A}]_{\mathrm{i}} &= 5.583~M\\ [\mathrm{B}]_{\mathrm{i}} &= 0.753~M \end{align*}

Which way will the reaction proceed?

Solution

\begin{align*} Q &= \dfrac{[B]^2}{[A]} \\[1ex] &= \dfrac{(0.753)^2}{5.583}\\[1ex] &= 0.102 \end{align*}

Since Q < K, the reaction will proceed to the right.

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