7.7 Calculating Equilibrium Concentrations · 7.8 Representations of Equilibrium

7.7 Calculating Equilibrium Concentrations · 7.8 Representations of Equilibrium

At equilibrium, a reaction mixture is not “finished”: particles continue colliding and reacting in both directions, while the macroscopic concentrations remain constant. The central calculation is therefore to connect a balanced equation, concentration changes, and the equilibrium constant.

Learning Objective 7.7: The student is able to calculate equilibrium concentrations by applying stoichiometry and the equilibrium constant expression.
Essential Knowledge 7.7.A: The equilibrium constant expression relates equilibrium concentrations of reactants and products.
Learning Objective 7.8: The student is able to use representations to describe the relationships between the concentrations of reactants and products in an equilibrium system.
Essential Knowledge 7.8.A: Representations of equilibrium must agree with the balanced equation and its stoichiometric relationships.

From a balanced equation to an equilibrium expression

For the general reaction

aA+bB\rightleftharpoons cC+dD

the concentration-based equilibrium constant is

$$ K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}. $$

The exponents are the stoichiometric coefficients from the balanced equation. Pure solids and pure liquids are omitted because their effective concentrations remain constant.

For example,

Sr(OH)_2(s)\rightleftharpoons Sr^{2+}(aq)+2OH^-(aq)

has

$$ K_{sp}=[Sr^{2+}][OH^-]^2. $$

The solid $Sr(OH)_2$ does not appear. If a saturated solution contains $[Sr^{2+}]=0.043\ \mathrm{M}$, the balanced equation requires

$$ [OH^-]=2(0.043\ \mathrm{M})=0.086\ \mathrm{M}. $$

Thus,

$$ K_{sp}=(0.043)(0.086)^2=3.18\times10^{-4}. $$

Misconception check — “The ions dissolve in a $1:1$ ratio.” They do not. Each formula unit produces one $Sr^{2+}$ ion and two $OH^-$ ions. A particulate model must therefore show twice as many hydroxide ions as strontium ions.

Calculating equilibrium concentrations with an ICE relationship

An ICE relationship records initial concentration, change, and equilibrium concentration. The change row comes directly from stoichiometry; the equilibrium row combines the initial amount with the unknown change.

Consider

N_2O_4(g)\rightleftharpoons 2NO_2(g)

with initial concentrations $[N_2O_4]_0=1.00\ \mathrm{M}$ and $[NO_2]_0=0$, and $K_c=0.36$.

Species	Initial	Change	Equilibrium
$N_2O_4$	$1.00$	$-x$	$1.00-x$
$NO_2$	$0$	$+2x$	$2x$

Substitute the equilibrium concentrations into the expression:

$$ 0.36=\frac{[NO_2]^2}{[N_2O_4]} =\frac{(2x)^2}{1.00-x}. $$

Solving gives $x=0.258$ (the negative mathematical root is chemically impossible). Therefore,

$$ [N_2O_4]_{\mathrm{eq}}=1.00-0.258=0.742\ \mathrm{M} $$

and

$$ [NO_2]_{\mathrm{eq}}=2(0.258)=0.517\ \mathrm{M}. $$

Equilibrium check:

$$ K_c=\frac{(0.517)^2}{0.742}=0.360\approx0.36. $$

The result is self-consistent. Notice that $NO_2$ changes by $2x$, not $x$; losing that coefficient produces an incorrect equilibrium composition even if the algebra is flawless.

The reaction quotient predicts the net direction

The reaction quotient, $Q$, has the same concentration form as $K$, but it uses concentrations that are not necessarily at equilibrium:

$$ Q_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}. $$

Compare $Q$ with the already-known value of $K$:

• If $Q<K$, the mixture contains too much reactant relative to equilibrium, so the net reaction proceeds toward products.
• If $Q>K$, the mixture contains too much product, so the net reaction proceeds toward reactants.
• If $Q=K$, the system is at equilibrium.

For the $N_2O_4/NO_2$ system, suppose the current concentrations are $[N_2O_4]=0.80\ \mathrm{M}$ and $[NO_2]=0.20\ \mathrm{M}$:

$$ Q_c=\frac{(0.20)^2}{0.80}=0.050. $$

Because $0.050<0.36$, the net change is toward products: more $N_2O_4$ dissociates to form $NO_2$ until $Q$ rises to $K$.

Representations must preserve stoichiometry

A correct equilibrium representation can be symbolic, graphical, particulate, or numerical, but all representations must tell the same chemical story. For $N_2O_4\rightleftharpoons2NO_2$, a particle diagram must represent each $N_2O_4$ molecule as one unit and each dissociation event as two $NO_2$ molecules; a $1:1$ product ratio contradicts the equation.

A specific stress can temporarily make $Q\ne K$. For example, adding $N_2O_4$ increases the denominator of

$$ Q_c=\frac{[NO_2]^2}{[N_2O_4]}, $$

so $Q$ becomes less than $K$. The system responds by shifting toward $NO_2$, restoring $Q=K$; the value of $K$ itself changes only if temperature changes.

Retrieval check: For

A(s)\rightleftharpoons2B(aq),

write the equilibrium expression, state the concentration change if $[A]$ dissolves by $x$, and predict the direction when $Q>K$.
Answer: $K=[B]^2$; $[B]$ changes by $+2x$; the system shifts toward reactants.