12/14 Chemical Equilibrium

The Nature of Chemical Equilibrium

A reversible reaction can proceed in both directions. For the general reaction

reactants form products in the forward direction, while products reform reactants in the reverse direction. At dynamic equilibrium,

The concentrations remain constant because the two rates are equal, not because the particles stop reacting. The concentrations of reactants and products do not have to be equal, and equilibrium does not require that all reactants be consumed.

Equilibrium can be reached from either direction. Starting with reactants produces products; starting with products produces reactants. At a fixed temperature, both paths can lead to the same equilibrium composition.

Takeaway: Equilibrium is a dynamic balance of rates, not a static state or necessarily an equal mixture.

Writing and Interpreting Equilibrium Constants

For the balanced reaction

the concentration-based equilibrium expression is

The brackets represent equilibrium molar concentrations, and the exponents are the stoichiometric coefficients in the balanced equation. For gases, a pressure-based expression can be written using partial pressures:

Pure solids and pure liquids are omitted because their activities remain effectively constant.

For example, for

the expression is

For

the pure solids are omitted, giving

The magnitude of the equilibrium constant indicates the favored composition:

• If $K\gg 1$, products are favored.

• If $K\ll 1$, reactants are favored.

• If $K\approx 1$, appreciable amounts of both sides are present.

The value of $K$ does not indicate how quickly equilibrium is reached. If a reaction is reversed, $K_{\text{reverse}}=1/K_{\text{forward}}$. If every coefficient is multiplied by $n$, the new constant is $K_{\text{new}}=K_{\text{original}}^n$. When reactions are added, their equilibrium constants are multiplied.

Takeaway: Write the equilibrium expression from the balanced equation, omit pure solids and liquids, and distinguish equilibrium composition from reaction speed.

Using the Reaction Quotient

The reaction quotient uses the same form as the equilibrium constant but can be calculated for a mixture that is not yet at equilibrium. For

Compare $Q$ with $K$ to determine the direction of the net reaction:

• If $Q<K$, the reaction proceeds toward products.

• If $Q=K$, the system is at equilibrium.

• If $Q>K$, the reaction proceeds toward reactants.

For

If $[\ce{NO2}]=0.100\,\mathrm{M}$ and $[\ce{N2O4}]=0.0500\,\mathrm{M}$, then

When $K_c=0.200$, $Q_c=K_c$, so the mixture is already at equilibrium. If instead $[\ce{NO2}]=0.200\,\mathrm{M}$, then

Because $Q_c>K_c$, the mixture contains too much product relative to equilibrium, and the net reaction proceeds in reverse to form more $\ce{N2O4}$.

Takeaway: Calculate $Q$ from the current composition, then compare it with $K$ before predicting a shift.

Predicting Equilibrium Shifts

Le Châtelier’s principle provides a qualitative way to predict how an equilibrium responds to a disturbance. The same reasoning can be expressed quantitatively: a concentration or pressure change alters $Q$, and the system shifts until $Q=K$ again at the unchanged temperature.

For the Haber reaction,

concentration changes produce these effects:

• Adding $\ce{N2}$ or $\ce{H2}$ shifts the equilibrium toward products.

• Removing $\ce{NH3}$ shifts the equilibrium toward products.

• Adding $\ce{NH3}$ shifts the equilibrium toward reactants.

• Removing a reactant shifts the equilibrium toward reactants.

For gases, decreasing volume increases pressure and favors the side with fewer moles of gas. The Haber reaction has four moles of gas on the reactant side and two moles on the product side, so decreasing volume shifts it toward $\ce{NH3}$, while increasing volume shifts it toward the reactants. If both sides have the same total number of gas moles, a volume change does not shift the equilibrium. Adding an inert gas at constant volume also does not shift the equilibrium because the reacting species’ partial pressures do not change.

Temperature changes are different because they change $K$. Treat heat as a reactant in an endothermic reaction:

Increasing temperature shifts an endothermic reaction toward products. For an exothermic reaction, heat is a product, so increasing temperature shifts the reaction toward reactants.

A catalyst increases both the forward and reverse reaction rates. It helps the system reach equilibrium sooner but does not change the equilibrium position or the value of $K$.

Takeaway: Concentration, pressure, and volume changes alter the position of equilibrium without changing $K$ at constant temperature; temperature changes $K$, and catalysts change only the rate of approach.

Calculating Equilibrium Concentrations with ICE

The ICE method organizes an equilibrium calculation into Initial, Change, and Equilibrium stages. Consider

with $K_c=0.0200$ and initial concentrations

Because two moles of $\ce{HI}$ decompose for every one mole of $\ce{H2}$ and $\ce{I2}$ formed, let the reaction change be represented by $x$:

• Initial: $[\ce{HI}]=0.500$, $[\ce{H2}]=0$, and $[\ce{I2}]=0$.

• Change: $[\ce{HI}]=-2x$, $[\ce{H2}]=+x$, and $[\ce{I2}]=+x$.

• Equilibrium: $[\ce{HI}]=0.500-2x$, $[\ce{H2}]=x$, and $[\ce{I2}]=x$.

Substitute the equilibrium concentrations into

to obtain

Solving gives $x=0.0551\,\mathrm{M}$. Therefore,

Check the result:

The calculated concentrations reproduce the given equilibrium constant, so the result is consistent. Any mathematically possible solution that produces a negative concentration must be rejected.

Takeaway: Stoichiometric coefficients determine the change row, and the final answer should always be checked by substituting the equilibrium concentrations back into the expression for $K$.

Approximations and a Reliable Workflow

In some calculations, a change variable can be treated as small compared with an initial concentration. For example, if the equilibrium expression contains $0.500-2x$, it may sometimes be approximated as $0.500$ when the change is less than approximately $5\%$ of the initial amount.

Estimate the relative change with

The approximation is acceptable only if the resulting change is sufficiently small. It must be checked after solving. If the change exceeds about $5\%$, solve the exact equation, often using the quadratic formula. Do not apply the approximation automatically.

A reliable workflow is:

1. Balance the chemical equation.

2. Write the correct equilibrium expression, omitting pure solids and liquids.

3. Calculate $Q$ if the initial mixture may not be at equilibrium.

4. Compare $Q$ and $K$ to identify the direction of change.

5. Set up an ICE table.

6. Use stoichiometric coefficients to express every change with one variable.

7. Substitute the equilibrium concentrations into the expression for $K$.

8. Solve for the variable and reject any result that gives a negative concentration.

9. Check the result by recalculating $K$.

10. Report appropriate significant figures and units when required.

Takeaway: Approximation can simplify an equilibrium calculation, but only a post-solution check can justify it.