**Determining Equilibrium Quantities from Initial Conditions **

The quantities (concentrations or partial pressures) at equilibrium can be determined if a set of initial conditions is provided. Initial, the system may not be at equilibrium (\(Q \ne K\)) and the values will shift accordingly to reach equilibrium (since a chemical reaction proceeding toward equilibrium is thermodynamically favored). Stoichiometric relationships between reactants and products are included in determining the amount by which each component changes.

**Example 1.**

For the reaction below, there is initially \(1.0\text{ atm}\) of \(\require{mhchem}\ce{NO}\) and \(0.50\text{ atm}\) \(\require{mhchem}\ce{O2}\) present. What is the equilibrium partial pressure of \(\require{mhchem}\ce{NO2}\)?

\(\require{mhchem}\ce{2 NO(g) + O2(g) <=>2 NO2(g)}\;\;\;\; K = 1.3 \times 10 ^{-5}\)

A. \(1.0\text{ atm}\)

B. \(1.3 \times 10 ^{-3} \text{ atm}\)

C. \(2.6 \times 10 ^{-3} \text{ atm} \)

*Solution*

C. \(2.6 \times 10 ^{-3} \text{ atm} \)

Initially, \(Q = 0\), since there are only reactants present (and no products). Since \(Q < K\), the partial pressures of reactants must decrease (by \(-2x \) and \( -x\), respectively, for \(\require{mhchem}\ce{NO}\) and \(\require{mhchem}\ce{O2}\)) and the partial pressure of the product must increase (by \( +2x\)).

At equilibrium, \(K = \frac{P^2_{NO_2}}{P^2_{NO} \times P_{O_2}} = \frac{(2x)^2}{(1.0-2x)^2(0.50-x)}\)

Since the equilibrium constant is small, the value \(x\) is small relative to the initial quantities.

\(K = \frac{(2x)^2}{(1.0-2x)^2(0.50-x)} \simeq \frac{(2x)^2}{(1.0)^2(0.50)} = 1.3\times10^{-5}\)

\(x = 1.27\times 10^{-3} \text{ atm}\)

At equilibrium, \(P_{NO_2} = 2x = 2(1.27 \times 10 ^{-3}) = 2.55 \times 10 ^{-3} \text{ atm}\)

**Example 2.**

For the reversible reaction \(\require{mhchem}\ce{A + B <=> C + D}\), if \(Q > K\) initially, what will happen to the \(\require{mhchem}\ce{[A]}\) as the reaction approaches equilibrium?

A. increase

B. decrease

C. no change

*Solution*

A. increase

If \(Q > K\), the reaction consumes products and generates reactants in order to reach equilibrium. Since \(\require{mhchem}\ce{A} \) is a reactant, its concentration will increase.

**Example 3.**

For the reaction \(\require{mhchem}\ce{2 NO2(g) <=> N2O4(g)}\), \(K = 169\). If the initial concentrations are \(\require{mhchem}\ce{[NO2]} = 0.010 \text{ M}\) and \(\require{mhchem}\ce{[N2O4]} = 0.50 \text{ M}\), which statement is true?

A. The reaction is at equilibrium, and no change will occur.

B. The concentration of \(\require{mhchem}\ce{N2O4}\) will increase.

C. The concentration of \(\require{mhchem}\ce{NO2}\) will increase.

*Solution*

C. The concentration of \(\require{mhchem}\ce{NO2}\) will increase.

Solve for \(Q \) using the initial concentrations.

\(Q = \frac{\require{mhchem}\ce{[N2O4]}}{[NO_2]^2} = \frac{0.50}{0.010^2} = 5000\)

Since \(Q > K\) (\(5000 > 169\)), the concentration of product (\(\require{mhchem}\ce{N2O4}\)) will decrease and the concentration of reactant (\(\require{mhchem}\ce{NO2}\)) will increase.