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






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






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.









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.