**Calculating Equilibrium Constant **

The equilibrium constant, K, for a reaction at equilibrium is calculated as the ratio of products to reactants, each raised to respective coefficients. For example, for the chemical reaction \(\require{mhchem}\ce{aA + bB <=> cC + dD}\), the equilibrium constant based on molar concentration is formatted as \(K =\frac{[A]^a[B]^b}{[C]^c[D]^d}\), where each concentration is the concentration of the species at equilibrium. Remember that there is no net change in the concentrations at equilibrium. Only the equilibrium values can be used with the equilibrium constant; at other points in the reaction, the more general reaction quotient is used.

**Example 1.**

The diagram shows an equilibrium mixture for the reaction \(\require{mhchem}\ce{A <=> B}\), where each circle represents \(1 \text{ M}\). What is the value of \(K \) for this reaction?

A. \(3\)

B. \(0.33\)

C. \(1\)

*Solution*

B. \(0.33\)

\(K = \frac{[B]}{[A]} = \frac{2}{6} = 0.33\)

**Example 2.**

For the reaction \(\require{mhchem}\ce{2A(g) + B(g) <=> 3C(g)}\), the equilibrium concentrations are\( \require{mhchem}\ce{[A]} = 0.25\text{ M}\), \(\require{mhchem}\ce{[B]} = 0.10\text{ M}\), and \(\require{mhchem}\ce{[C]} = 0.010\text{ M}\). What is the equilibrium constant, \(K\)?

A. \(0.0016\)

B. \(0.40\)

C. \(25,000\)

*Solution*

A. \(0.0016\)

\(K = \frac{[C]^3}{[A]^2\times [B]} = \frac{(0.010)^3}{(0.25)^2\times0.10} = 0.0016\)

**Example 3.**

At equilibrium for the following reaction \(\require{mhchem}\ce{N2(g) + 3H2(g) <=> 2NH3(g)}\), the partial pressures are \(P_{N_2} = 2.00\text{ atm}\), \(P_{H_2} = 1.00\text{ atm}\), and \(P_{NH_3} = 3.00\text{ atm}\). What the equilibrium constant, \(K\)?

A. \(1.50\)

B. \(3.00\)

C. \(4.50\)

*Solution*

C. \(4.50\)

\(K = \frac{P^2_{NH_3}}{P_{N_2}\times P^3_{H_2}} = \frac{3.00^2}{2.00\times(1.00)^3} = 4.50\)