Chemical reactions often reach a state of chemical equilibrium, a dynamic process where the rates of the forward and reverse reactions are equal. At equilibrium, the concentrations of reactants and products remain constant. The equilibrium constant (\(K\)) quantifies the ratio of products to reactants at this steady state for a specific temperature. For gaseous reactions, the equilibrium constant is expressed in terms of the partial pressures of the gases involved, designated as \(K_p\). The \(K_p\) value is fixed for a given reaction at a particular temperature and offers insight into the extent of the reaction.
Understanding the \(K_p\) Expression
The expression for \(K_p\) is a mathematical ratio derived from the balanced chemical equation. The partial pressures of the products form the numerator, and the partial pressures of the reactants form the denominator. For a generic reversible reaction involving gases, such as \(aA + bB \rightleftharpoons cC + dD\), the \(K_p\) expression is \(K_p = \frac{P_C^c P_D^d}{P_A^a P_B^b}\). Here, \(P\) represents the partial pressure of each gaseous species at equilibrium, and the superscripts (\(a, b, c, d\)) are the stoichiometric coefficients.
Each partial pressure is raised to the power of its corresponding coefficient, reflecting the reaction’s stoichiometry. Only gaseous species are included in the \(K_p\) expression.
Pure solids and pure liquids are omitted from the expression because their effective pressure remains constant and is incorporated into the constant’s overall value. For example, in the thermal decomposition of calcium carbonate (\(\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)\)), the \(K_p\) expression simplifies to \(K_p = P_{\text{CO}_2}\), since both \(\text{CaCO}_3\) and \(\text{CaO}\) are solids. A large \(K_p\) value indicates that the reaction favors the formation of products at equilibrium.
Solving \(K_p\) Directly from Partial Pressures
The most direct method for determining \(K_p\) is to measure the partial pressures of all gaseous components at equilibrium and substitute these values into the \(K_p\) expression. The partial pressure of a single gas is the pressure it would exert if it occupied the total volume alone. This value is commonly calculated by multiplying the gas’s mole fraction by the total system pressure at equilibrium.
Consider the synthesis of ammonia (the Haber process): \(\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)\). The \(K_p\) expression is \(K_p = \frac{P_{\text{NH}_3}^2}{P_{\text{N}_2} P_{\text{H}_2}^3}\). To solve for \(K_p\), the equilibrium partial pressures for each gas must first be established, typically measured in atmospheres (atm) or bars.
Suppose a reaction vessel at a specific temperature reaches equilibrium with the following measured partial pressures: \(P_{\text{N}_2} = 0.50 \text{ atm}\), \(P_{\text{H}_2} = 1.50 \text{ atm}\), and \(P_{\text{NH}_3} = 0.80 \text{ atm}\). Substitute these pressures into the equilibrium expression, applying the stoichiometric coefficients as exponents.
The calculation proceeds: \(K_p = \frac{(0.80)^2}{(0.50) \cdot (1.50)^3}\). The numerator is \(0.64\). The denominator is calculated as \((1.50)^3 = 3.375\), which is multiplied by \(0.50\), resulting in \(1.6875\).
Dividing the numerator by the denominator yields \(K_p\): \(\frac{0.64}{1.6875} \approx 0.38\). Although individual partial pressures have units, the resulting equilibrium constant \(K_p\) is generally reported as a unitless quantity. This value provides a snapshot of the product-to-reactant ratio at equilibrium for that specific temperature.
Converting Between \(K_c\) and \(K_p\)
If the equilibrium constant is known in terms of molar concentrations (\(K_c\)), \(K_p\) can be calculated using the relationship derived from the ideal gas law: \(K_p = K_c(RT)^{\Delta n}\). This formula provides an indirect method for solving for \(K_p\) without needing individual partial pressures.
The variables in this equation require specific units for a correct result. \(R\) is the ideal gas constant, typically \(0.08206 \text{ L} \cdot \text{atm}/\text{mol} \cdot \text{K}\). \(T\) must be the absolute temperature of the system, expressed in Kelvin (K), which is obtained by adding \(273.15\) to the Celsius temperature.
The term \(\Delta n\) represents the change in the number of moles of gas during the reaction, calculated as the total moles of gaseous products minus the total moles of gaseous reactants. For the reaction \(\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)\), the moles of gaseous products is \(2\), and the moles of gaseous reactants is \(1 + 3 = 4\).
The calculation for the change in moles is \(\Delta n = 2 – 4 = -2\). If the total number of moles of gas remains the same (\(\Delta n = 0\)), then \(K_p\) is numerically equal to \(K_c\) because \((RT)^0\) equals \(1\). This conversion is useful when only concentration data is available for a gaseous equilibrium system.