At equilibrium, the rates of the forward and reverse reactions are equal, and the concentrations of reactants and products remain constant. This state of balance is quantified by the equilibrium constant, which indicates the ratio of products to reactants. Chemists define this ratio in two primary ways: the equilibrium constant based on concentration, \(K_c\), and the equilibrium constant based on partial pressures, \(K_p\). The \(K_c\) value uses the molar concentrations (Molarity) of the gaseous or dissolved species, whereas \(K_p\) is specifically used for gaseous systems and relies on the partial pressures of the gases involved.
The Mathematical Relationship Between \(K_p\) and \(K_c\)
The connection between the two equilibrium constants, \(K_p\) and \(K_c\), is expressed through a single, standardized formula that applies exclusively to systems involving gases. The relationship is always defined by the equation: \(K_p = K_c (RT)^{\Delta n}\). The conversion is only necessary and possible for reactions occurring in the gas phase where partial pressures can be measured. If the change in the number of moles of gas is zero, the term \((RT)^{\Delta n}\) simplifies to one, and \(K_p\) becomes equal to \(K_c\).
Decoding the Variables
The symbol \(R\) represents the Universal Gas Constant. Its value must be selected to align with the pressure units typically used in \(K_p\) calculations, which are usually atmospheres. Therefore, the value \(R = 0.08206 \text{ L} \cdot \text{atm} / (\text{mol} \cdot \text{K})\) is the standard choice for this conversion.
The variable \(T\) is the absolute temperature of the reaction system, and it must always be expressed in Kelvin (K). If the temperature is provided in Celsius, it must be converted to Kelvin by adding \(273.15\) to the Celsius value before substituting it into the equation.
The exponent \(\Delta n\) (pronounced “delta n”) is the change in the number of moles of gaseous species during the reaction. This value is determined by subtracting the total number of moles of gaseous reactants from the total number of moles of gaseous products, as derived from the balanced chemical equation. Only species specifically noted as gases (g) are included in this count, while solids (s), liquids (l), and aqueous species (aq) are ignored.
Why the Ideal Gas Law Governs the Conversion
The fundamental reason a simple mathematical relationship exists between the two constants stems from the Ideal Gas Law, \(PV = nRT\). The Ideal Gas Law, \(PV = nRT\), directly relates the physical properties of an ideal gas. The concentration constant \(K_c\) is based on molar concentration, which is moles divided by volume (\(n/V\)). By algebraically rearranging the Ideal Gas Law to \(P = (n/V)RT\), one can see that the partial pressure of a gas is directly proportional to its molar concentration at a constant temperature. Substituting this concentration-pressure relationship for every gaseous component in the \(K_p\) expression and simplifying the resulting terms eventually leads to the \(K_p = K_c (RT)^{\Delta n}\) formula.
Applying the Conversion: A Guided Example
To illustrate the conversion process, consider the Haber process, a well-known industrial reaction for the synthesis of ammonia: \(\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)\). Assume the reaction reaches equilibrium at a temperature of \(400^\circ\text{C}\) and the concentration-based equilibrium constant is \(K_c = 0.5\). The objective is to find \(K_p\).
The first step is to calculate \(\Delta n\), the change in the number of moles of gas. The product side has two moles of \(\text{NH}_3(g)\), and the reactant side has one mole of \(\text{N}_2(g)\) and three moles of \(\text{H}_2(g)\), totaling four moles of reactants. Therefore, \(\Delta n = 2 \text{ (products)} – 4 \text{ (reactants)} = -2\).
Next, the temperature must be converted to Kelvin: \(T(\text{K}) = 400^\circ\text{C} + 273.15 = 673.15 \text{ K}\). Now, all known values can be substituted into the conversion formula \(K_p = K_c (RT)^{\Delta n}\).
Using \(K_c = 0.5\), \(R = 0.08206 \text{ L} \cdot \text{atm} / (\text{mol} \cdot \text{K})\), \(T = 673.15 \text{ K}\), and \(\Delta n = -2\), the equation becomes \(K_p = 0.5 \cdot (0.08206 \cdot 673.15)^{-2}\). The product of \(R\) and \(T\) inside the parentheses is calculated first: \(0.08206 \cdot 673.15 \approx 55.24\). The equation then simplifies to \(K_p = 0.5 \cdot (55.24)^{-2}\). Calculating the term raised to the power of \(-2\) yields \(K_p = 0.5 \cdot (0.0003276)\). The final calculation results in \(K_p \approx 0.00016\).