Chemical reactions that can proceed in both the forward and reverse directions are known as reversible reactions. When the rate of the forward reaction becomes exactly equal to the rate of the reverse reaction, the system has achieved a state called chemical equilibrium. At this point, the concentrations of reactants and products remain constant, although the reactions are still occurring at a molecular level. To quantify the position of this equilibrium, chemists use an equilibrium constant, which expresses the ratio of products to reactants. For systems involving gases, this constant is defined in terms of the partial pressures of the gaseous species and is designated as \(K_p\). The magnitude of \(K_p\) indicates the extent to which a reaction favors the formation of products at a specific temperature.
Defining the Equilibrium Constant Based on Pressure
The equilibrium constant \(K_p\) is mathematically defined as the ratio of the partial pressures of the products to the partial pressures of the reactants, with each pressure value raised to the power of its stoichiometric coefficient from the balanced chemical equation. For a general reversible reaction involving gaseous species, where \(aA + bB \rightleftharpoons cC + dD\), the expression for \(K_p\) is written as \(K_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b}\). In this formula, \(P_X\) represents the partial pressure of species \(X\), and the lowercase letters \(a\), \(b\), \(c\), and \(d\) are the coefficients from the balanced equation. (4 sentences)
The partial pressure of a gas is the pressure that the individual gas would exert if it alone occupied the entire volume of the container. Only gaseous species are included in the \(K_p\) expression. The partial pressures of pure solids or pure liquids are treated as constant and are not included in the calculation. \(K_p\) is dependent only on temperature and remains constant regardless of the initial amounts of reactants or products. (4 sentences)
Direct Calculation Using Known Partial Pressures
The most direct way to calculate \(K_p\) is by substituting the equilibrium partial pressures of all gaseous reactants and products into the \(K_p\) expression. This calculation is only possible once the system has reached equilibrium, meaning the partial pressures are no longer changing. The first step involves writing the balanced chemical equation and then formulating the correct \(K_p\) expression based on the stoichiometry. (3 sentences)
Consider the synthesis of ammonia, known as the Haber process, which is represented by the equation \(\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)\). The \(K_p\) expression for this reaction is written as \(K_p = \frac{(P_{\text{NH}_3})^2}{(P_{\text{N}_2}) (P_{\text{H}_2})^3}\). The partial pressure of ammonia is squared, and the partial pressure of hydrogen is cubed, corresponding to their coefficients in the balanced equation. (3 sentences)
Imagine a specific reaction vessel at a certain temperature where the equilibrium partial pressures are measured to be \(P_{\text{N}_2} = 0.094\ \text{atm}\), \(P_{\text{H}_2} = 0.039\ \text{atm}\), and \(P_{\text{NH}_3} = 0.003\ \text{atm}\). These measured values are then substituted directly into the \(K_p\) expression to find the numerical value of the constant. Calculation requires consistent pressure units, such as atmospheres (atm) or pascals (Pa), for all components. (3 sentences)
Plugging in the values yields \(K_p = \frac{(0.003)^2}{(0.094) (0.039)^3}\), resulting in a \(K_p\) value of approximately \(1.62\) for this specific temperature. Although partial pressures are typically measured in units like atmospheres, \(K_p\) is generally reported as a unitless value. (2 sentences)
This method is straightforward when all equilibrium partial pressures are provided. If only initial pressures or total pressure are known, the change in pressure based on the stoichiometry of the reaction must be calculated first to determine the equilibrium partial pressures of all species before the final substitution into the \(K_p\) expression can occur. (2 sentences)
Calculating \(K_p\) from the Concentration Constant \(K_c\)
When the equilibrium constant is known in terms of molar concentrations, \(K_c\), it is possible to convert this value to \(K_p\) using a specific thermodynamic relationship. This conversion is necessary because \(K_c\) is calculated using molarity (\(\text{mol}/\text{L}\)), while \(K_p\) uses pressure units. The relationship connecting the two constants is \(K_p = K_c(RT)^{\Delta n}\). (3 sentences)
In this equation, \(R\) represents the ideal gas constant, which is typically used as \(0.08206\ \frac{\text{L} \cdot \text{atm}}{\text{mol} \cdot \text{K}}\) when pressures are in atmospheres. The variable \(T\) is the absolute temperature of the reaction system, which must always be expressed in Kelvin. The term \(\Delta n\) is the change in the total number of moles of gas during the reaction. (3 sentences)
The value of \(\Delta n\) is calculated by taking the sum of the stoichiometric coefficients of the gaseous products and subtracting the sum of the stoichiometric coefficients of the gaseous reactants. For the reaction \(aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g)\), \(\Delta n\) is calculated as \((c+d) – (a+b)\). This calculation is performed using the balanced equation, and only species in the gaseous state are counted for the mole change. (3 sentences)
If a reaction has an equal number of moles of gaseous products and reactants, then \(\Delta n\) equals zero, which simplifies the conversion formula significantly. Since any term raised to the power of zero equals one, the relationship becomes \(K_p = K_c(RT)^0\), which means \(K_p\) is numerically equal to \(K_c\). Conversely, if \(\Delta n\) is positive, \(K_p\) will be greater than \(K_c\), and if \(\Delta n\) is negative, \(K_p\) will be less than \(K_c\). (3 sentences)