# 8.3 The Law of Mass Action The **Law of Mass Action** was proposed by chemists C.M. Guldberg and P. Waage in 1864 to describe the state of chemical equilibrium. It states that: > *"the rate at which a substance reacts is proportional to its active mass, and the rate of a chemical reaction is proportional to the product of the active masses of the reacting substances."* * **Active Mass**: For dilute solutions, this is the molar concentration of the reactants and products, expressed in moles per decimeter cubed ($mol\,dm^{-3}$). * The rate of a reaction is proportional to the product of the molar concentrations of each reactant, with each concentration raised to a power equal to its stoichiometric coefficient in the balanced chemical equation. --- ## Deriving the Equilibrium Constant ($K_c$) Consider a general, reversible gaseous reaction: $$ aA_{(g)} + bB_{(g)} \rightleftharpoons cC_{(g)} + dD_{(g)} $$ Where: * $A, B, C, D$ are the chemical species. * $a, b, c, d$ are their stoichiometric coefficients in the balanced equation. According to the Law of Mass Action: 1. **Rate of the Forward Reaction ($R_f$)**: The rate at which reactants A and B form products C and D. $$ R_f \propto [A]^a [B]^b $$ $$ R_f = k_f [A]^a [B]^b $$ where $k_f$ is the rate constant for the forward reaction. 2. **Rate of the Reverse Reaction ($R_r$)**: The rate at which products C and D revert to reactants A and B. $$ R_r \propto [C]^c [D]^d $$ $$ R_r = k_r [C]^c [D]^d $$ where $k_r$ is the rate constant for the reverse reaction. 3. **At Equilibrium**: The rate of the forward reaction equals the rate of the reverse reaction. $$ R_f = R_r $$ $$ k_f [A]^a [B]^b = k_r [C]^c [D]^d $$ 4. **The Equilibrium Constant ($K_c$)**: Rearranging the equation gives the ratio of the rate constants, which is a new constant called the **equilibrium constant, $K_c$**. $$ \frac{k_f}{k_r} = \frac{[C]^c [D]^d}{[A]^a [B]^b} $$ $$ K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b} $$ This is known as the **equilibrium constant expression**. The square brackets $[...]$ denote the molar concentration of the species at equilibrium. ## Conditions for Equilibrium (Features of $K_c$) * $K_c$ applies *only* at the equilibrium state. The subscript 'c' indicates that concentrations are in $mol\,dm^{-3}$. * $K_c$ is dependent on **temperature** but independent of the initial concentrations of reactants or products. * The expression for $K_c$ is derived directly from the balanced chemical equation: * Products are in the **numerator**. * Reactants are in the **denominator**. * Each concentration is raised to the power of its stoichiometric coefficient. * The magnitude of $K_c$ indicates the **position of equilibrium**: * If **$K_c > 1$**: The concentration of products is greater than reactants at equilibrium. The equilibrium lies to the right (favors products). * If **$K_c < 1$**: The concentration of reactants is greater than products at equilibrium. The equilibrium lies to the left (favors reactants). ## Equilibrium Expressions in Different Units For reactions involving gases, equilibrium can be expressed in terms of partial pressures ($K_p$), number of moles ($K_n$), or mole fractions ($K_x$). ### 1. Partial Pressure ($K_p$) Based on Henry's Law, at a constant temperature, the partial pressure of a gas is directly proportional to its molar concentration. For the general reaction $aA_{(g)} + bB_{(g)} \rightleftharpoons cC_{(g)} + dD_{(g)}$: $$ K_p = \frac{P_C^c \times P_D^d}{P_A^a \times P_B^b} $$ Where $P_A, P_B, P_C, P_D$ are the partial pressures of the gases at equilibrium. The relationship between $K_p$ and $K_c$ is: $$ K_p = K_c(RT)^{\Delta n} $$ * $\Delta n = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) = (c+d) - (a+b)$ * $R$ is the ideal gas constant ($0.0821 \, dm^3\,atm\,K^{-1}\,mol^{-1}$). * $T$ is the absolute temperature in Kelvin (K). ### 2. Number of Moles ($K_n$) $$ K_n = \frac{n_C^c \times n_D^d}{n_A^a \times n_B^b} $$ Where $n_A, n_B, n_C, n_D$ are the number of moles of each species at equilibrium. The relationship between $K_p$ and $K_n$ is: $$ K_p = K_n \left( \frac{P}{n} \right)^{\Delta n} $$ * $P$ is the total pressure of the mixture at equilibrium. * $n$ is the total number of moles of all species at equilibrium. ### 3. Mole Fraction ($K_x$) $$ K_x = \frac{x_C^c \cdot x_D^d}{x_A^a \cdot x_B^b} $$ Where $x_A, x_B, x_C, x_D$ are the mole fractions of each species. The relationship between $K_p$ and $K_x$ is: $$ K_p = K_x (P)^{\Delta n} $$ --- ## Worked Examples ### Writing Equilibrium Constant Expressions **Problem Solving Strategy:** 1. Write the products in the numerator and reactants in the denominator within square brackets. 2. Raise each concentration to the power of its coefficient from the balanced chemical equation. **Example 8.1: Haber Process** $$ N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} $$ **Expression:** $$ K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} $$ **Example 8.2: Carbon Monoxide Oxidation** $$ CO_{(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons CO_{2(g)} $$ **Expression:** $$ K_c = \frac{[CO_2]}{[CO][O_2]^{\frac{1}{2}}} $$ ### Calculating $K_c$ from Equilibrium Concentrations **Example 8.3** The following equilibrium concentrations were observed for the reaction at $500^{\circ}C$. Calculate $K_c$. $$ A_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{(g)} $$ 1. **Given Values:** * $[A_2] = 0.399 \, M$ * $[B_2] = 1.197 \, M$ * $[AB] = 0.203 \, M$ 2. **Apply the Formula:** $$ K_c = \frac{[AB]^2}{[A_2][B_2]} $$ 3. **Calculation:** $$ K_c = \frac{(0.203)^2}{(0.399)(1.197)} $$ $$ K_c = \frac{0.041209}{0.4776} $$ $$ K_c = 0.086 $$ ### Calculating $K_p$ and $K_c$ from Partial Pressures **Example 8.4** The following reaction was studied at $25^{\circ}C$. Calculate its $K_p$ and $K_c$. $$ 2NO_{(g)} + Cl_{2(g)} \rightleftharpoons 2NOCl_{(g)} $$ Equilibrium partial pressures were found to be: * $P_{NOCl} = 1.2 \, atm$ * $P_{NO} = 5.0 \times 10^{-2} \, atm$ * $P_{Cl_2} = 3.0 \times 10^{-1} \, atm$ **Solution:** 1. **Write the $K_p$ expression:** $$ K_p = \frac{(P_{NOCl})^2}{(P_{NO})^2 (P_{Cl_2})} $$ 2. **Substitute partial pressures to find $K_p$:** $$ K_p = \frac{(1.2)^2}{(5.0 \times 10^{-2})^2 (3.0 \times 10^{-1})} $$ $$ K_p = \frac{1.44}{(2.5 \times 10^{-3})(0.3)} = \frac{1.44}{7.5 \times 10^{-4}} $$ $$ K_p = 1.9 \times 10^3 $$ 3. **Calculate $\Delta n$:** $$ \Delta n = (\text{moles of products}) - (\text{moles of reactants}) = 2 - (2+1) = -1 $$ 4. **Write the expression relating $K_p$ and $K_c$:** $$ K_p = K_c (RT)^{\Delta n} $$ 5. **Substitute known values and solve for $K_c$:** * $K_p = 1.9 \times 10^3$ * $R = 0.0821 \, dm^3\,atm\,K^{-1}\,mol^{-1}$ * $T = 25^{\circ}C + 273 = 298 \, K$ * $\Delta n = -1$ $$ 1.9 \times 10^3 = K_c (0.0821 \times 298)^{-1} $$ $$ 1.9 \times 10^3 = \frac{K_c}{24.46} $$ $$ K_c = (1.9 \times 10^3) \times 24.46 $$ $$ K_c = 4.65 \times 10^4 $$ --- ## Possible Questions/Answers **Q1:** Write the $K_c$ expressions for the following reactions after balancing them: * (i) $SO_{2(g)} + O_{2(g)} \rightleftharpoons SO_{3(g)}$ * (ii) $NH_{3(g)} + O_{2(g)} \rightleftharpoons NO_{(g)} + H_2O_{(g)}$ * (iii) $CH_{4(g)} + H_2O_{(g)} \rightleftharpoons CO_{(g)} + H_{2(g)}$ **A1:** * (i) Balanced: $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$ $$ K_c = \frac{[SO_3]^2}{[SO_2]^2 [O_2]} $$ * (ii) Balanced: $4NH_{3(g)} + 5O_{2(g)} \rightleftharpoons 4NO_{(g)} + 6H_2O_{(g)}$ $$ K_c = \frac{[NO]^4 [H_2O]^6}{[NH_3]^4 [O_2]^5} $$ * (iii) Balanced: $CH_{4(g)} + H_2O_{(g)} \rightleftharpoons CO_{(g)} + 3H_{2(g)}$ $$ K_c = \frac{[CO][H_2]^3}{[CH_4][H_2O]} $$ **Q2:** Give the balanced equations that correspond to the following equilibrium expressions: * (i) $K_c = \frac{[CH_3OH]}{[CO][H_2]^2}$ * (ii) $K_c = \frac{[N_2][H_2O]^2}{[NO]^2 [H_2]^2}$ **A2:** * (i) $CO_{(g)} + 2H_{2(g)} \rightleftharpoons CH_3OH_{(g)}$ * (ii) $2NO_{(g)} + 2H_{2(g)} \rightleftharpoons N_{2(g)} + 2H_2O_{(g)}$