What is the mathematical derivation of the Law of Mass Action and the equilibrium constant (Kc) for a general reversible chemical reaction?

Here is a mathematically rigorous, step-by-step breakdown of the quantitative and theoretical core topics for Class 10 Chemistry (BISE Faisalabad syllabus). --- ### **1. Mathematical Derivation of the Law of Mass Action & Equilibrium Constant ($K_c$)** Consider a general reversible chemical reaction where reactants $A$ and $B$ react to form products $C$ and $D$, with stoichiometric coefficients $a, b, c,$ and $d$: $aA + bB \underset{k_r}{\overset{k_f}{\rightleftharpoons}} cC + dD$ #### **Step 1: Application of the Law of Mass Action** According to Guldberg and Waage's Law of Mass Action, the rate of a chemical reaction is directly proportional to the product of the active masses (molar concentrations) of the reacting substances, raised to powers equal to their respective stoichiometric coefficients. * **Rate of Forward Reaction ($R_f$):** $R_f \propto [A]^a [B]^b \implies R_f = k_f [A]^a [B]^b$ *(where $k_f$ is the rate constant for the forward reaction).* * **Rate of Reverse Reaction ($R_r$):** $R_r \propto [C]^c [D]^d \implies R_r = k_r [C]^c [D]^d$ *(where $k_r$ is the rate constant for the reverse reaction).* #### **Step 2: Dynamic Equilibrium Condition** At dynamic equilibrium, the rate of the forward reaction equals the rate of the reverse reaction ($R_f = R_r$): $k_f [A]^a [B]^b = k_r [C]^c [D]^d$ #### **Step 3: Algebraic Rearrangement for $K_c$** Rearranging the constants to one side and active concentrations to the other: $\frac{k_f}{k_r} = \frac{[C]^c [D]^d}{[A]^a [B]^b}$ Since the ratio of two constants ($\frac{k_f}{k_r}$) is itself a constant, defined as the **Equilibrium Constant ($K_c$)**: $\mathbf{K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}}$ --- ### **2. General Unit Derivation Rules for $K_c$** The dimensional unit of $K_c$ depends on the change in the number of moles ($\Delta n = n_{\text{products}} - n_{\text{reactants}}$): $\text{Units of } K_c = \left(\text{mol}\cdot\text{dm}^{-3}\right)^{\Delta n}$ #### **Case A: Equal moles ($\Delta n = 0$)** $\text{Example: } \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$ $K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} \implies \frac{(\text{mol}\cdot\text{dm}^{-3})^2}{(\text{mol}\cdot\text{dm}^{-3})(\text{mol}\cdot\text{dm}^{-3})} = \text{No Units}$ #### **Case B: Unequal moles ($\Delta n \neq 0$)** $\text{Example: } \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$ $\Delta n = 2 - (1 + 3) = -2$ $K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} \implies \frac{(\text{mol}\cdot\text{dm}^{-3})^2}{(\text{mol}\cdot\text{dm}^{-3})(\text{mol}\cdot\text{dm}^{-3})^3} = \frac{1}{(\text{mol}\cdot\text{dm}^{-3})^2} = \mathbf{\text{mol}^{-2}\cdot\text{dm}^6}$ --- ### **3. Mathematical Framework of the $\text{pH}$ Scale and Auto-ionization of Water** Pure water undergoes auto-ionization at $25^\circ\text{C}$: $\text{H}_2\text{O}(l) \rightleftharpoons \text{H}^+(aq) + \text{OH}^-(aq)$ The equilibrium expression is: $K_c = \frac{[\text{H}^+][\text{OH}^-]}{[\text{H}_2\text{O}]}$ Since $[\text{H}_2\text{O}]$ is virtually constant ($55.5 \text{ M}$), we define the **Ion-Product Constant of Water ($K_w$)**: $K_w = K_c [\text{H}_2\text{O}] = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14} \text{ M}^2 \quad (\text{at } 25^\circ\text{C})$ #### **Derivation of $\text{pH} + \text{pOH} = 14$** Taking the negative logarithm ($-\log_{10}$) on both sides of the $K_w$ expression: $-\log_{10}(K_w) = -\log_{10}\left([\text{H}^+][\text{OH}^-]\right)$ Applying the logarithmic product rule $\log(a \cdot b) = \log a + \log b$: $-\log_{10}(K_w) = \left(-\log_{10}[\text{H}^+]\right) + \left(-\log_{10}[\text{OH}^-]\right)$ By definition, $\text{p}X = -\log_{10}[X]$: $\text{p}K_w = \text{pH} + \text{pOH}$ Since $K_w = 10^{-14}$, $\text{p}K_w = -\log_{10}(10^{-14}) = 14$: $\mathbf{\text{pH} + \text{pOH} = 14}$ --- ### **4. Stoichiometric Reactions in Solvay's Process (Metallurgy)** The industrial preparation of sodium carbonate ($\text{Na}_2\text{CO}_3$) involves strict chemical balance: 1. **Ammoniation of Brine:** $\text{NH}_3 + \text{H}_2\text{O} + \text{CO}_2 \rightarrow \text{NH}_4\text{HCO}_3$ 2. **Precipitation of Sodium Bicarbonate:** $\text{NH}_4\text{HCO}_3 + \text{NaCl} \rightarrow \text{NaHCO}_3\downarrow + \text{NH}_4\text{Cl}$ 3. **Calcination (Thermal Decomposition):** $2\text{NaHCO}_3 \xrightarrow{\Delta} \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2\uparrow$ 4. **Recovery of Ammonia ($\text{NH}_3$):** $\text{CaO} + \text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2$ $2\text{NH}_4\text{Cl} + \text{Ca(OH)}_2 \rightarrow 2\text{NH}_3\uparrow + \text{CaCl}_2 + 2\text{H}_2\text{O}$