The approach to the following equilibrium was observed kinetically from both directions: PtCl4^2- + — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the Forward and Backward Rate Constants The given net rate equation: \[-\frac{d[\ce{PtCl4^{2-}}]}{dt} = (3.9 \times 10^{-5}\text{ s}^{-1})[\ce{PtCl4^{2-}}] - (2.1 \times 10^{-3}\text{ L mol}^{-1}\text{s}^{-1})[\ce{Pt(H2O)Cl3^-}][\ce{Cl^-}]\] Comparing with the standard form $-\frac{d[\ce{PtCl4^{2-}}]}{dt} = k_f[\ce{PtCl4^{2-}}] - k_b[\ce{Pt(H2O)Cl3^-}][\ce{Cl^-}]$: * $k_f = 3.9 \times 10^{-5}\text{ s}^{-1}$ (forward aquation, first-order) * $k_b = 2.1 \times 10^{-3}\text{ L mol}^{-1}\text{s}^{-1}$ (backward complexation, second-order) Step 2 - Find the Equilibrium Constant for the Forward Reaction (\(K_h\)) At equilibrium, net rate = 0: \[k_f[\ce{PtCl4^{2-}}] = k_b[\ce{Pt(H2O)Cl3^-}][\ce{Cl^-}]\] The equilibrium constant for the forward aquation reaction: \[K_h = \frac{k_f}{k_b} = \frac{3.9 \times 10^{-5}}{2.1 \times 10^{-3}} \approx 0.0185\text{ mol L}^{-1}\] Step 3 - Calculate \(K_{\text{eq}}\) for Complexation (Reverse Reaction) The complexation reaction (binding of 4th Cl$^-$) is the reverse: \[\ce{Pt(H2O)Cl3^- + Cl^- <=> PtCl4^{2-} + H2O}\] \[K_{\text{eq}} = \frac{1}{K_h} = \frac{k_b}{k_f} = \frac{2.1 \times 10^{-3}}{3.9 \times 10^{-5}} \approx \boxed{53.8\text{ L mol}^{-1}}\] Step 4 - Evaluate Each Option * **(A) 53.8 mol L$^{-1}$**: Incorrect. Units are inverted. Complexation has two reactants → one product, so units are L mol$^{-1}$. * **(B) 0.018 mol L$^{-1}$**: Incorrect. This is $K_h$ (forward aquation), not the complexation constant. * **(C) 53.8 L mol$^{-1}$**: Correct. Reciprocal of $K_h$ with correct units for complexation. * **(D) 0.018 L mol$^{-1}$**: Incorrect. Wrong numerical value (mixing $K_h$ value with complexation units). \[\boxed{\text{C}}\]