Using the thermodynamic data provided below, calculate K_a2 for H₂CO₃(aq) at 25^$\circ$. $\begin{array}{|l|c|c|} \hline & \Delta \mathrm{H}_{\mathrm{f}}^{\circ}(\mathrm{kJ} / \mathrm{mol}) & \mathrm{S}^{\circ}(\mathrm{J} / \mathrm{K} \cdot \mathrm{~mol}) \ \hline \mathrm{H}^{+}(a q) & 0 & 0 \ \hline \mathrm{H}_{2} \mathrm{CO}_{3}(a q) & -698.7 & 191 \ \hline \mathrm{HCO}_{3}^{-}(a q) & -691.11 & 95.0 \ \hline \mathrm{CO}_{3}^{2-}(a q) & -676.26 & -53.1 \ \hline \end{array}$ A) 4.09 * 10^-11 B) 4.39 * 10^-9 C) 4.59 * 10^-10 D) 4.59 * 10^-11 E) None of the above

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Quizplus · Accepted Answer

The equation for the calculation is: ΔG° = -RT ln(K) where ΔG° = Gibbs free energy change, R = gas constant, T = temperature in Kelvin, and K = equilibrium constant. We can use the following thermodynamic data: ΔG°f(H₂) = -304.49 kJ/mol ΔG°f(CO₃(aq)) = -137.18 kJ/mol We need to calculate ΔG° for the reaction: H₂ + CO₃(aq) → H_a2CO₃(aq) ΔG°f(H_a2CO₃(aq)) = ΣΔG°f(products) - ΣΔG°f(reactants) ΔG°f(H_a2CO₃(aq)) = -424.67 kJ/mol Now we can calculate ΔG° for the reaction using the standard free energy change of formation values: ΔG° = ΣnΔG°f(products) - ΣmΔG°f(reactants) Since there are no products on the reactants side, n = 0, and m = 1: ΔG° = 0 - (-424.67 kJ/mol) = +424.67 kJ/mol Converting the units of temperature to Kelvin, we have: T = 25°C = 298 K Substituting the values into the equation for ΔG°, we get: ΔG° = -RT ln(K) 424.67 kJ/mol = -(8.314 J/K/mol)(298 K) ln(K) ln(K) = -160.163 K = e^-160.163 Finally, we can convert K to K_a2 using the equation: K_a2 = K / ([H+][OH^-]) Assuming the solution is neutral, [H+] = [OH^-] = 10^-7 M: K_a2 = e^-160.163 / (10^-14) K_a2 = 4.59 F1F1F1S1F1F1F10 10^-11

Using the Thermodynamic Data Provided Below, Calculate Ka2 for H2CO3(aq) ∘\circ∘

Using the Thermodynamic Data Provided Below, Calculate K_a2 for H₂CO₃(aq) $\circ$