Use the data given to calculate the value of Κ for the reaction at 25°C

AgCl(s) Ag+(aq) + Cl− (aq)

AgCl(s) Ag+(aq) Cl− (aq)
S° (J K−1 mol−1) +96.2 +72.68 +56.4
ΔH°f (kJ/mol) −127.07 +105.58 −167.2

Question

Use the data given to calculate the value of Κ for the reaction at 25°C

AgCl(s)  Ag+(aq) + Cl− (aq)

AgCl(s)	Ag+(aq)	Cl− (aq)
S° (J K−1 mol−1)	+96.2	+72.68	+56.4
ΔH°f (kJ/mol)	−127.07	+105.58	−167.2

BlackTom AI · Accepted Answer

To tackle the problem, begin by identifying the reaction of interest: AgCl(s) ⇄ Ag+(aq) + Cl−(aq). We need the equilibrium constant K at 25 °C, which is linked to the standard Gibbs energy change via K = exp(-ΔG°/RT).

First, compute the standard enthalpy change for the reaction from the given standard enthalpies of formation (ΔHf°): ΔHrxn° = [ΔHf°(Ag+(aq)) + ΔHf°(Cl−(aq))] − ΔHf°(AgCl(s)). Substituting the values: ΔHrxn° = [(+105.58) + (−167.2)] − (−127.07) kJ/mol = (−61.62) + 127.07 = +65.45 kJ/mol.

Next, determine the standard entropy change ΔSrxn° from the standard molar entropies (S°): ΔSrxn° = [S°(Ag+(aq)) + S°(Cl−(aq))] − S°(AgCl(s)). That is ΔSrxn° = (72.68 + 56.4) − 96.2 J/K/mol = 129.08 − 96.2 = +32.88 J/K/mol, which equals 0.03288 kJ/K/mol.

Now assemble ΔG° at 25 °C (298 K) using ΔG° = ΔHrxn° − TΔSrxn°: ΔG° = 65.45 kJ/mol − (298 K)(0.03288 kJ/K/mol) ≈ 65.45 − 9.79 ≈ 55.66 kJ/mol.

Proceed to compute RT at 298 K with R = 8.314 J/mol·K = 0.008314 kJ/mol·K: RT = 0.008314 × 298 ≈ 2.475 kJ/mol.

Finally, evaluate K from K = exp(-ΔG°/RT) = exp(-55.66 / 2.475) ≈ exp(-22.5). A negative exponent of about 22.5 yields a very small number, roughly on the order of 10^−10. Numerically, exp(-22.5) ≈ 1.7 × 10^−10.

Thus, the calculated value for K at 25 °C is a small equilibrium constant on the order of 10^−10, consistent with the dissolution of AgCl to Ag+ and Cl− in water.

Use the data given to calculate the value of Κ for the reaction at 25°C AgCl(s) Ag+(aq) + Cl− (aq) AgCl(s) Ag+(aq) Cl− (aq) S° (J K−1 mol−1) +96.2 +72.68 +56.4 ΔH°f (kJ/mol) −127.07 +105.58 −167.2单项选择题

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