Calculate the standard enthalpy of formation ($\Delta_f H^\circ$) of aqueous $\mathrm{Ag}^+$ ions. G

Question

Calculate the standard enthalpy of formation ($\Delta_f H^\circ$) of aqueous $\mathrm{Ag}^+$ ions. Given: $\Delta_f H^\circ(\mathrm{Cl}^-_{(aq)}) = -167 	ext{ kJ/mol}$, enthalpy of precipitation of $\mathrm{AgCl}_{(s)} = -65 	ext{ kJ/mol}$, and $\Delta_f H^\circ(\mathrm{AgCl}_{(s)}) = -127 	ext{ kJ/mol}$.

Professor PC Thomas Classes · Answer

(a) : The precipitation reaction is $\mathrm{Ag}^+_{(aq)} + \mathrm{Cl}^-_{(aq)} ightarrow \mathrm{AgCl}_{(s)}$. The enthalpy of this reaction (crystallization/precipitation) is given by $\Delta H_{ppt} = \Delta_f H^\circ(\mathrm{AgCl}) - [\Delta_f H^\circ(\mathrm{Ag}^+) + \Delta_f H^\circ(\mathrm{Cl}^-)]$. Substituting the given values: $-65 = -127 - [\Delta_f H^\circ(\mathrm{Ag}^+) + (-167)]$. Rearranging the equation: $-65 = -127 - \Delta_f H^\circ(\mathrm{Ag}^+) + 167$. Thus, $\Delta_f H^\circ(\mathrm{Ag}^+) = 167 - 127 + 65 = 105 	ext{ kJ/mol}$.