The enthalpy of hydration of AgBr is -819 kJ.mol^-1 at 25 $\circ$ C.Given that the hydration enthalpy of Br^- is -309 kJ.mol^-1,calculate the enthalpy of hydration of Ag⁺ ions.

Question

Quizplus · Accepted Answer

The enthalpy of hydration of AgBr is the sum of the enthalpies of hydration of Ag⁺ and Br⁻ ions. Therefore, the enthalpy of hydration of Ag⁺ is calculated as -819 kJ/mol - (-309 kJ/mol) = -510 kJ/mol.

The Enthalpy of Hydration of AgBr Is -819 KJ ∘\circ∘ CGiven That the Hydration Enthalpy of Br- Is -309 KJ

The Enthalpy of Hydration of AgBr Is -819 KJ $\circ$ CGiven That the Hydration Enthalpy of Br^- Is -309 KJ