Enthalpy for the reaction
${Ag}^{+} (a q) + {Br}^{-} (a q) ⟶ AgBr (s) is - 90 k J$
Magnitude of enthalpy of formation of ${Ag}^{+} (aq)$ and ${Br}^{-} (aq)$ are in the ratio $5 : 6 .$ Formation of ${Ag}^{+}$ $(aq)$ is an endothermic process whereas formation of ${Br}^{-}$ is an exothermic process. Enthalpy of formation of $AgBr$ is $- 110 kJ / mole$ . Calculate the enthalpy of formation of ${Ag}^{+} (aq)$ in kJ/mol

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Detailed Solution

${Ag}^{+} (a q) + {Br}^{-} (a q) ⟶ AgBr (s); Δ H = - 90 kJ$

$\frac{Δ H_{{fAg}^{+} (a q)}^{\circ}}{Δ H_{f {Br}^{-} (aq)}^{\circ}} = \frac{5}{6} = \frac{- 5}{6} . . . . . . . (i)$

$Δ H_{f AgBr}^{\circ} = - 110 kJ {mol}^{- 1}$

$- 90 = - 110 - (Δ H_{f {Ag}^{+}}^{\circ} + Δ H_{f {Br}^{-}}^{\circ}) . . . . . (ii)$

$Solving eqns. (i) and (ii), we get$

$Δ H_{f A g^{+}}^{\circ} = 100 kJ {mol}^{- 1}$

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