Question

Enthalpy for the reaction \[\ce{Ag^+ (aq) + Br^- (aq) -> AgBr(s)}\] is −84.54 kJ. Magnitude of enthalpy of formation of Ag⁺ (aq) and Br⁻ (aq) are in the ratio 8 : 9. Formation of Ag⁺ (aq) is an endothermic process whereas formation of Br⁻ (aq) is an exothermic process. Enthalpy of formation of AgBr is −99.54 kJ/mol. The enthalpy of formation of Ag⁺ (aq) is ______ kJ/mol.

Options

• 50

• 100

• 120

• 280

Answer 1

Enthalpy for the reaction \[\ce{Ag^+ (aq) + Br^- (aq) -> AgBr(s)}\] is −84.54 kJ. Magnitude of enthalpy of formation of Ag⁺ (aq) and Br⁻ (aq) are in the ratio 8 : 9. Formation of Ag⁺ (aq) is an endothermic process whereas formation of Br⁻ (aq) is an exothermic process. Enthalpy of formation of AgBr is −99.54 kJ/mol. The enthalpy of formation of Ag⁺ (aq) is 120 kJ/mol.

Explanation:

\[\ce{Ag^+ (aq) + Br^- (aq) -> AgBr(s)}\];

ΔH = −84.54 kJ .....(i)

\[\ce{Ag(s) + aq -> Ag^+ (aq)}\];

ΔH = 8x kJ .....(ii)

\[\ce{1/2 Br2(l) + aq -> Br^- (aq)}\];

ΔH = −9x kJ .....(iii)

\[\ce{Ag(s) + 1/2 Br2 (l) -> AgBr(s)}\];

ΔH = −99.54 kJ ....(iv)

From (i) + (ii) + (iii) – (iv),

(–84.54) + 8x + (–9x) – (–99.54) = 0

∴ x = 15

⇒ `Δ_"f""H"_(Ag^+"(aq)")`= 8x = 120 kJ/mol