Answer to Question #204800 in Chemistry for Ankush

Question

1. From the following enthalpy changes, calculate the value of ∆H° for the reactions:

Xe (g) + F2  (g) 🡪 XeF2 (s) ∆H° = -123  kJ

Xe (g) + 2F2  (g) 🡪 XeF4 (s) ∆H° = -262 kJ

XeF2 (s) + F2 (g) 🡪 XeF4 (s)

Answer 1

(i) Xe (g) + F₂ (g) 🡪 XeF₂ (s) ∆H°₁ = -123 kJ

(ii) Xe (g) + 2F₂ (g) 🡪 XeF₄ (s) ∆H°₂ = -262 kJ

Reverse (i)

(iii) XeF₂ (s) 🡪 Xe (g) + F₂ (g) ∆H°₃ = +123 kJ

Add (iii) and (ii)

(iv) XeF₂ (s) + F₂ (g) 🡪 XeF₄ (s) ∆H°_net = ∆H°₂ + ∆H°₃ = -262 kJ + 123 kJ = -139kJ

Hence, the ∆H°_net for the reaction will be: -139kJ