Answer to Question #204800 in Chemistry for Ankush

Question #204800
  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)




1
Expert's answer
2021-06-09T12:13:48-0400

(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




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