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Question: For the given reaction H<sub>2</sub>(g) + F<sub>2</sub>(g) → 2HF(g) ∆H° = – 124 kcal H<sub>2</sub...

For the given reaction

H2(g) + F2(g) → 2HF(g) ∆H° = – 124 kcal

H2(g) → 2H(g) ∆H° = 104 kcal

F2(g) → 2F(g) ∆H° = 37.8 kcal

then the value of ∆H0 for H(g) + F(g) → HF(g) is

A

142 kcal

B

– 132.9 kcal

C

132 kcal

D

134 kcal

Answer

– 132.9 kcal

Explanation

Solution

Now for the reaction,

H2(g) + F2(g) → 2HF(g);

∆H0 = – 124 kcal

∆H0 = ΣB.E. (reactants) – ΣB.E of (products)

or, – 124 = ∆HH–H + ∆HF–F –2∆HH–F = 104 + 37.8 – 2∆HH–F

∴ 2∆H0H–F = 104 + 37.8 + 124 = 265.8 kcal

Bond energy of H–F = 265.82\frac{265.8}{2} = 132.9 kcal

∴∆H0 for the given reaction = – 132.9 kcal