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Question: \(\Delta H\)for the reaction, \(OF_{2} + H_{2}O \rightarrow O_{2} + 2HF\) (B.E. of \(O - - F,O - - ...

ΔH\Delta Hfor the reaction, OF2+H2O→O2+2HFOF_{2} + H_{2}O \rightarrow O_{2} + 2HF (B.E. of

O−−F,O−−H,H−−FO - - F,O - - H,H - - F and O = O are 44, 111, 135 and

119 kcal mol−1mol^{- 1} respectively)

A

– 222 Kcal

B

– 88 kcal

C

– 111kcal

D

– 79 kcal

Answer

– 79 kcal

Explanation

Solution

: ΔH=∑B.E.R−∑B.E.P\Delta H = \sum B.E._{R} - \sum B.E._{P}

= (2×44+2×111)-(119 + 2×135) = - 79 kcal