The bond energies of `H--H` , `Br--Br` and `H--Br` are `433,, 192` and `364KJmol^(-1)` respectively. The `DeltaH^(@)` for the reaction
`H_(2)(g)+Br_(2)(g)rarr2HBr(g)` is

Options

A. `- 261 kJ`
B. `+ 103 kJ`
C. `+ 261 kJ`
D. `- 103 kJ`

Answer

D

Explanation

If the reaction is carried out in the gas phase only then the reaction enthalpy `(Delta_(r) H^(@))` can be calculated by the following relationship:
`Delta_(r) H^(@) = sum "Bond enthalpies"_("reactants") - sum "Bond enthalpies"_("product")`
` = [Delta_(H - H) H^(@) + Delta_(Br - BR) H^(@)] - (2 Delta_(H - Br)H^(@))`
`(433 + 192) - (2) (364)`
`= (625) - (728) = - 103 kJ mol^(-1)`