Given, `H_(2)(g)+Br_(2)(g)rarr2HBr(g),DeltaH_(1)^(0)` and standard enthalpy of condensaion of bromine is `DeltaH_(2)^(0)`, standard enthalpy of formation of `HBr` at `25^(@)C` is

Options

A. `DeltaH_(1)^(0)//2`
B. `DeltaH_(1)^(0)//2+DeltaH_(2)^(0)`
C. `DeltaH_(1)^(0)//2-DeltaH_(2)^(0)`
D. `(DeltaH_(1)^(0)-DeltaH_(2)^(0))//2`

Answer

D

Explanation

`H_(2)(g)+Br_(2)(g)rarr2HBr(g) " "Delta H=DeltaH_(1)^(o)`......(i)
`Br_(2)(g)rarrBr_(2)(l)" " Delta H=Delta H_(2)^(o)`.....(ii)
`[eq_(1)-eq_(2)]`
`H_(2)(g)+Br_(2)(l)rarr 2HBr(g)" " Delta H=Delta H_(1)^(o)-Delta H_(2)^(o)`
Required equation , `(1)/(2)H_(2)(g)+(1)/(2)Br_(2)(l) rarr HBr(g) " " Delta H=[(Delta H_(1)^(o)-Delta H_(2)^(o))/(2)]`