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

Options

A. `DeltaH_(1)^(@)//2`
B. `DeltaH_(1)^(@)//2+DeltaH^(@)2`
C. `DeltaH_(1)^(@)-DeltaH_(2)^(@)//2` s
D. `(DeltaH_(1)^(@) - DeltaH_(2)^(@))//2`

Answer

D

Explanation

`H_(2)(g)+Br_(2)(g)rarr2HBr(g),DeltaH=DeltaH_(1)^(@)`
`Br_(2)(g)rarrBr_(2)(l),DeltaH=DeltaH_(2)^(@)`
Subtracting equation (ii) from equation (i), we get
`H_(2)(g)+Br(l)rarr2HBr(g),DeltaH=Delta_(1)^(@)-DeltaH_(2)^(@)`
Required equation, `(1)/(2)H_(2)(g)+(1)/(2)Br_(2)(l)rarrHBr(g)` , `Delta=[(DeltaH_(1)^(@)=DeltaH_(2)^(@))/(2)]`