`NH_(3)(g)+3Cl_(2)toNCl_(3)(g)+3HCl(g),DeltaH_(1)`
`N_(2)(g)+3H_(2)(g)to2NH_(3)(g),DeltaH_(2)`
`H_(2)(g)+Cl_(2)(g)to 2HCl(g),DeltaH_(3)`
The heat of formation of `NCl_(3)`(g) in the terms of
`DeltaH_(1),DeltaH_(2),DeltaH_(3)` is :

Options

A. `DeltaH_(f)=-DeltaH_(1)+(DeltaH_(2))/(2)-(3)/(2)DeltaH_(3)`
B. `DeltaH_(f)=DeltaH_(1)+(DeltaH_(2))/(2)-(3)/(2)DeltaH_(3)`
C. `DeltaH_(f)=DeltaH_(1)+(DeltaH_(2))/(2)-(3)/(2)DeltaH_(3)`
D. None of the above

Answer

D

Explanation

To find `1//2N_(2)+3//2Cl_(2)toNCl_(3)`, `DeltaH_(f)=?`
`(1)/(2)N_(2)+(3)/(2)H_(2)toNH_(3)`, `(-DeltaH_(2))/(2)`……..`(i)`
`NH_(3)+3Cl_(2)toNCl_(3)+3HCl`, `-DeltaH_(1)`………`(ii)`
`3HCl to (3)/(2)Cl_(2)+(3)/(2)H_(2)`, `(-3DeltaH_(3))/(2)`……….......`(iii)`
Adding `(i)`, `(ii)` and `(iii)`
`1//2N_(2)+3//2Cl_(2)toNCl_(3)`
`DeltaH=(-DeltaH_(2))/(2)-DeltaH_(1)-(3DeltaH_(2))/(2)`