The standard enthalpy of formation of `NH_(3)` is `-46.0KJ mol^(-1)` . If the enthalpy of formation of `H_(2)` from its atoms is `-436KJ mol^(-1)` and that of `N_(2)` is `-712KJ mol^(-1)` , the average bond enthalpy of `N-H` bond in `NH_(3)` is

Options

A. `+1056kJ" "mol^(-1)`
B. `-1102kJ" "mol^(-1)`
C. `-964kJ" "mol^(-1)`
D. `+352kJ" "mol^(-1)`

Answer

D

Explanation

`BE(N-=N)=-712kJ" "mol^(-1)`
`BE(H-H)=-436kJ" "mol^(-1)`
Let B.E. of `(N-H)=x" kJ "mol^(-1)`
`DeltaH=sum(B.E.)_("Reactants")-sum(B.E.)_("products")`
`-46=[(1)/(2)(712)+(3)/(2)(436)]-3x`
`x=352kJ" "mol^(-1)`