The standard enthalpy of formation of NH3 is -46.0 kJ `mol^(-1)`. If the enthalpy of formation of `H_2` from its atoms is -436 kJ mol and that of `N_2` is -712 kJ mol, the average bond enthalpy of N - H bond in `NH_3` is

Options

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

Answer

C

Explanation

`(1)/(2)N_(2)+(3)/(2)H_(2)rarr NH_(3) , Delta H=-46 kJ mol^(-1)`
`2H rarr H_(2) , Delta H =-436 kJ mol^(-)`
`2N rarr N_(2), Delta H-712 kJ mol^(-1)`
eq (ii) and (iii) suggests that the bond dissociation energies of `H_(2)` and `N_(2)` into atoms are 436 and 712 kJ `mol^(-1)`
`therefore` For eq(i)
`Delta H= Sigma(B.E.)_("Reactants")-Sigma(B.E.)_("Products")`
`-46=[(1)/(2)xx712+(1)/(2)xx436]-3xx(B.E.)_(N-H)`
`therefore (B.E.)_(N-H)=(1056)/(3)=352 kJ mol^(-1)`