Question

The standard enthalpy of formation of `NH_(3)` is `-46.0kJmol^(1-)`. If the enthalpy of formation of `H_(2)` from its atoms is `- 436 kJ mol^(-1)` and that of N2 is `- 712 kJ mol^(-1)`, the average bond enthalpy of N - H bond in `NH_(3)` is
A. `-1102 kJ mol^(-1)`
B. `-964 kJ mol^(-1)`
C. `+352 kJ mol^(-1)`
D. `+1056 kJ mol^(-1)`

Answer 1

Correct Answer - C
`(1)/(2)N_(2)+(3)/(2)H_(2)rarrNH_(3)`
`DeltaH=H_(f)(NH_(3))-(1)/(2)H_(f)(N_(2))-(3)/(2)H_(f)(H_(2))`
`-46=H_(f)(NH_(3))-(1)/(2)(-712)-(3)/(2)xx(-436)`
`H_(f)(NH_(3))=-46-356-654=-1056 KJ` enthalpy of formation of `NH_(3)`=-1056 KJ
Average bond enthalpy of N - H bond
`= (1056)/(2)=352`