1.

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

Answer»

`-1102 kJ mol^(-1)`
`-964 kJ mol^(-1)`
`+352 kJ mol^(-1)`
`+1056 kJ mol^(-1)`

Solution :`(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`


Discussion

No Comment Found

Related InterviewSolutions