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The standard enthalpy of formation of NH...

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

A

`-964 KJ mol^(-1)`

B

`+352 KJ mol^(-1)`

C

`+1056 KJ mol^(-1)`

D

`-1102 KJ mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
B
`({:(N_(2)(g) + (3)/(2) H_(2)(g)rarr NH_(3),:,DeltaH_(f)^(@)=-46.0 "KJ mol"^(-1)),(2H(g)rarrH_(2)(g),,,Delta_(f)^(@)=-436 "KJ mol"^(-1)),(2N(g)rarrN_(2)(g),,,DeltaH_(f)^(@)=-712"KJ mol^(-1)),(NH_(3)(g) rarr (1)/(2)N_(2)(g)+(3)/(2)H_(2)(g),,,DeltaH =+46),((3)/(2)H_(2)rarr3H,,,DeltaH=+436xx(3)/(2)),((1)/(2)N_(2)rarrN,,,DeltaH=+712xx(1)/(2)):})/`
`(NH_(3)(g) rarr N(g)+ 3H(g) " "," "DeltaH =+ 1056 "KJ mol^(-1)`
Average bond enthaply of N-H bond `=(1056)/(3) =+352 "KJ mol^(-1)`
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