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The standard enthalpy of formation of NH...

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

A

`− 1102kJ mol^(−1)`

B

`− 964 kJ mol^(−1)`

C

`+ 352kJ mol^(−1)`

D

`+ 1056kJ mol^(−1)`

Text Solution

Verified by Experts

The correct Answer is:
C
`1/2 N_2 + 3/2 H_2 to NH_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 = -1056KJ`
enthalpy of formation of `NH_3 = - 1056 KJ`
Average bond enthalpy of `N - H` bond
` = 1056/3 = 352`
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