Question

Calculate the enthalpy of formation of ammonia from the following bond enegry data:
`(N-H) bond = 389 kJ mol^(-1), (H-H) bond = 435 kJ mol^(-1)`, and `(N-=N)bond = 945.36kJ mol^(-1)`.

Answer 1

`N-=N+3(H-H)to2underset(H)underset(|)overset(H)overset(|)(N)-H," "DeltaH=?`
`DeltaH=[DeltaH_((N-=N))+3xxDeltaH_((H-H))]-[6DeltaH_((N-H))]`
`=945.36+3xx435.0-6xx389.0=-83.64kJ`
Heat of formation of `NH_(3)=(DeltaH)/(2)=-(83.64)/(2)=41.82` kJ `mol^(-1)`