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Calculate the enthalpy of formation of a...

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)`.

A

`-"41.82 kJ mol"^(-1)`

B

`+"83.64 kJ mol"^(-1)`

C

`-"945.36 kJ mol"^(-1)`

D

`-"833 kJ mol"^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
`N_(2)+3H_(2)rarr 2NH_(3)`
`DeltaH=DeltaH(N-=N)+3xxDeltaH(H-H)-2xx3DeltaH(N-H)`
`=945.36+3xx435.0-6xx389.0=-"83.64 kJ"`
`"Heat of formation of "NH_(3)=(-83.64)/(2)=-"41.82 kJ/mol"`
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