InterviewSolution
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InterviewSolution
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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)`. |
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Answer» `N-= N +3 (H-H) rarr underset(H)underset(|)overset(H)overset(|)(2N)-H,DeltaH^(Theta) =`? `DeltaH^(Theta) = [DeltaH^(Theta)underset((N-=N))+3xxDeltaH_((H-H))^(Theta)]-[6xxDeltaH_((N-H))^(Theta)]` `= 945.36 +3 xx 435.0 - 6 xx 389.0 =- 83.64 kJ` Heat of formation of `NH_(3) = (DeltaH^(Theta))/(2) =- (83.64)/(2)` `=- 41.82kJ mol^(-1)` |
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