MHT CET202122 Sep 2021Evening ShiftChemistryThermodynamics (C)Actual
What is enthalpy of formation of NH ₃ if bond enthalpies are as ( N N )=941 ~kJ ,( H - H )=436 ~kJ ,( ~N - H )=389 ~kJ ?
Options
- A-84.5 ~kJ
- B-21.25 ~kJ
- C-42.5 ~kJ
- D-63.45 ~kJ
Correct answer
C. -42.5 ~kJ
Step-by-step solution
aligned & 1 2 ~N _ 2( ~g ) + 3 2 H _ 2( ~g ) NH _ 3( ~g ) & H _ reacion = H _ f ( NH ₃ ) = 1 2 BE _ ( NaN ) + 3 2 BE ( H - H )-3 BE _ ( N - H ) & H _ f ( NH ₃ ) = 1 2 941+ 3 2 436-3 389 & =470.5+654-1167 & =-42.5 ~kJ aligned