NTA Abhyas JEE Main2020ChemistryThermodynamics (C)Practice
When the following reaction was carried out in a bomb calorimeter, Δ U is found to be − 740 .0 kJ/mol of NH 2 CN(s) (s) at 300 K . N H 2 C N ( s ) + 3 2 O 2 g → N 2 g + C O 2 g + H 2 O ( l ) Calculate ΔH 300 K for the reaction.
Options
- A-738.75 kJ
- B+738.75 kJ
- C-824.75 kJ
- D-919.57 kJ
Correct answer
A. -738.75 kJ
Step-by-step solution
ΔH   =   ΔU   +   Δn g RT ΔH = − 740000 + 0.5 × 8.3 × 300 = − 738755   J