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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

  1. A-738.75 kJ
  2. B+738.75 kJ
  3. C-824.75 kJ
  4. 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

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