99 Percentile Qs Bank for JEE MainChemistryThermodynamics (C)
At 60 ° C , dinitrogen tetroxide is fifty percent dissociated. Find its standard free energy change at this temperature and one atmosphere. [ Given: log 1 . 33 = 0 . 1239 ]
Options
- A  -650 J . mol   - 1
- B- 830   J . mol - 1
- C- 790   J . mol - 1
- D- 875   J · mol - 1
Correct answer
C. - 790   J . mol - 1
Step-by-step solution
The reaction showing the dissociation will be: N 2 O 4 ( s ) ⇌ 2 NO 2 ( g ) t = 0 1   mol 0   mol t = t eq 0 . 5   mol 1   mol Clearly, from this equation we can say that the total number of moles is 0 . 5 + 1 = 1 . 5   mol Hence, the partial pressure for the molecule will be P N 2 O 4   =   0 . 5 1 . 5 × 1 atm   =   1 3   atm P NO 2   =   1 1 . 5 × 1 atm   = 2 3   atm Now, according to the law of chemical equilibrium we know that