TestHub
TestHub

Chemistry - Thermodynamics - II Question with Solution | TestHub

ChemistryThermodynamics - IIFree Energy and EquilibriumMedium2 minPYQ_2022
ChemistryMediummatching list

2O3g3O2g

At 300 K, ozone is fifty percent dissociated. The standard free energy change at this temperature and 1 atm pressure is -......J mol-1. (Nearest integer)

[Given: ln 1.35=0.3 and R=8.3 J K-1 mol-1]

Answer:
747
Solution:

2O3g3O2g

Initially 1 mole 0

1-0.5 32×0.5

0.5 mole 1.52=0.75 mole

Kp=PO23PO32=0.751.2530.51.252=353252

=0.630.42=0.2160.16=1.35

ΔG°=-RTlnKp

=-8.3×300ln1.35

=-8.3×300×0.3

=-747J/mole

Stream:JEESubject:ChemistryTopic:Thermodynamics - IISubtopic:Free Energy and Equilibrium
2mℹ️ Source: PYQ_2022

Doubts & Discussion

Loading discussions...