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Question

For the equilibrium N2O42NO2
(GoN2O4)298K=100kJmol1 and (GoNO2)298K=50kJmol1.
When 5 mole/litre of each is taken, calculate the value of G in kJ mol1 for the reaction at 298 K.(nearest integer)

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Solution

N2O42NO2;G0N2O4=100kJ mol1conc.at t=055G0NO2=50kJ mol1

G0 for reaction =(2×G0NO2)G0N2O4=(2×50)100=0
G=G0+2.303RTlogQ
Now, G=0+2.303×8.314×103×298 log525
=+3.988 kJ mol14 kJ mol1

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