Q.

At 57°C, gaseous dinitrogen tetroxide is 50 % dissociated. Calculate the standard free energy change per mole of N2O4 (g) at this temperature and at 1 atm.

(R=8.3JK-1 mol-1 , In 10=2.3,log2=0.3,log 3=0.48)

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a

-756 J mol-1

b

-856 J mol-1

c

-656 J mol-1

d

None of these

answer is A.

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Detailed Solution

Given Temperature T1=570C+2730C=330K

Gaseous dinitrogen tetroxide=50 % dissociated

R=8.3JK1mol1 ln 10=2.3, log2=0.3, log3=0.48

Standard free energy, ΔG0=?

The Reaction is: 

                                N2O4g2NO2(g)

Initial  (t = 1)                     1                    0 

At equilibrium (t = eq)    1-a                2a

Total moles   =1α+2α=1+α

a = 0.5  and P = 1 atm

P  N2O4=1α1+αPT

=10.51+0.5×1

=0.51.5atm

PNO2=2α1+αPT

=2×0.51+0.5×1=11.5atm

KP=PNO22PN2O4

=11.520.51.5

=43

Gibbs energy formula, 

ΔG0=2.303nRTlogKP

=2.303×1×8.3×330×log43

=2.303×8.3×3300.60.48

=756Jmol1

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