At 90 degree celsius the vapour density of N2O4 is 24.8(H=1).calculate the degree of dissociation at this temperature.
Answer (1)
Student Expert
12th Aug, 2019
Dear Pawan,
N2O4 ---------NO2.
Initially N2O4 has conc.= 1, then at equilibrium N2O4 has conc.1-X.
And NO2 has conc. =2x
initially we have taken N2O4 =1
Molar mass =92 .V.D at eq.=24.8 M.w.=60.
n initial/ n final =M final/M initial
1/1-X+2X=60/92
1/1+X=60/92
1+X=92/60
X=92/60 -1
X=32/60
X=0.53
percentage dissociation=0.53 X 100=53
All the best
Hope this helps
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