heat of dissociation

a given sample of N2O4 in a closed vessel shows 20% dissociation in NO2 at 300K and 1atm,find its heat of dissociation

2 Answers

1
Vivek ·

anyone?

1
satan92 ·

It should be heat of dissociation per mole

here ΔH=VΔP as V and T remains constant

further
PV=nRT

so
VΔP=ΔnRT

N2O4 => 2NO2
c(1-x) 2cx

initial n=c
final n=c(1-x) +2cx=c(1+x) where x=1/5

so Δn=cx=1/5 as c=1(per mole)

so ΔH=1/5*8.314*300

=498.8 J

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