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