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

anyone?

It should be heat of dissociation per mole

here Î”H=VÎ”P as V and T remains constant

further
PV=nRT

so
VÎ”P=Î”nRT

N₂O₄ => 2NO₂
c(1-x) 2cx

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

so Î”n=cx=1/5 as c=1(per mole)

so Î”H=1/5*8.314*300

=498.8 J

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