IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry - Chemical Equilibrium-3

joyfrancis

Calculate vapour density of phosphine at 300K and 2.5atm when it shows 40% dissosiation to attain equilibrium.

4PH3

P₄ + 6H2

taruntanuj007

it seasy calculate moles of each susbstance at equilibrium then calculate the molecular mass of mix at equilibrium and then finally calculate its vapour density by dividing the mass by two

catch_arnnie

is "300K and 2.5atm" the initial condition or equilibriun condition?

KAB

Use the equation

=(D-d)/((n-1)d)

Where ,

is the dissociation constant

D is the vapour density of phosphine before dissociation

d is the vapour density of phosphine after dissociation.

n is the no. of moles of gaseous products obtained by the dissociation of 1 mole of reactant.

Here PH₃

(1/4)P₄+(3/2)H₂

Here D=(31+3)/2=34/2=17 , d=? ,

=0.4 , n=1/4+3/2=1.75

So using

=(D-d)/((n-1)d) we get,

17-d=0.3d

or d=17/1.3

or d=13.07

So vapour density of phosphine at 300K and 2.5atm is 13.07

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