IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry

spandana

at 27^o c 4.6 grams of N₂O₄ undergoes decomposition and exerts an equilibrium pressure of 1.6 atm when vapourised in a 1 lit flask.the fraction of N₂O_{4 dissociated is} [N₂O_4^---->2NO2]reversible reaction.

taruntanuj007

N₂O₄

2 NO₂

0.05 -

0.05-x 2x

so 0.05 -x are the moles of N₂O₄left at equilibirum

in a 1L vessel now its equilibrium pressure is 1.6 atm

PV = nRT

1.6 * 1 = (0.05-x) 0.0821 * 300

0.05 - x = 0.065

here 's the problem i think sumdata is wrong since x is not negative x = - 0.015

khushi

how can the answer be negative

akku

The answer is not negative .

I think in the last step

the moles corresponding to eqb pressure=.05-x+2x=.05+x=.065

which implies that x=.015

fraction of N2O4 DISSOCIATED=.015/.05=.3 ANSWER

smriti.mathur

Moles of N ₂O₄ = 4.6 / 82

PV = nRT

1.6 * 1 = n * 0.082 * 300

n = 16 / 82*3

N₂ O₄

2NO₂

(1-x) 2x

If a is the initial conc.

a(1-x) 2ax

Total moles = 4.6/82 * (1+ x) = 16 / 82*3

1 + x = 1.16

x = 0.16

Therefore, % dissociation = 0.16 * 100 = 16%

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