Question

A mixture of Nitrogen and Hydrogen (1 : 3 mole ratio) is at an initial pressure of 200 atm. If 20% of the mixture reacts by the time equilibrium is reached, the equilibrium pressure of the mixture is

Answer 1

Answer:

Thank you for asking this question. Here is your answer:

The reaction for this would be:

N2 +3H2 = 2NH3

Weight ratio is 28:6 which is 14:6

75 gm N2 present (14/17)×75

and H2 present (3/17)×75 gm

So N2 is present

(14/17)×(75/28)=2.35 mole

=nN2 and H2 present (3/17)×(75/2)

= 7.05 mole =nH2

Kp = { (nNH3)^2 / (nN2)×(nH2)^3} × {P/£n}^-2

= {(1.17)^2 / (2.35) × (7.05)^3} × {200/10.57}^-2

=4.64×10^-6 (atm^-2)

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