Question

A mixture of nitrogen and hydrogen is at an initial pressure of 200 atm if 20 percent of mixture reacts by time equilibrium is reached the equilibrium pressure of mixture is

Answer 1

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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