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tubakhan4833

31.01.2021

Chemistry

1.5 atm is the pressure of dinitrogen tetroxide after equilibrium is reached the second time.

Explanation:

$2NO_2(g)\rightleftharpoons N_2O_4(g)$

initially

4.9 atm 0

At equilbrium:

(4.9-2p) p

Pressure of nitrogen dioxide gas at equilibrium = 2.7 atm

So, (4.9-2p) = 2.7 atm

p = 1.1 atm

The value of equilibrium constant K_p :

$K_p=\frac{p_{N_2O_4}}{(p_{NO_2})^2}$

$K_p=\frac{p}{(4.9-2p)^2}$

$K_p=\frac{1.1 atm}{(4.9-2\times 1.1)^2}=0.1509$

After addition of 1.2 atm of nitrogen dioxide gas more to the flask,will the reestablish an equilibrium in flask.

$2NO_2(g)\rightleftharpoons N_2O_4(g)$

(2.7+1.2) atm 1.1 atm

At second equilibrium

(2.7+1.2-2x) atm (1.1+x) atm

The expression of K_p :

$K_p=\frac{p_{N_2O_4}}{(p_{NO_2})^2}$

$0.1509=\frac{(1.1+x)}{(2.7+1.2-2x)^2}$

Solving for x:

x = 0.3827 atm

The pressure of dinitrogen tetroxide after equilibrium is reached the second time:

(1.1+x) atm = (1.1+0.3827) atm =1.4827 atm ≈ 1.5 atm

1.5 atm is the pressure of dinitrogen tetroxide after equilibrium is reached the second time.

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