Question

Nitrogen dioxide is one of the many oxides of nitrogen (often form another form of NOx, dinitrogen tetroxide A chemical engineer studying this reaction fils a 500. mL flask at 7.9 °C with 4.9 atm of nitrogen dioxide gas. He thèn raises the temperature considerably and when the mixture has come to equilibrium determines that it contains 2.7 atm of nitrogen dioxide gas The engineer then adds another 1.2 atm of nitrogen dioxide, and allows the mixture to come to equilibrium again. Calculate the pressure of dinitrogen tetroxide after equilibrium is reached the second time. Round your answer to 2 significant digits collectively called·N ' that are of interest to atmospheric chemistry. It can eact with ter to 囲 atm □-10 I Don't Know Submit

Answer 1

Answer:

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.