Question

Nitrogen dioxide is one of the many oxides of nitrogen (often collectively called "NOx") that are of interest to atmospheric chemistry. It can react with itself to form another form of NOx, dinitrogen tetroxide. A chemical engineer studying this reaction fills a 2.0 L flask at 25C with 3.0 atm of nitrogen dioxide gas. He then raises the temperature considerably, and when the mixture has come to equilibrium determines that it contains 2.1 atm of nitrogen dioxide gas. The engineer then adds another 1.5 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.

Answer 1

Answer:

Partial pressure of dinitrogen tetroxide after equilibrium is reached the second time is 0.82 atm.

Explanation:

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

Initially

3.0 atm                 0

At equilibrium

(3.0-2p)                 p

Equilibrium partial pressure of $NO_2=2.1atm=3.0-2p$

p = 0.45 atm

The value of equilibrium constant wil be given by :

$K_p=\frac{p_{N_2O_4}}{(p_{NO_2})^2}=\frac{p}{(3.0-2p)^2}$

$K_p=\frac{0.45}{(2.1)^2}=0.10$

After addition of 1.5 atm of nitrogen dioxide gas equilibrium reestablishes it self :

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

After adding 1.5 atm of $NO_2$:

(2.1+1.5) atm                0.45 atm

At second equilibrium:'

(3.6-2P)                     (0.45+P)

The expression of equilibrium can be written as:

$K_p=\frac{p'_{N_2O_4}}{(p'_{NO_2})^2}$

$0.10=\frac{(0.45+P)}{(3.6-2P)^2}$

Solving for P:

P = 0.37 atm

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

= (0.45+P) atm = (0.45 + 0.37 )atm = 0.82 atm

Partial pressure of dinitrogen tetroxide after equilibrium is reached the second time is 0.82 atm.