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 125.L tank with 35.mol of nitrogen dioxide gas. When the mixture has come to equilibrium she determines that it contains 8.0mol of nitrogen dioxide gas. The engineer then adds another 12.mol of nitrogen dioxide, and allows the mixture to come to equilibrium again. Calculate the moles of dinitrogen tetroxide after equilibrium is reached the second time.

Answer 1

Answer:

The moles of dinitrogen tetroxide after equilibrium is reached the second time is 18.78 moles.

Explanation:

$2NO_2\rightleftharpoons N_2O_4$

Initially 35.0 mol

Eq'm     8.0 mol                        13.5 mol

Moles of nitrogen dioxide = 35.0 mol

Moles of nitrogen dioxide at equilibrium = 8.0 mol

According to reaction, 2 moles nitrogen oxide gives 1 mol of dinitrogen tetraoxide. Then 27 mol of nitrogen dioxde will give:

$\frac{1}{2}\times 27 mol=13.5 mol$

Moles of dinitrogen tetroxide = 13.5 mol

Concentration of nitrogen dioxide at equilibrium,$[NO_2] =\frac{8.0 mol}{125.0 L}$

Concentration of dinitrogen tetradioxide at equilibrium,$[N_2O_4] =\frac{13.5 mol}{125.0 L}$

Equilibrium constant of reaction:

$K_c=\frac{[N_2O_4]}{[NO_2]^2}$

$K_c=\frac{\frac{13.5 mol}{125.0 L}}{(\frac{8.0 mol}{125.0 L})^2}=26.37$

Now, adds another 12.0 moles of nitrogen dioxide.

$2NO_2\rightleftharpoons N_2O_4$

Initially 35.0 mol

Eq'm     8.0 mol                        13.5 mol

After adding 12 moles of nitrogen dioxide the moles

    (8.0 mol+12 mol)

Again after attaining equilibrium second time:

     (20.0 -x)mol                       (13.5+x/2) mol

Concentration of nitrogen dioxide at second equilibrium,$[NO_2] =\frac{(20.0 -x)mol}{125.0 L}$

Concentration of dinitrogen tetradioxide at second equilibrium,$[N_2O_4] =\frac{(13.5 +x/2)mol}{125.0 L}$

$K_c=\frac{\frac{(13.5+x/2)}{125 L}}{(\frac{(20.0 mol -x)}{125 L})^2}$

$26.37=\frac{(13.5+x/2)\times 125}{(20.0-x)^2}$

on solving for x:

we get x = 10.56

Moles of dinitrogen tetroxide after equilibrium is reached the second time:

$(13.5 + \frac{x}{2})mol = 13.5 + (\frac{10.56}{2}) mol= 18.78 mol$