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:

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

Answer 2

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

What is dinitrogen tetroxide?

$\rm N_2O_3$ is a chemical compound used in the synthesis of chemicals.

The reaction is:

$\rm 2NO_2 <--> N_2O_4$

Now, the initial moles are

35. mol  of nitrogen dioxide gas

At equilibrium, = 8.0mol of nitrogen dioxide gas.

2 moles nitrogen oxide gives 1 mol  of dinitrogen tetra oxide.

Then 27 mol of nitrogen dioxide will give:

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

Mass of dinitrogen tetroxide = 13.5 mol

Concentration of nitrogen dioxide at equilibrium = $\dfrac{8.0 \;mol}{125\;L}$

Concentration of dinitrogen tetra dioxide at equilibrium = $\dfrac{13.5\;mol}{125\;mol}$

Equilibrium constant of reaction

$Kc = \dfrac{N_2O_4}{[NO_2]^2}\\\\\\Kc=\dfrac{\dfrac{13.5\;mol}{125\;mol}}{[\dfrac{8.0 \;mol}{125\;L}]^2} = 26.37$

Add 12.0 mol to the equilibrium value 8.0 mol

   (8.0 mo l+ 12 mol)

Initially               35.0 mol

At equilibrium    8.0 mol                        13.5 mol

Attaining equilibrium second time

$(20.0 -x)mol \;\;\;\;\; (13.5+\dfrac{x}{2}) mol$

Concentration of nitrogen dioxide at second equilibrium = $\rm NO_2 = \dfrac{(20.0-x)mol}{125.o\;L}$

Concentration of dinitrogen tetra dioxide at second equilibrium =

$\rm N_2O_4 = \dfrac{ (13.5+\dfrac{x}{2}) mol}{125.o\;L}$

Now the value of Kc is

$26.37 =\dfrac{\rm \dfrac{ (13.5+\dfrac{x}{2}) mol}{125.o\;L}\\\\}{\rm \dfrac{(20.0-x)mol}{125.o\;L}]^2} = 10.56$

we get x = 10.56

Moles of dinitrogen tetroxide after equilibrium is reached the second time

$(13.5+\dfrac{x}{2}) mol = (13.5+\dfrac{10.05}{2}) mol= 18.78\;mol$

Thus, the value of dinitrogen tetroxide after equilibrium reached the second time is 18.78 mol.

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