What are the concentrations of NO2 and N2O4 in an equilibrium reaction?

[Mathman23]

Hi I have simplified my last question in the hope of that someone, can help med answer it.

The following equality reaction:

$2 NO_{2} \leftrightharpoons N_{2} O_{4}$

Where the total number of mole's of the two substances in equilibrium is
2,6 x 10^2 mol.

The reaction takes place in a container with a volume of 0,50 Liters.

This means that the total concentration of the two substances are
0,052 mol/liter.

I need to calculate the concentration of both $[NO_2]$ and
$[N_{2} O_{4}]$

2 NO_2 <----> N_2 O_4
------------------------------
int | 0,052 + x | 0,052 + x
change | x | x
equi | 0,052 + x - (2x) | 0,052 + x - (2x)

 

[chem_tr]

Okay, let me say that the total moles are 3x, and $NO_2$ is 2x while $N_2O_4$ is x, which is equal to 8,67.10^-3 moles.

There you can do the calculation to find the corresponding concentrations of the gases in 0.50 liters of a container.

 

[Mathman23]

Hello,

Thank You for Your answer.

My assumption is accurate then ?

$K_{c} = \frac{[N_2 O_4]}{[NO_2]^2} = \frac{(0.052 + x - (2x))}{(0.052 +x - (2x))^2} = ?$

Sincerely
Fred

Okay, let me say that the total moles are 3x, and $NO_2$ is 2x while $N_2O_4$ is x, which is equal to 8,67.10^-3 moles.

There you can do the calculation to find the corresponding concentrations of the gases in 0.50 liters of a container.