Equality reaction

  • Thread starter Mathman23
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  • #1
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Hi I have simplified my last question in the hope of that someone, can help med answer it.

The following equality reaction:

[itex]2 NO_{2} \leftrightharpoons N_{2} O_{4}[/itex]

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 [itex] [NO_2] [/itex] and
[itex] [N_{2} O_{4}] [/itex]

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)
 

Answers and Replies

  • #2
chem_tr
Science Advisor
Gold Member
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Okay, let me say that the total moles are 3x, and [itex]NO_2[/itex] is 2x while [itex]N_2O_4[/itex] 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.
 
Last edited:
  • #3
254
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Hello,

Thank You for Your answer.

My assumption is accurate then ?

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

Sincerely
Fred


chem_tr said:
Okay, let me say that the total moles are 3x, and [itex]NO_2[/itex] is 2x while [itex]N_2O_4[/itex] 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.
 

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