A piston contains CHCl3 N2O4 and NO2 Two phases are presentS

A piston contains CHCl3, N2O4, and NO2. Two phases are present

Solution

ANSWER

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Let\'s assume the reaction is taking place in a 1.00 L volume.

because of that, we know there is 0.140 mol of NO2. 57% of it goes away:

0.140 times 0.57 = 0.0798 mol (this amount goes away, I\'ll come back to it later.)

How much NO2 remains:

0.140 mol minus 0.0798 mol = 0.0602 mol

So, the [NO2] at equilibrium is 0.0602 M

Now, 2 NO2 are used up to make 1 N2O4, so do this:

0.0798 mol divided by 2 = 0.0399 mol (that\'s the amount of N2O4 made)

and it concentration is 0.0399 M

Kc = 0.0399 / 0.0602^2 = 11.0

N2O4(g)<--> 2 NO2(g)

1.30 moles of N2O4 are placed in a 5.60 L container at 100