For the equilibrium: N204(g) = 2N02(g) K = 0.36 at 100°C
a sample of 0.25 mol N2O4 is allowed to dissociate and come to equilibrium
in a 1.5 L flask at 100°C. What are the equilibrium concentrations of NO2
and N204?​

NO

2

]=0.12M

Explanation:

Hello,

In this case, for the undergoing chemical reaction at equilibrium, the law of mass action for the equilibrium statement turns out:

Kc=\frac{[NO_2]^2}{[N_2O_4]}Kc=

[N

2

O

4

]

[NO

2

]

2

Next, by introducing the reaction extent xx (ICE methodology) we can write:

Kc=\frac{(2x)^2}{[N_2O_4]_0-x}Kc=

[N

2

O

4

]

0

−x

(2x)

2

Whereas the equilibrium constant is 0.36 and the initial concentration of dinitrogen tetroxide is 0.100 M:

0.36=\frac{(2x)^2}{0.100M-x}0.36=

0.100M−x

(2x)

2

Therefore, solving such quadratic equation we obtain two values of xx :

\begin{gathered}x_1=-0.15M\\x_2=0.06M\end{gathered}

x

1

=−0.15M

x

2

=0.06M

Clearly, the solution is 0.06M since the reaction extent must not be negative, thereby, the equilibrium concentration nitrogen dioxide turns out:

[NO_2]=2*0.06M[NO

2

]=2∗0.06M

[NO_2]=0.12M[NO

2

]=0.12M

Best regards.

explanation:

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i believe the answer is three water molecules