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At $t = 0$ we have 10 moles of $\ce{NO2}$ and 10 moles of $\ce{N2O4}$ satisfying this reaction in a 0.75 liter tank (the barrier is closed at this time):

$$\ce{N2O4 <=> 2NO2}$$

The system is in thermal contact with the environment.

At time t=t0 the barrier is released and the gases are free to expend to the other tank ($V_2=1$ liter)

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The question is what will be the concentration of $\ce{NO2}$ when they reach equilibrium.

What I tried is calculating the chemical equilibrium before based on the concentration quotient $\text{K}_c$ (13.33) and then to use it as reference after the system reaches equilibrium. I got a weird result (negative number) Where have I gone wrong?

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  • $\begingroup$ You Kc value is correct , the answer comes positive ( on solving the quadratic which I hope you got ). So I guess there is some calculation mistake on your part . Can you include your steps ? $\endgroup$
    – Rajesh
    Mar 7, 2015 at 12:45

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This is a pretty straight forward equilibrium problem in disguise. If we assume the initial state represents equilibrium, we can determine the equilibrium constant by plugging in the concentrations to the law of mass action. We can get the concentrations because we have the number of moles and the volume of the container.

$$\ce{N2O4 <=> 2NO2}$$

$$K_c=Q_{eq}=\dfrac{[\ce{NO2}]^2}{[\ce{N2O4}]}$$

Okay, now we let the gas expand. Being in thermal contact with the surroundings is important, since it means the temperature will not change (which means the equilibrium constants does not change). Now we will find that we are not at equilibrium any more. We will probably find that $Q_c < K_c$, so equilibrium needs to shift toward $\ce{NO2}$.

Now, we can set up the ICE (initial, change, equilibrium) table:

          [N2O4]           [NO2]
I         5.71 M           5.71 M
C         -x               +2x
E         5.71 -x          5.71 + 2x

Substitute into the law of mass action at equilibrium:

$$K_c = \dfrac{(5.71+2x)^2}{5.71-x}=13.33$$

Rearrange and solve for $X$. Being a quadratic equation, you should get two roots (two solutions for $x$), and as is always the case for an equilibrium problem, one of them will be nonsense. For example, you could get one value of $x$ to be larger than the initial concentrations!

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  • $\begingroup$ This being a homework question , shouldn't you have waited for OP to add the working/steps/calculation ? I even commented this above . $\endgroup$
    – Rajesh
    Mar 8, 2015 at 17:27
  • $\begingroup$ I think it's fine. The asker showed effort, Ben focused on the problem description and the buildup to reach the answer, and the actual final computations were not performed, so the asker still has work to do on their part. $\endgroup$
    – Nicolau Saker Neto
    Mar 8, 2015 at 17:53
  • $\begingroup$ I guess the build-up was not required just the last paragraph would have been great . Anyway nice answer . $\endgroup$
    – Rajesh
    Mar 8, 2015 at 17:59
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    $\begingroup$ Even though the OP may only need the last paragraph, someone else looking at this question may need a little more. $\endgroup$
    – Ben Norris
    Mar 8, 2015 at 18:35

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