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As an example, let's assume that we have a gas $\ce{P(g)}$ inside a chamber and when the temperature inside it increases to a certain value there is an equilibrium that prevails inside the system.

So for the sake of argument, let us assume that this equilibrium is:

$$\ce{P(g) <=> Q(g) + R (g)},$$

And let's assume the figure below shows that chamber when equilibrium is established:

enter image description here

What I want to know is, what are the possible changes that might occur if I add an inert gas $\ce{I(g)}$ to this chamber?

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Quoting from the Wikipedia page on Le Chatelier's principle, we have to distinguish two cases:

  1. The total volume is constant:

    Adding an inert gas into a gas-phase equilibrium does not result in a shift of the equilibrium in this case. This is due to the fact that adding an inert gas does not change the partial pressures of the other gases in the container. While it is true that the total pressure of the system increases, the total pressure does not have any effect on the equilibrium constant; rather, it is a change in partial pressures that will cause a shift in the equilibrium.

  2. The total volume is not constant:

    If, however, the volume is increased, the partial pressures of all gases will decrease which results in a shift of the equilibrium towards the side with the greater number of moles of gas.

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  • $\begingroup$ Could you explain why the partial pressure will not change upon the addition of an inert gas in when the volume is constant $\endgroup$ Jun 11, 2015 at 7:38
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    $\begingroup$ The only additional pressure in the container will come from the inert gas. The pressures exerted by the reactants/products have no way to get larger or smaller (To calculate the partial pressures, use P = nRT/V, using the molar amount of the individual gas. You'll see that because your volume is constant, as is your temperature, the partial pressure must remain equal). $\endgroup$ Jun 11, 2015 at 12:37
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    $\begingroup$ The partial pressure of a gas is the hypothetical pressure of that gas if it alone occupied the volume of the mixture at the same temperature. So, if the volume is constant in your problem, the partial pressure $p_i=n_iRT/V$ is constant. $\endgroup$ Jun 11, 2015 at 12:47
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    $\begingroup$ I'm pretty sure this answer is strictly correct only for ideal gases. With strongly interacting molecules, at sufficiently high pressures, or at sufficiently low temperatures, I would think adding moles to the system would start to directly affect the equilibrium constant. $\endgroup$
    – hBy2Py
    Jun 11, 2015 at 13:37
  • $\begingroup$ Of course! you're right. The answer is valid for ideal gases. $\endgroup$ Jun 11, 2015 at 13:48

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