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The likelihood is that I'm misunderstanding what's going on here.

Consider the reaction $\ce{A <=> B}$, where $K_\mathrm{c}=1$.

  1. Initially, the system at equilibrium, where $[\ce A]=\pu{1M}$ and $[\ce B]=\pu{1M}$.

  2. The system is then disturbed to increase $[\ce A]$ by $\pu{1M}$. Instantaneously, $[\ce A]=\pu{2M}$ and $[\ce{B}]=\pu{1M}$. Le Chatelier's principle suggests that the equilibrium position moves to the right to minimise the increase in $[\ce A]$.

  3. The system then returns to equilibrium, such that $[\ce A]=\pu{1.5M}$ and $[\ce{B}]=\pu{1.5M}$.

Clearly, the relative amounts of $[\ce A]$ and $[\ce B]$ have not changed, so the equilibrium position has not changed, which is in direct contradiction with Le Chatelier's principle.

I think my misunderstanding may lie in one of the following:

  • My definition of equilibrium position (the relative amounts of reactants and products at equilibrium).
  • Between which numbers the equilibrium position is considered to move, let me elaborate on this: the relative amount of $[\ce A]$ and $[\ce B]$ have not changed between 1 and 3 ($1/1=1.5/1.5$), but they have changed between 2 and 3 ($1/2\ne 1.5/1.5$), so I see how the equilibrium position could be said to have moved right from step 2 to step 3 as the amount of B increases and the amount of A decreases.
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    $\begingroup$ In order for you to get maximum returns on any answer, I think you should tell us the definition of Le Châtelier's principle that you are using. $\endgroup$ – Zhe Oct 26 '17 at 15:54
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    $\begingroup$ "If a change is made to a system in equilibrium, the position of the equilibrium will move to minimise that change" @Zhe $\endgroup$ – Rational Function Oct 26 '17 at 15:56
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    $\begingroup$ That description is qualitative and not quantitative. But certainly, you increased the concentration of $\ce{A}$ by $1\ \mathrm{M}$, but the final increase was significantly less than that. $\endgroup$ – Zhe Oct 26 '17 at 16:53
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    $\begingroup$ You say "the relative amount of [A] and [B] have not changed between 1 and 2 (1/1=1.5/1.5)". Do you mean between 1 and 3? $\endgroup$ – owjburnham Oct 26 '17 at 19:24
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    $\begingroup$ You misunderstand the principle, It doesn't say K will change, it says if A changes B will change so and so in order to keep K constant (applied to the given situation). $\endgroup$ – Greg Oct 27 '17 at 0:17
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Le Chatelier's Principle does not directly say what happens to concentration ratios. Nor does it directly compare conditions after a disturbance with those before. It says what happens only after a disturbance has been applied.

Here, the disturbance is adding one mole of A per liter. When equilibrium is re-established you have only an additional one-half mole of A per liter, the rest having reacted to form B. The initial addition of A has been partially offset, which is exactly what Le Chatelier's Principle says.

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    $\begingroup$ So Le Chatelier's Principle applies between the instant a change is made to a system in equilibrium until the moment it reaches equilibrium once again, not from an initial state of equilibrium to reaching equilibrium once again? @Oscar Lanzi $\endgroup$ – Rational Function Oct 26 '17 at 15:59
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    $\begingroup$ Exactly. Hope this helps. $\endgroup$ – Oscar Lanzi Oct 26 '17 at 16:13
  • $\begingroup$ And in terms of the equilibrium position in my example: the equilibrium position moves to the right between the instant [A] is increased and the moment equilibrium is reached with [A]=[B]=1.5M. Between the initial and final states of equilibrium, the equilibrium position has not moved, since the chemicals are present in the same relative amounts? In general, equilibrium position does not move between states of equilibrium, it moves between states of instantaneous disturbance and return to equilibrium? (Sorry if this is unclear) @Oscar Lanzi $\endgroup$ – Rational Function Oct 26 '17 at 16:29
  • $\begingroup$ Basically that is how a reaction works. You have to start out of equilibrium, or disturb the system out of equilibrium, then the reaction will move back to equilibrium. It cannot move directly from one equilibrium state to another without something to set it off. $\endgroup$ – Oscar Lanzi Oct 26 '17 at 19:46
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You are simply slightly misapplying Le Chatelier. The principle is used to predict the reaction of a system to changes in the external conditions, i.e. temperature or pressure, not to you wilfully changing the internal composition of the system.

The general idea (system minimises the change) is of course right, but at constant conditions, the law of mass action allows a direct quantitative predicition, so why bother with Le Chatelier?

Anyway when the system reaches equillibrium after your addition of compound A, the relative change in the system is already zero again, so, following Le Chatelier, the change in the equillibrium constant would also be nil. No contradiction at all.

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