Control of a reaction with Le Chatelier's principle before reaching equilibrium?

[grburst]

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Le Chatelier's principle tells you how a system will react once it is perturbed from equilibrium. What can you say about perturbations of systems that are not yet at equilibrium?

Let's say you have a reaction 2A --> 3B + 4C that is not yet at equilibrium. So, 2 moles of A make 3 moles of B and 4 moles of C.

If I increase the pressure of the vessel while the reaction is proceeding, does this decrease the rate of the forward reaction/increase the rate of the reverse? One expects that in response to increased pressure, 3B and 4C will recombine to make 2A. Similarly, if I decrease pressure, does that speed up the forward reaction?

 

[AureatEagle]

Le Chatelier's principle tells you how a system will react once it is perturbed from equilibrium. What can you say about perturbations of systems that are not yet at equilibrium?

Let's say you have a reaction 2A --> 3B + 4C that is not yet at equilibrium. So, 2 moles of A make 3 moles of B and 4 moles of C.

If I increase the pressure of the vessel while the reaction is proceeding, does this decrease the rate of the forward reaction/increase the rate of the reverse? One expects that in response to increased pressure, 3B and 4C will recombine to make 2A. Similarly, if I decrease pressure, does that speed up the forward reaction?

No, you cannot accurately predict any shifts for a system that is not yet at equilibrium. For a really simple example, consider if you had only A present in your reaction flask and no B or C formed yet at all. Even if you increase the pressure slightly, the reaction will still shift to the right to form some B and C since none of it is there at all (assuming K isnt infinitely tiny, of course).

 

[grburst]

No, you cannot accurately predict any shifts for a system that is not yet at equilibrium. For a really simple example, consider if you had only A present in your reaction flask and no B or C formed yet at all. Even if you increase the pressure slightly, the reaction will still shift to the right to form some B and C since none of it is there at all (assuming K isnt infinitely tiny, of course).

That's a really nice counterexample. Thanks!