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I am having issues understanding the consequence(s) of a decrease in pressure on a chemical reaction.

I have understood that when we increase the pressure, the side having more number of moles wants its moles to go to the other side where it is more empty(this is how I learnt it). But I fail to understand why if we decrease the pressure by increasing volume the side having less moles goes to the side having more moles?

Also in the case of adding inert gases why does the equilibrium shift in the direction in which larger no of moles of gas are formed?

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  • $\begingroup$ Why does it want to increase the pressure and the number of moles? Won't it want a balance? $\endgroup$ – Jai Mahajan Aug 23 '14 at 18:59
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The system doesn't "want" anything. This is a major misconception with Le Chatelier's principle.

Think about the following reversible reaction, everything in the gas phase:

$$\ce{A + B <=> C}$$

We'll simplify the rate law of the forward reaction to

$$\text{rate} = K_\mathrm{fwd}[\ce{A}][\ce{B}]$$

The reverse rate law is

$$\text{rate} = K_\mathrm{rev}[\ce{C}]$$

If the volume is doubled, what happens to the rate of the forward reaction? It is quartered. What happens to the forward reaction? It is halved. So, the reaction "shifts to reactants".

If the volume is halved, what happens to the forward rate? It is quadrupled. The reverse rate is doubled. The reaction "shifts to products".

What happens when an inert gas is added? Since the inert gas appears in neither the forward nor the reverse rate law, nothing happens.

I'm afraid you have learned some misconceptions.

Le Chatelier's principle was developed before much of what is studied in a general chem course today was known. It has been used to manipulate the outcome of reactions. All of the predictions made by Le Chatelier's principle can be made by understanding how reactions work at the particle level. Le Chatelier's Principle is from the macroscopic perspective and leads many students to generate misconceptions.

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  • $\begingroup$ What happens if we add inert gas at constant pressure. I think there should be 3 cases delta n>=or<0. Please explain to me that also. Thanks $\endgroup$ – Jai Mahajan Aug 23 '14 at 20:13
  • $\begingroup$ How can the inert gas be added at constant pressure? $\endgroup$ – Brinn Belyea Aug 23 '14 at 20:23
  • $\begingroup$ Well that's what my books says. $\endgroup$ – Jai Mahajan Aug 23 '14 at 20:26
  • $\begingroup$ If inert gas is added then to maintain pressure constant,volume is increased. Hence equilibrium will shift in the direction in which larger number of moles of gas are formed $\endgroup$ – Jai Mahajan Aug 23 '14 at 20:27
  • $\begingroup$ The inert gas has no effect on the equilibrium, hence the word inert. Ignore the inert gas in any problem. $\endgroup$ – Brinn Belyea Aug 23 '14 at 20:31

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