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# Kc HELP!!!

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1. Could anyone help me understand the following question:

2NO2(g) --> N2O4(g) (delta H=-57)

The experiment is repeated but the pressure in the container is doubled.
Explain in terms of Kc the effect on the concentrations of N2O4 when the mixture has reached equilibrium.

Kc does not change
[NO2] increases more than [N2O4]
equlibrium shifts to right to restore Kc

Why does Kc shift equilibrium rather than le Chatelier?
Why does [NO2] increase more than [N2O4]?
If [NO2] increases more than [N2O4], then why does equilibrium shift right?
Has it got anything to do with the fact that [NO2] is squared in Kc equation?

2. It probably doesn't say that Le Chatelier shifts the equilibrium because it's more of a rule of thumb than an actual explanation of why things happen. You could figure out the direction of the shift using it, but using the Kc helps you to figure out the numeral change, while Le Chatelier just points you to the right direction.

Think of it this way; at the beginning, you have 2 moles of NO2 and 1 mole of N2O4. When the pressure is changed, let's say you end up with 1,8 moles of NO2 and 1,1 moles of N2O4. If the change in pressure was brought about with a change in volume, say from 2 liters to 1, at first the concentration of NO2 would be 1 and after the pressure increase it would be 1,8; while with N2O4 it would first be 0,5 and then 1,1.

If you think of it mathematically, since we often use numbers like 0,2 or 0,05 as concentrations, squaring a concentration actually makes the number smaller, so the [NO2] has to increase more than [N204] just to keep Kc constant.

This didn't answer your question fully, but I hope it helps at least a bit!
3. That really does help thank you very much!

But now I'm wondering what would you do if nothing is squared in the Kc equation? Would you then revert back to number of gaseous molecules to decide what increases more (e.g. the side with fewer gaseous molecules must increase more or something like that?)
4. (Original post by alar)
That really does help thank you very much!

But now I'm wondering what would you do if nothing is squared in the Kc equation? Would you then revert back to number of gaseous molecules to decide what increases more (e.g. the side with fewer gaseous molecules must increase more or something like that?)
So you mean like a situation where y (g) + z (g) -----> x (g)? Yeah I'd say it's a useful strategy to start with. If you're working from conditions where [y] = [z], then it's basically the same situation as before and you can treat the equation as if just one term was squared. When the concentrations aren't the same, I guess you'd have to work out the actual increases mathematically, but I'm sure no one will ask you that without giving more info.
5. Okay that makes sense, thanks so much!
6. (Original post by alar)
Could anyone help me understand the following question:

2NO2(g) --> N2O4(g) (delta H=-57)

The experiment is repeated but the pressure in the container is doubled.
Explain in terms of Kc the effect on the concentrations of N2O4 when the mixture has reached equilibrium.

Kc does not change
[NO2] increases more than [N2O4]
equlibrium shifts to right to restore Kc

Why does Kc shift equilibrium rather than le Chatelier?
Why does [NO2] increase more than [N2O4]?
If [NO2] increases more than [N2O4], then why does equilibrium shift right?
Has it got anything to do with the fact that [NO2] is squared in Kc equation?

There is a crucial mistake in your mark scheme - the [NO2] should be squared - the concentrations of each change in exactly the same way if the pressure is changed but the square of [NO2] increases more than [\N2O4} when pressure increasd
7. (Original post by GDN)
There is a crucial mistake in your mark scheme - the [NO2] should be squared - the concentrations of each change in exactly the same way if the pressure is changed but the square of [NO2] increases more than [\N2O4} when pressure increasd
Cheers that makes everything clearer!

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