5G.1C

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Jasmine W 1K
Posts: 6
Joined: Sat Sep 07, 2019 12:18 am

5G.1C

Postby Jasmine W 1K » Thu Jan 09, 2020 12:06 am

c) If one starts with a higher pressure of reactant, the equilibrium constant will be larger.
I don't really understand why this is false, because I thought that pressure affects the equilibrium constant for gases. Can someone please explain why this is false?

Sanjana K - 2F
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Joined: Sat Sep 07, 2019 12:17 am
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Re: 5G.1C

Postby Sanjana K - 2F » Thu Jan 09, 2020 12:29 am

If you start with a higher reactant pressure, you'll have a lower reactant volume (because increasing pressure tends to compress volume) so the equilibrium will actually shift left to make up for that loss in volume, according to Le Chatelier's Principle. So, with the reaction shifting left, the equilibrium constant will actually be smaller

ng1D
Posts: 30
Joined: Wed Sep 18, 2019 12:17 am

Re: 5G.1C

Postby ng1D » Thu Jan 09, 2020 12:41 am

The equilibrium constant will remain the same because it is not affected by the amount of reactants or products. If there is a higher pressure of reactant, the pressure of the products will change so that it equals the equilibrium constant, which remains constant.

Gabriella Bates 2L
Posts: 87
Joined: Thu Jul 11, 2019 12:15 am

Re: 5G.1C

Postby Gabriella Bates 2L » Thu Jan 09, 2020 1:02 pm

The equilibrium constant is not affected by the initial partial pressures of the reactant due to Le Chatlier's principle. This principle states that when stress is placed on the system (such as increasing the partial pressure of a reactant), then the equilibrium will shift in order to counteract that stress. Therefore, in this case, the equilibrium will favor the reaction with the least moles of gaseous molecules in order to relieve the stress of the high reactant partial pressure, and will proceed in this direction until the equilibrium pressures are reached. The only factor that will affect the value of the equilibrium constant is temperature.


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