Example in Monday's Lecture


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Michelle Lee 2E
Posts: 64
Joined: Thu Jul 27, 2017 3:01 am

Example in Monday's Lecture

Postby Michelle Lee 2E » Tue Nov 21, 2017 10:19 am

Solids and Liquids are not included in the Equilibrium Constant equation so in the class example Ca(OH)2 (s) ---> Ca2+ (aq) + 2OH- (aq), if we looked at the reverse reaction, would it just be the inverse? In chemical equilibrium, does this mean that this chemical equation will not change in the slightest?

Caitlin Mispagel 1D
Posts: 47
Joined: Tue Oct 10, 2017 7:13 am

Re: Example in Monday's Lecture

Postby Caitlin Mispagel 1D » Tue Nov 21, 2017 10:56 am

The equilibrium constant would be 1/([Ca2+][OH-]^2). The concentrations will remain the same at equilibrium whether the reaction is a forward reaction or reverse reaction.

mayasinha1B
Posts: 70
Joined: Fri Sep 29, 2017 7:04 am

Re: Example in Monday's Lecture

Postby mayasinha1B » Tue Nov 21, 2017 12:01 pm

If you calculate K (forwards) for the forward reaction, the backwards reaction has a K (backwards) value that is the inverse of the K (forwards). This is because you calculate using a ratio of the concentration of reactants to the power of stoichiometric coefficients and products to the power of their stoichiometric coefficients. These values do not change when switching a reaction. The only thing that changes is the reactants become products and products become reactants, which, if you look at the K equation, works out to be the inverse.

Salman Azfar 1K
Posts: 50
Joined: Thu Jul 13, 2017 3:00 am

Re: Example in Monday's Lecture

Postby Salman Azfar 1K » Tue Nov 21, 2017 4:55 pm

With regards to what you asked about the equation changing, it literally will not change at all. You pretty much just switch the sides the reactants and products are on and that's it, and as mentioned above, the new equilibrium constant is just the inverse of the old one.


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