Winter 2016 Final Question 2A


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Karla_Coronado_1J
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Winter 2016 Final Question 2A

Postby Karla_Coronado_1J » Thu Mar 16, 2017 10:27 pm

In the solution to this question there is the following part:

K = PPCL5/PPCl3PCl2 = 2x/x2

How do we get the 2x/x2? I know this is from Chem 14A, but a quick refresher would be very helpful.

Thanks in advance!

Chem_Mod
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Re: Winter 2016 Final Question 2A

Postby Chem_Mod » Thu Mar 16, 2017 10:51 pm

I will cover this in detail in my 4pm review tomorrow in Moore 100.

Vincent Tse 2B
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Re: Winter 2016 Final Question 2A

Postby Vincent Tse 2B » Fri Mar 17, 2017 12:09 am

You might want to review setting up an ICE chart: since we're dealing with pressure, one would consider the initial partial pressures, the change in partial pressures, and the resulting partial pressures at equilibrium.

In this problem, however, we are only given information at equilibrium, so we can disregard the I-C part of ICE. The "E" part is more important because we are dealing with equilibrium anyway (note: you should have calculated K using deltaG0 = -RTln(K) before the next steps ).

We're told the partial pressures at equilibrium in terms such that each of the reactants' partial pressure is equal to half of that of the product's; no numerical values are given, so assign variable x.

If the reactants are half of the product's partial pressure, let's call the partial pressure of one reactant x and make the partial pressure of the product 2x; this makes the relationship true.

Now, we set up our equilibrium ratio of K = PRODUCTS/REACTANTS in terms of partial pressure. This will then be: K = 2x/(x)*(x) which is also K = 2x/x^2. (there is x*x because there are two reactants, which each having partial pressure x)

Knowing what K is, you can solve for x and the resulting value will be the partial pressure for the REACTANTS (since both are equal PCl3 = Cl2 = x). Multiply x by 2 to get the partial pressure of PCl5.

Hopefully this helps!


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