7th edition 5G.5

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Saachi_Kotia_4E
Posts: 68
Joined: Fri Sep 28, 2018 12:23 am

7th edition 5G.5

Postby Saachi_Kotia_4E » Tue Jan 08, 2019 10:37 pm

I am working through the homework for this section and I dont understand how to do 5G.5, especially part c. Can someone explain the logic behind solving this problem?

Nicolette_Canlian_2L
Posts: 77
Joined: Fri Sep 28, 2018 12:25 am
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Re: 7th edition 5G.5

Postby Nicolette_Canlian_2L » Wed Jan 09, 2019 1:00 am

I second this!! Also, how did you figure out part a and b? How do we determine equilibrium by looking at the molecules?

Helen Mejia 1I
Posts: 67
Joined: Fri Sep 28, 2018 12:24 am

Re: 7th edition 5G.5

Postby Helen Mejia 1I » Wed Jan 09, 2019 9:14 am

Nicolette_Canlian_2L wrote:I second this!! Also, how did you figure out part a and b? How do we determine equilibrium by looking at the molecules?


For part a, you will see that the diatomic molecules are decreasing. It goes from 11 to 8 to 5. It has reached equilibrium in flask #3 because in flask #4 it has the same amount of diatomic molecules which is 5. No change occurred after flask #3.
For part b, 6 X2 molecules out of 11 X2 molecules have decomposed at equilibrium. So to find the percentage it will be (6/11) x 100= 54.5%

taywebb
Posts: 60
Joined: Fri Sep 28, 2018 12:15 am

Re: 7th edition 5G.5

Postby taywebb » Wed Jan 09, 2019 10:10 am

Here's what I could gather from looking at the solutions manual. We first have to recognize that the decomposition of X2 has an equilibrium equation of X2 <-> 2X, making the partial pressure ratio 1:2. Since the initial pressure is 0.10 bar, we must multiply it by the fraction of molecules that have not already decomposed (1-0.545) to get the partial pressure for X2 (the reactant/denominator). We then have to find the partial pressure for the products/numerator, which is the initial pressure multiplied by the fraction of X2 molecules that have decomposed into X (0.545) and then that multiplied by 2 because the decomposition of X2 results in 2 moles of X. To then find K we just use our normal equation K = (Px)^2/(Px2) and plugin (0.109)^2 (partial pressure of product) /(0.0455) (partial pressure of reactant) to get 0.26 as the final answer.


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