5l.29

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Jina Kwon
Posts: 100
Joined: Fri Aug 09, 2019 12:15 am

5l.29

Postby Jina Kwon » Wed Jan 15, 2020 7:59 pm

I keep getting the answer wrong to this question: At 25 8C, K 5 3.2 3 10234 for the reaction 2 HCl(g) ∆
H2(g) 1 Cl2(g). If a reaction vessel of volume 1.0 L is filled with
HCl at 0.22 bar, what are the equilibrium partial pressures of HCl,
H2, and Cl2?

claudia_1h
Posts: 111
Joined: Fri Aug 09, 2019 12:16 am

Re: 5l.29

Postby claudia_1h » Wed Jan 15, 2020 8:10 pm

Hi, not sure where you might be going wrong but I can try to explain how I did it. I set up an ICE chart where HCl changes by -2x and the other two gases change by +x. Then, because K is very small (it is less than 10^-3), I set it equal to x^2/(0.22)^2. We can do this because because K is so small, the initial value is going to change very little relative to itself. Then, solving for x, you get 3.9 x 10^-18, and can go back to your ICE chart to see how to plug it in and get the equilibrium values. Hope that helps!

Katie Kyan 2K
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Joined: Fri Aug 09, 2019 12:16 am
Been upvoted: 1 time

Re: 5l.29

Postby Katie Kyan 2K » Wed Jan 15, 2020 8:15 pm

In this question you would put 0.22 bar into the ICE box for initial pressure of HCl. To find the equilibrium partial pressures you need to use the equilibrium constant for the decomposition of HCl which is given as 3.2x10^-34. After completing the ICE box you should get:
3.2x10^-34 = x^2/(0.22-2x)^2

Since the equilibrium constant is less than 10^-3, we can assume that x is not significant enough to affect the equilibrium pressures. Therefore, we can write:
3.2x10^-34 = x^2/0.22^2
Solving for x will tell us that the pressures of H2 and Cl2 at equilibrium are 3.9x10^-18 bar and the equilibrium pressure of HCl is 0.22 bar.

Annie Chantasirivisal_4G
Posts: 114
Joined: Wed Sep 18, 2019 12:21 am

Re: 5l.29

Postby Annie Chantasirivisal_4G » Wed Jan 15, 2020 8:20 pm

Hi! The way I did it is by setting up the ice table, which will end up with HCl having 0.22-2x, and H2 and Cl2 each having x at equilibrium. Then, you plug those values into where they would go in the equilibrium constant ((PH2)(PCl2))/((PHCl)^2) which gives you x^2/(0.22-2x)^2 is equal to K, 3.2 x 10^-34.
Since K is very small (smaller than our threshold of 10^-3), you can assume that the change in the initial HCl partial pressure from the HCl partial pressure at equilibrium is going to be very small, so you can approximate that HCl partial pressure will remain as 0.22 bar. Continuing from that approximation your equation would now look like 3.2x10^-34(0.22)^2=x^2 once you cross multiply, and solving for x would give you x= +/- 3.9 x 10^-18. Negative partial pressures do not exist, thus the x we will use in positive 3.9 x 10^-18, which means our H2 and Cl2 partial pressures at equilibrium will be 3.9 x 10^-18 bar.

To recap:
HCl=0.22 bar
H2=3.9 x 10^-18 bar
Cl2=3.9 x 10^-18 bar

Hope this helps!

Prasanna Padmanabham 4I
Posts: 111
Joined: Thu Jul 25, 2019 12:17 am

Re: 5l.29

Postby Prasanna Padmanabham 4I » Thu Jan 16, 2020 10:16 am

I have a followup question:

Why did you all decide to set x = to change in equilibrium for H2 or Cl2, and 2X = change in HCL instead of saying x = change in HCL and 0.5X = change in H2 and Cl2?

I did it the second way and I got the wrong answer. I'm trying to understand conceptually why my way is wrong.

Thank you!

Annie Chantasirivisal_4G
Posts: 114
Joined: Wed Sep 18, 2019 12:21 am

Re: 5l.29

Postby Annie Chantasirivisal_4G » Thu Jan 16, 2020 8:51 pm

Prasanna Padmanabham 4I wrote:I have a followup question:

Why did you all decide to set x = to change in equilibrium for H2 or Cl2, and 2X = change in HCL instead of saying x = change in HCL and 0.5X = change in H2 and Cl2?

I did it the second way and I got the wrong answer. I'm trying to understand conceptually why my way is wrong.

Thank you!


Technically you should still get the same answer as the molar ratios are right (2:1:1 and 1:0.5:0.5 are the same thing) but personally I feel like using whole numbers rather than decimals is a lot easier bc there's less of a mix-up when you have to multiply and expand into the quadratic equation... I think you might have made an error multiplying or adding at some point? Sorry I can't be of help much without seeing your work.


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