Module: Equilibrium Part 3 Question 17

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Hannah Lee 2F
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Joined: Thu Jul 11, 2019 12:15 am

Module: Equilibrium Part 3 Question 17

Postby Hannah Lee 2F » Mon Jan 06, 2020 9:41 pm

17. If the initial amounts of CO and H2O were both 0.100 M, what will be the amounts of each reactant and product at equilibrium for the following reaction? Keq = 23.2 at 600K
CO (g) + H2O (g) ⇌ CO2 (g) + H2 (g)

After using ICE, for my quadratic equation, I solved it from:
x^2 / (0.100 - x)^2 = 23.2
23x^2 - 4.64x + 0.232 = x^2
22x^2 - 4.64x + 0.232 = 0
However, using the quadratic formula, x turns out to be 0.126, which doesn't make sense because then the equilibrium concentration will be a negative number. Can someone tell me what I'm doing wrong?

Justin Vayakone 1C
Posts: 110
Joined: Sat Sep 07, 2019 12:19 am

Re: Module: Equilibrium Part 3 Question 17

Postby Justin Vayakone 1C » Mon Jan 06, 2020 10:22 pm

If you can solve the problem, without the x^2 approximation, then do it that way. In this case, making x^2 equal to 0 is too much of an approximation. This is how I solved for x: (in file attachment)
Edit: Click on the image to see it right side up
Attachments
ICETableMath.JPG
Last edited by Justin Vayakone 1C on Mon Jan 06, 2020 10:23 pm, edited 1 time in total.

Charisse Vu 1H
Posts: 101
Joined: Thu Jul 25, 2019 12:17 am

Re: Module: Equilibrium Part 3 Question 17

Postby Charisse Vu 1H » Mon Jan 06, 2020 10:22 pm

Remember that when solving a quadratic equation, x will have two values. The other value that you have not yet solved for will probably give you the right answer to the question! (Your quadratic equation is correct, you just need to solve for the other x).

MAC 4G
Posts: 121
Joined: Wed Sep 18, 2019 12:16 am

Re: Module: Equilibrium Part 3 Question 17

Postby MAC 4G » Mon Jan 06, 2020 10:27 pm

I believe you did the problem right, at least that's how I did it. I believe you want the positive number because negative concentrations are unphysical, or in other words, you want a positive number to make chemical sense.


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