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Module: Equilibrium Part 3 Question 17

Posted: Mon Jan 06, 2020 9:41 pm
by Hannah Lee 2F
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?

Re: Module: Equilibrium Part 3 Question 17

Posted: Mon Jan 06, 2020 10:22 pm
by Justin Vayakone 1C
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

Re: Module: Equilibrium Part 3 Question 17

Posted: Mon Jan 06, 2020 10:22 pm
by Charisse Vu 1H
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).

Re: Module: Equilibrium Part 3 Question 17

Posted: Mon Jan 06, 2020 10:27 pm
by MAC 4G
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.