2 Solutions to Quadratic Equations

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Najia Saleem 2G
Posts: 41
Joined: Wed Nov 18, 2020 12:20 am

2 Solutions to Quadratic Equations

Postby Najia Saleem 2G » Sun Jan 10, 2021 3:51 pm

When doing the quadratic equation for problem #17 in module 3 of the AVF topics, I found that I got two positive solutions. I have pasted the problem below, and I was wondering if we get two positive solutions do we choose the least amount as the value of X? This is what I found the answers did. For reference, the two solutions I got were: 0.126 and 0.0828

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)

A. [CO] = [H2O] = 0.0172 M, [CO2] = [H2] = 0.0828 M

B. [CO] = [H2O] = 0.0172 M, [CO2] = [H2] = 0.0199 M

C. [CO] = [H2O] = 0.0801 M, [CO2] = [H2] = 0.0828 M

D. [CO] = [H2O] = 0.0801 M, [CO2] = [H2] = 0.0199 M

Christine Ma 3L
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Re: 2 Solutions to Quadratic Equations

Postby Christine Ma 3L » Sun Jan 10, 2021 4:02 pm

Because the equilibrium concentrations of CO and H2O are 0.100-X (according to the ICE table), X cannot equal 0.126 or else the concentrations would be negative. For problems like these, I would look at the two solutions and see which one makes sense since one of the solutions will result in a negative equilibrium concentration.

Edward Tang 1k
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Re: 2 Solutions to Quadratic Equations

Postby Edward Tang 1k » Sun Jan 10, 2021 4:11 pm

There are two ways to tackle the issue of getting two positive numbers when solving a quadratic.

First one is more mathematically complicated and sometimes impossible to do if the numbers aren't too nice. When you set up the question I believe you got something like (X^2)/((0.1-X)^2)=23.2, with X representing the change in concentration. When you have something raised to the power of two, everything becomes positive. And obviously in chemistry we can't have negative concentrations. So usually when I see anything raised to the second power I would eliminate the square but taking the square root of both side of the equation, which in this case leaving you with X/(0.1-X)=sqrt(23.2). When you solve this, I believe you will only get one answer, which is 0.0828.

The other one would make more conceptual sense. After solving the the quadratic without taking the sqrt, you have 0.126 and 0.0828. If you did have X=0.126, then the reactants would be negative since it's concentration at equilibrium is expressed by 0.1-X, and 0.1-0.126 would be negative concentration which doesn't make sense. So the right answer is 0.0828.

RitaThomas_3G
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Re: 2 Solutions to Quadratic Equations

Postby RitaThomas_3G » Sun Jan 10, 2021 6:15 pm

The way that I would approach a problem like this would be to initially look at the ICE table, where we can see that one of the concentrations is equal to 0.1-x. Logically, we can see that with an x value of 0.126, the concentration would be negative. Therefore, this positive x value would not make sense.

Najia Saleem 2G
Posts: 41
Joined: Wed Nov 18, 2020 12:20 am

Re: 2 Solutions to Quadratic Equations

Postby Najia Saleem 2G » Sun Jan 10, 2021 7:52 pm

Christine Ma 3L wrote:Because the equilibrium concentrations of CO and H2O are 0.100-X (according to the ICE table), X cannot equal 0.126 or else the concentrations would be negative. For problems like these, I would look at the two solutions and see which one makes sense since one of the solutions will result in a negative equilibrium concentration.


Thank you for the simple explanation! I now understand why 0.126 would not be an appropriate answer.

Izamary Marquez 2H
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Re: 2 Solutions to Quadratic Equations

Postby Izamary Marquez 2H » Sun Jan 10, 2021 8:01 pm

Christine Ma 3L wrote:Because the equilibrium concentrations of CO and H2O are 0.100-X (according to the ICE table), X cannot equal 0.126 or else the concentrations would be negative. For problems like these, I would look at the two solutions and see which one makes sense since one of the solutions will result in a negative equilibrium concentration.


Exactly! My TA said that there will always only be one answer. If you get a negative and a positive, it will be the positive value. If you get two positive values, there will more than likely be one that doesn't work because it is larger than the given concentration. I am not sure if we will ever see a problem where we get two solutions that could both work.


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