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The following question comes from General Chemistry: Principles and Modern Applications, 11th ed. by Pettruci.

Exercise 15-33. A mixture consisting of 0.150 mol H2 and 0.150 mol I2 is brought to equilibrium at 445 °C, in a 3.25 L flask. What are the equilibrium amounts of H2, I2, and HI? $$\ce{H2(g) + I2(g) <=> 2HI(g)}\qquad K_c=50.2\text{ at }445\,\mathrm{^\circ C}$$

First, I found the concentrations of H2 and I2: $$ \begin{aligned} \phantom{}[\ce{H2}]=[\ce{I2}]&=\frac{0.150\text{ mol}}{3.25\text{ L}} \\ &=0.0461538\text{ M} \end{aligned} $$

Setting up the ICE table:

$$ \begin{array}{|l|c c c c c|} \hline &\ce{H2}&+&\ce{I2}&\rightleftharpoons&\ce{2HI} \\ \hline \text{Initial}&0.0461538&&0.0461538&&0 \\ \text{Change}&-x&&-x&&+2x \\ \text{Equilibrium}&0.0461538-x&&0.0461538-x&&2x \\ \hline \end{array} $$ Then, $$ K_c=50.2=\frac{(2x)^2}{(0.0461538-x)(0.0461538-x)}. $$ Most solutions I've seen for similar problems involve taking the square root of both sides of this equation. However, they only take the principal root, effectively ignoring a second valid solution. Here, I'll do the proper solution for finding $x$: $$ \begin{aligned} \\ 50.2&=\left(\frac{2x}{0.0461538-x}\right)^2 \\ \sqrt{50.2}&=\pm\frac{2x}{0.0461538-x} \end{aligned} \\ \begin{aligned} 7.0852&=\frac{2x}{0.0461538-x}&&\text{or}&7.0852&=-\frac{2x}{0.0461538-x} \\ 0.327009-7.0852x&=2x&&&0.327009-7.0852x&=-2x \\ 0.327009&=9.0852x&&&0.327009&=5.0852x \\ x&=0.0359936&&&x&=0.064306 \end{aligned} $$ As you can see, there are two positive (and therefore, valid) solutions for $x$. However, only one of the solutions results in a positive $[\ce{H2}]_\mathrm{eq}$: $$ \begin{aligned} \phantom{}[\ce{H2}]_\mathrm{eq}=0.0461538-x \end{aligned} \\ \begin{aligned} \phantom{}[\ce{H2}]_\mathrm{eq}&=0.0461538-0.0359936&&\text{or}&[\ce{H2}]_\mathrm{eq}&=0.0461538-0.064306 \\ &=0.0101602\text{ M}&&&&=-0.0181522\text{ M} \end{aligned} $$ And $x=0.0359936$ does indeed lead to the correct answer according to the textbook.

My question is: was it just a coincidence that I was able to eliminate one of the possible solutions for $x$? In other words, is there a scenario where I could end up with two possible $[\ce{H2}]_\mathrm{eq}$s? If not, what forbids this?

Edit: To reiterate, I am aware that one of the solutions is to be eliminated for being "non-physical". My question is whether it is possible for both solutions to be physical.

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    $\begingroup$ See this which is almost the same problem in reverse, and it's not a dupe of what you're asking. In short, in any case when you get a non-physical answer (such as a negative molarity) when solving an equation for more than one root, discarding it as you have is the correct thing to do. $\endgroup$ – Todd Minehardt Mar 6 at 3:46
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    $\begingroup$ I'm not sure how to prove it, but I think that a system with more than one equilibrium condition would imply very bad things, possibly all the way up to a perpetual motion device. Having two equilibrium concentrations effectively means that even after infinite time, a system still hasn't decided on its ground state. $\endgroup$ – Nicolau Saker Neto Mar 6 at 5:48
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In other words, is there a scenario where I could end up with two possible [solutions]?

No, there is always a single solution. The reaction quotient Q assumes values from zero (no products) to infinity (no reactants). It varies monotonically with the extent of reaction (the x in the ICE table). So there is always a solution (because Q covers the full range of possible equilibrium constant), and it is always unique (because for a given set of starting values, every value of Q is unique - follows from the monotonic variation).

The only exception is a process where none of the concentration vary with x (when all species are pure solids or liquids). In this case, Q takes a single value (except for running out of reactant or product). If that value is equal to K, x can take a range of values. If Q is different from K, there is no solution, and there won't be any equilibrium (i.e. the reaction would go to completion).

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An ICE table is moderately complicated, so it ends up seeming a bit mysterious when you get two solutions. But it's really not that mysterious at all. In fact, this sort of thing is pretty common, since it happens anytime you have two variables related to each other through a square. Let's use a very simple example to illustrate this.

Suppose you are given the following problem:

The area of a square is $\pu{25 m2}$. What is the length of each side?

To solve this, we use $A = l^2$, where $A$ is the area, and $l$ is the length of each side. Hence:

$$l = \sqrt{\pu{25 m2}} = \pu{\pm 5 m}$$

Now clearly a side can't have a negative length — it's non-physical. So we choose $l = \pu{+5 m}$ instead of $l = \pu{-5 m}.$ Nothing mysterious about that. I.e., what you were encountering was merely a property of squares, not a property specific to ICE tables.

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  • $\begingroup$ I understand that the second solution can be eliminated for being non-physical. I was wondering if there was a situation where both answers are physical and neither can be eliminated. $\endgroup$ – Brandon Mar 6 at 5:32
  • $\begingroup$ @Brandon I was mostly responding to your question "was it just a coincidence that I was able to eliminate one of the possible solutions for x?", since the question seemed to imply that there was something unusual (a "coincidence") going on to give you just one solution, and I was trying to explain that it wasn't a coincidence, but rather a natural outcome of having squares. Having said that, I will try to give some thought to addressing your other question. $\endgroup$ – theorist Mar 6 at 6:30

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