Can ICE tables give two solutions? What does this mean physically?

Question

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.

Answer 1

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).

Answer 2

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.