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Consider the chemical reaction:

$$ \mathrm{ClO_3} + \text I_2 \to \mathrm{Cl} + \mathrm {IO_3} $$

When given the reaction in a test, I did not know whether I was meant to balance it by manipulating the coefficients or to balance it using chemical concepts like adding various ions to the reaction. My teacher claims I am supposed to be able to "tell" which approach to use and not use chemical concepts if it can be solved mathematically.

Suppose we place coefficients on each chemical species: from left to right, $a$, $b$, $c$, and $d$. We can then create the following system to model this reaction:

$a = d$ (for O)

$a=c$ (for Cl)

2b = d (For I)

This question's answers, although beyond my grade level, have been useful. In an answer by user Sophie Alpert, she says

If [a set of equations that can be used to balance a chemical equation] has no solution where the coefficients are nonzero, then you know that the equation cannot be balanced.

My question is: how do you determine whether these sets of equations have solutions, without actually finding said solutions?

Fouad Saffar
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  • I think that we need not only look for nonzero solutions, but for positive solutions. That is, unless you allow for moving one chemical from one side to the other. Cf. $xH_2O+yH_2=zO_2$ which has a nonzero solution $x=2,y=-2,z=1$. (Luckily we can swap hydrogen to the other side.) –  Dec 07 '21 at 23:46
  • Correct. Edited. – Fouad Saffar Dec 07 '21 at 23:49
  • To rephrase what Sophie Alpert said: since all the coefficients are integers or fractional, you can arbitrarily normalize them by setting (say) $a$ to 1 (we're interested in proportions, not the actual values). If you end up with (say) $b = \frac{1}{2}$ and you want integers, multiply all coefficients by 2 at the end. Now that $a = 1$, simply solve your linear system for $b$, $c$ and $d$. – Charlie Vanaret Dec 08 '21 at 00:10

1 Answers1

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So let's consider Sophie's method to solve your equation. I'm going to copy the equation here because I want to have all of the information readily available in the answer itself, without any need to consult with the original question.

$$ \mathrm{ClO}_3~+~\mathrm{I}_2~\rightarrow~\mathrm{Cl}~+~\mathrm{IO}_3 $$

There we go. Nice.

Now, according to Sophie's method, we will put some coefficients in front of the reactants and products. These coefficients correspond to our unknowns and finding them solves the problem. Sophie used capital roman letters but since the symbols for the elements can contain capital roman letters, I figured it would be best if we used lower-case greek letters just to be pretty clear what we are doing.

$$ \alpha\mathrm{ClO}_3~+~\beta\mathrm{I}_2~\rightarrow~\gamma\mathrm{Cl}~+~\delta\mathrm{IO}_3 $$

Now... I am taking your equation at face value but my chemical experience is already ringing some alarm bells. I feel like oxidation states are missing. For instance, should it be $\mathrm{Cl}^-$? Anyway, I am going to carry out with the method just to show what is up, but you should double-check the equation.

So the next step in Sophie's method is to assign an equation for each element. We have three elements in that equation. iodine (I), chlorine (Cl) and oxygen (O). So we will have three equations.

  • For I: $2\beta=\delta$
  • For Cl: $\alpha=\gamma$
  • For O: $3\alpha=3\delta$

Already we see that we have more unknowns than we have equations. This means that this particular system of equations is overdetermined and it has infinite solutions. If we had more equations than unknowns, that would be one way in which the chemical equation would be impossible to balance. We can reach that point if we have lots of different elements packed together into few compounds.

Moving on, we can rearrange the linear equations above to look like this:

  • $0\alpha + 2\beta + 0\gamma - \delta = 0$
  • $\alpha + 0\beta - \gamma + 0\delta = 0$
  • $3\alpha + 0\beta + 0\gamma - 3\delta = 0$

This is equivalent to the matrix equation $\mathbf{A}\vec{x}=\vec{0}$ if

$$ \mathbf{A} = \begin{bmatrix} 0 & 2 & 0 & -1 \\ 1 & 0 & -1 & 0 \\ 3 & 0 & 0 & -3\end{bmatrix} $$

and if

$$ \vec{x} = \begin{bmatrix} \alpha \\ \beta \\ \gamma \\ \delta \end{bmatrix} $$

That system will have a non-trivial solution if matrix $\mathbf{A}$ has a null space that contains more vectors than the trivial $\vec{0}$. Another way in which the equation can be impossible to balance, even if you have the proper number of elements and compounds, is if it leads to a matrix of this type that has only a trivial null-space. This one is harder to correlate with chemical information. I wouldn't be able to tell you whether there are rules of thumb we can use in this regard.

Finally. We know that atoms can't be transferred in fractions from one compound to the other during a chemical transformations, so there must be an integer number of atoms on both sides of the equations and they have to be the same for all elements. This puts a final constraint on the type of solution we can obtain. The equation might be impossible to balance even if our matrix $\mathbf{A}$ has a non-trivial null-space if none of the basis vectors in the nulls-space can be renormalized to integer components, which would correspond to a solution of integer stoichiometric coefficients.

These are methods that, as you have requested, tell you whether or not your equation is possible to balance without having to find the solutions. Except maybe for the last one. #cantBePerfect

urquiza
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  • I excluded oxidation states and states of matter to make it more of a maths question than a chemistry question. Although I am unfamiliar with matrices to this extent, this helped a lot. Thank you. – Fouad Saffar Dec 09 '21 at 00:02