# Can a chemical reaction have two balanced equations?

I was recently studying Xenon compounds and read this reaction:

$$\ce{6XeF4 + 12H2O -> 4Xe + 2XeO3 + 24HF + 3O2}$$

However this reaction can also be written as the following:

$$\ce{4XeF4 + 8H2O → 2Xe + 2XeO3 + 16HF + O2}$$

Now both of these are valid but wouldn't that create problems, like in quantitative analysis or in kinetics of the reaction?

• I was thinking about my statement in kinetics and now I realised the various species are independently reacting. Moreover an interesting pattern emerges if you keep adding the difference of these reactions to the first one indefinitely. Thanks to all of you for clearing this for me. – user34209 Aug 28 '16 at 7:12

Expand the second equation so it has the same amount of educt on the left side:

$$\ce{6XeF4 + 12H2O → 3Xe + 3XeO3 + 24HF + 1.5 O2}$$

Difference on the product side, compared to first equation : $\ce{-1 Xe + 1 XeO3 - 1.5 O2}$.

This excess of one $\ce{XeO3}$ can go

$$\ce{XeO3 → Xe + 1.5 O2}$$

, exactly the amount of xenon and oxygen that was missing compared to the first equation.

Your equations describe not one reaction, but (at least) two, which don't have to occur in a predefined proportion. It's not directly clear, and would involve an in-depth investigation, what exactly the elementar reactions are.

So you could make an infinite number of chemically correct, but quantiatively different equations for the hydrolysis of xenon flouride. All depends on the reaction conditions, probably.

• The amount of oxygen should be $\frac{3}{2}$, which comes from multiplying the OP second equation by ${6}{4}$. It seems to me at least. Apologize if I'm wrong. And I think you have answered his most important point (the post title): it has to be experimentally proven. Then if we take first as The reaction, then reaction two is not valid. (+1) – user43021 Jun 2 '18 at 15:06
• What amount of oxygen, and what experiment? The two equations predict twodifferent reactions! – Karl Jun 3 '18 at 14:05
• @santimirandarp The oxygen checks out as far as i can see. 12 atoms on the left, 12 on the right. If you think otherwise, please write down the corrected equation. – Karl Jun 3 '18 at 19:13

The first reaction is actually two reactions:

\begin{align} \ce{4 XeF4 + 8 H2O &→ 2 Xe + 2 XeO3 + 16 HF + O2} \\ \ce{2 XeF4 + 4 H2O &→ 2 Xe + 8 HF + 2 O2} \\ \hline \ce{6 XeF4 + 12 H2O &-> 4 Xe + 2 XeO3 + 24 HF + 3O2} \end{align}

• That is just one arbitrary possibility. – Karl Aug 28 '16 at 6:44
• I couldn't find any other linear combination of reactions that would give the same result (multiplying every stoichiometric factor in the above schema with the same number does not yield a new result). – aventurin Aug 28 '16 at 9:54
• It's the only way to split that specific equation into two other, specific reactions, of course. But the exact stochiometry of the reaction is not clear. – Karl Aug 28 '16 at 10:29
• This reaction can also be balanced with following factors: 5,10,3,2,20,2 – Rima Aug 28 '16 at 12:52
• This is just another linear combination of the actual two reactions, with both the first and second with factor 1. Above the factors are 1 and 2 respectively. – aventurin Aug 28 '16 at 13:13

If the coefficients that balance the chemical reaction are expressed as an ordered 6-tuple (a, b, c, d, e, f) where a is the coefficient of XeF4, b is the coefficient of H2O, etc., then from a purely mathematical point of view, the set of all possible coefficients are of the form:

(x/4, x/2, [x/12]+[2y/3], [x/6]-[2y/3], x, y)

where x+8y is a multiple of 12, x-4y is a multiple of 6 and x > 4y.

The mathematical reason for the existence of distinct ways of balancing is that the matrix that represents the reaction has two free variables when expressed in reduced row-echelon form.