5
$\begingroup$

I found this equation while testing my equation balancer: $$ \ce{a C7H6O3 + b C4H6O3 -> c C9H8O4 + d C2H4O2} $$

Obviously, $a = b = c = d = 1$ is a solution. However, if we were to bring in some maths:

$ \begin{cases} 7 a + 4b &= 9c + 2d\\ 6a + 6b &= 8c + 4d\\ 3a + 3b &= 4c + 2d\\ \end{cases}\\ \begin{cases} a &= \frac{11}{9}\alpha - \frac{2}{9}\beta\\ b &= \frac{1}{9}\alpha + \frac{8}{9}\beta\\ c &= \alpha\\ d &= \beta\\ \end{cases} $

This leaves us with many ways of balancing the equation:

  • when $\alpha = 1, \beta = 1$, $a = b = c = d = 1$; $\ce{C7H6O3 + C4H6O3 -> C9H8O4 + C2H4O2}$
  • when $\alpha = 1, \beta = 2$, $\begin{cases} a &= 7\\ b &= 17\\ c &= 9\\ d &= 18\\ \end{cases}$; $\ce{7 C7H6O3 + 17 C4H6O3 -> 9 C9H8O4 + 18 C2H4O2}$
  • when $\alpha = 2, \beta = 1$...

From what I know, this happens when the equation consists of 2 or more independent equations (as is the case here: More than one way of balancing a chemical equation). However, here the two equations are:

$$ \ce{11 C7H6O3 + C4H6O3 -> 9 C9H8O4} \\ \ce{-2 C7H6O3 + 8 C4H6O3 -> 9 C2H4O2} $$

What is concerning here is that the second equation has a negative stoichiometric coefficient. My question is, is a negative coefficient acceptable in an equation, and in this case which is the correct way to balance the equation?

Improve this question
$\endgroup$
6
  • $\begingroup$ Your equation was already balanced, so why would you try to get other solution? $\endgroup$
    – Mithoron
    Commented May 1, 2018 at 19:25
  • $\begingroup$ @Mithoron I guess I am being more a mathematician than a chemist here, but clearly there are other solutions, which makes me wonder if they too are valid solutions. $\endgroup$
    – Jingjie Yang
    Commented May 1, 2018 at 19:32
  • 6
    $\begingroup$ There is nothing special about a negative coefficient. If you prefer not to have it, transfer that term to the other side. $\endgroup$
    – Ivan Neretin
    Commented May 1, 2018 at 19:51
  • 4
    $\begingroup$ You may answer your question yourself. $\endgroup$
    – Ivan Neretin
    Commented May 1, 2018 at 20:17
  • 1
    $\begingroup$ @Jingjie YANG you are correct, negative coefficients just flip to be positive on the other side. $\endgroup$
    – Tyberius ♦
    Commented May 2, 2018 at 1:01

1 Answer 1

Reset to default
7
$\begingroup$

As @Ivan Neretin and @Tyberius pointed out, the negative coefficient of a reactant can be flipped to be positive by moving the reactant to the products' side.

So, in my case, the equations become $$\begin{align} \ce{11C7H6O3 + C4H6O3 &-> 9C9H8O4}\\ \ce{8C4H6O3 &-> 2C7H6O3 + 9C2H4O2} \end{align}$$

Now, since the original equation $\ce{C7H6O3 + C4H6O3 -> C9H8O4 + C2H4O2}$ consists of 2 independent equations which can be separately balanced and combined in an arbitrary proportion. See More than one way of balancing a chemical equation for a more detailed explanation.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.