5
$\begingroup$

I'm trying to use the null space method to balance the following equation: enter image description here. I obtained the following composition matrix: $$\begin{bmatrix} 3 & 8 & 1 & 12 & 4 & 2 \\ 1 & 0 & 0 & 1 & 0 & 0 \\ 4 & 4 & 3 & 40 & 3 & 1 \\ 0 & 2 & 1 & 3 & 1 & 0 \\ 0 & 1 & 0 & 12 & 0 & 0 \end{bmatrix}$$ where rows are in order H P O N Mo. I take the rref of this matrix, and augment it with one row of zeroes except the last element is 1. Taking the inverse of that matrix, I get $$\begin{bmatrix} \frac{1}{51} & \frac{4}{17} & \frac{35}{51} & -\frac{1}{51} & -\frac{56}{51} & 1\\ \end{bmatrix}.$$ After scaling the elements by 51 I would get the end result of coefficients: 1, 12, 35, 1, 56, 51 which does not at all balance the equation. A correct solution would be 1, 12, 21, 1, 21, 12.

I have used the exact same steps to balance other equations. What am I doing wrong?

$\endgroup$
4
$\begingroup$

Fortunately this issue was just caused by a small oversight. In the midst of so many coefficients, you accidentally skipped the second nitrogen atom in $\ce{NH4NO3}$, which changes the fifth element in the fourth row from 1 to 2. The correct composition matrix is then:

$$\begin{bmatrix} 3 & 8 & 1 & 12 & 4 & 2 \\ 1 & 0 & 0 & 1 & 0 & 0 \\ 4 & 4 & 3 & 40 & 3 & 1 \\ 0 & 2 & 1 & 3 & \color{red}{\textbf{2}} & 0 \\ 0 & 1 & 0 & 12 & 0 & 0 \end{bmatrix}$$

After complementing with the bottom row $\begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}$ and inverting the matrix, we obtain:

$$\frac{1}{96}\begin{bmatrix} 3 & 79 & 2 & -9 & -14 & -8 \\ 36 & -204 & 24 & -108 & -72 & -96 \\ 39 & -477 & 90 & -213 & -246 & -168 \\ -3 & 17 & -2 & 9 & 14 & 8 \\ -51 & 417 & -66 & 249 & 174 & 168 \\ 0 & 0 & 0 & 0 & 0 & 96 \end{bmatrix}$$

Finally, transposing the rightmost column and dividing its coefficients by 8, the result is $\begin{bmatrix} -1 & -12 & -21 & 1 & 21 & 12 \end{bmatrix}$, exactly as expected.

Thank you for the linked article! Hopefully others may find it useful. If you're into linear algebra applied to chemistry, I recommend you take a peek into chemical reaction network theory!

| improve this answer | |
$\endgroup$
  • $\begingroup$ Ah, yes. The classic one. That's easy to fix. Thank you very much $\endgroup$ – cethy Sep 8 at 14:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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