Balancing a chemical equation involving ethanol and hydrogen peroxide by setting up a vector equation

Question

I'm having trouble balancing a chemical equation for the combustion of ethanol with hydrogen peroxide to produce carbon dioxide and water. The equation is given as:

$$x_4 C _2 H _5 OH +x_3 H _2 O _2 \rightarrow x_2 H _2 O +x_1 CO _2$$

I know that the coefficients $x_1, x_2, x_3$, and $x_4$ represent the number of molecules needed to balance the equation, so that there are equal numbers of carbon, hydrogen, and oxygen atoms on both sides of the equation.

I tried setting up a vector equation that describes the numbers of atoms of each type present in the reaction, similar to the method used for balancing the combustion of propane equation given in the textbook. My equation is:

$$x_{4}\begin{pmatrix}2\\ 5\\ 1\end{pmatrix} +x_{3}\begin{pmatrix}0\\ 2\\ 2\end{pmatrix} =x_{2}\begin{pmatrix}0\\ 2\\ 1\end{pmatrix} +x_{1}\begin{pmatrix}1\\ 0\\ 2\end{pmatrix}$$

And since it would make more convenient if everything was moved onto one side of the equation, we have:

$$x_{4}\begin{pmatrix}2\\ 5\\ 1\end{pmatrix} +x_{3}\begin{pmatrix}0\\ 2\\ 2\end{pmatrix} -x_{2}\begin{pmatrix}0\\ 2\\ 1\end{pmatrix} -x_{1}\begin{pmatrix}1\\ 0\\ 2\end{pmatrix} =0$$

Which is the same as:

$$x_{4}\begin{pmatrix}2\\ 5\\ 1\end{pmatrix} +x_{3}\begin{pmatrix}0\\ 2\\ 2\end{pmatrix} +x_{2}\begin{pmatrix}0\\ -2\\ -1\end{pmatrix} +x_{1}\begin{pmatrix}-1\\ 0\\ -2\end{pmatrix} =0$$

But I'm not sure how to solve this system of equations to find the values of $x_1, x_2, x_3$, and $x_4$. The textbook mentions row reduction and the use of free variables, but I'm not confident in my ability to apply those concepts to this problem. Can someone please help me with this? Also, if you have any tips for balancing chemical equations in general, I would appreciate it. Thank you!

Answer 1

Balancing most chemical equations should not require vectors; I would suggest a simpler technique for balancing these types of equations.

$$\_\_ \; C_2H_5OH + \_\_ \; H_2O_2 \longrightarrow \_\_ \; H_2O + \_\_ \; CO_2$$

First, start with elements that only show up once on each side of the equation, and assign variables to balance those. In your equation, the only element that qualifies would be $C$.

$$\underline{\; \alpha \;} C_2H_5OH + \_\_ \; H_2O_2 \longrightarrow \_\_ \; H_2O + \underline{\; 2\alpha \;} CO_2$$

Now assign variables to all the remaining coefficients:

$$\underline{\; \alpha \;} C_2H_5OH + \underline{\; \beta \;} H_2O_2 \longrightarrow \underline{\; \gamma \;} H_2O + \underline{\; 2\alpha \;} CO_2$$

Next, write equations to balance all the elements which weren't already balanced in the first step (in our case, $C$).

$$\begin{align} H: && 6\alpha + 2\beta &= 2\gamma \\[2mm] O: && \alpha + 2\beta &= \gamma + 4\alpha \end{align}$$

Simplifying the two equations yields the system

$$\begin{align} 3\alpha + \beta - \gamma &= 0 \\[2mm] 3\alpha -2\beta + \gamma &= 0 \end{align}$$

Since there are only two equations, we will only be able to eliminate one of the variables [see note below]; in this case the easiest would be to eliminate $\gamma$ by adding the two equations, which gives:

$$6\alpha - \beta = 0$$

Now choose the smallest natural numbers to satisfy the equation; clearly in this case that would be $\alpha = 1$ and $\beta = 6$, which then leads to $\gamma = 9$. Thus the balanced equation is

$$ C_2H_5OH + 6 \, H_2O_2 \longrightarrow 9 \, H_2O + 2 \, CO_2$$

Occasionally, solving for the final variable(s) might yield fractional solution(s), but then you can just multiply all values by the LCD. For example, in this case, if we had eliminated $\alpha$ instead, we would have gotten the equation $3\beta - 2\gamma = 0$; the lowest values would be $\beta = 2$ and $\gamma = 3$, from which we would get $\alpha = \frac{1}{3}$, but then multiplying all values by $3$ yields the correct solution.

Note: Chemical equations will generally lead to systems that have at least one degree of freedom, i.e. the number of equations will be one lower than the number of variables. [I believe all equations should yield systems that have exactly one degree of freedom, at least I've never seen any that didn't; but I don't know enough chemistry to guarantee this.] This means that you should be able to eliminate all but two of the variables.

Answer 2

First, correct the typo in the vector associated with $x_4$ coefficient $$x_4\pmatrix{2\\ 6\\ 1}$$ Next, concatenate the vectors to write the equation in matrix form $$\pmatrix{ -1 & \;\;0 & 0 & 2 \\ \;\;0 & -2 & 2 & 6 \\ -2 & -1 & 2 & 1 \\ }\cdot\pmatrix{x_1\\x_2\\x_3\\x_4} = \pmatrix{0\\0\\0} $$ Setting $x_1=1$ reduces the number of unknowns to match the number of equations and allows the use standard Linear Algebra to solve for the unknown coefficients. $$\pmatrix{ \;\;0 & 0 & 2 \\ -2 & 2 & 6 \\ -1 & 2 & 1 \\ }\cdot\pmatrix{x_2\\x_3\\x_4} = \pmatrix{1\\0\\2} \quad\implies\quad \pmatrix{x_2\\x_3\\x_4} = \pmatrix{4.5\\3.0\\0.5} \\ $$ Include $x_1$ again and multiply by $2$ to clear the fractions.
This generates a set of integer coefficients $$ \pmatrix{x_1\\x_2\\x_3\\x_4} = \pmatrix{2 \\ 9 \\ 6 \\ 1} \\\\ $$

Update

Another way to approach the problem is to note that $$Ax=0$$ means that $x$ lies in the nullspace of $A.\:$
Such vectors can be calculated using the pseudoinverse denoted by $A^+$ $$x = \left(I-A^+A\right)w$$ where $w$ is an arbitrary vector.

In this case $$\eqalign{ A &= \pmatrix{ -1 & \;\;0 & 0 & 2 \\ \;\;0 & -2 & 2 & 6 \\ -2 & -1 & 2 & 1 \\ } \qquad A^+ &= \frac{1}{122}\pmatrix{ -58 &\;\;24 &-30 \\ \;\;44 &-14 &-13 \\ -52 &\;\;11 &\;\;32 \\ \;\;32 &\;\;12 &-15} }$$