Lecture 13 - Applications to Chemistry

Learning Objective

• Use linear algebra to balance chemical equations

Balancing Equations

In chemical reactions, chemicals are combined to produce other chemicals. For example, this chemical equations represents the reaction when propane (\( \rm C_3 \rm H_8 \)) is burned in the presence of oxygen (\(\rm O_2\)) to produce water (\(\rm H_2 \rm O\)) and carbon dioxide (\(\rm C \rm O_2\)): \[ \rm C_3 \rm H_8 + \rm O_2 \rightarrow \rm H_2 \rm O + \rm C \rm O_2 \]

The atoms are neither created nor destroyed by the reaction, just rearranged. If there are 10 million carbon atoms before the reaction, then there will still be 10 million carbon atoms after the reaction. Our goal is to find coefficients for the chemical equation to represent the ratios of the various chemicals before and after the reaction.

Specifically, we are looking to find integer values \( x_1, x_2, x_3 \), and \( x_4 \) so that the number of atoms of each element on the left-hand side of the equation equals the number of atoms of that element on the right-hand side: \[ {\color{red} x_1}\ {\rm C_3 H_8} + {\color{red} x_2}\ {\rm O_2} \rightarrow {\color{red} x_3}\ {\rm H_2 O} + {\color{red} x_4}\ {\rm C O_2} \]

We will set up one algebraic equation for each different element that appears in the chemical equation. In this example, we have three elements: Carbon, Hydrogen, and Oxygen. Each side of each algebraic equation represents the number of atoms on that side of the chemical equation:

Now we solve this system of equations in the normal way. Moving all the variables to the left-hand side gives the homogeneous system \[ \begin{eqnarray*} 3x_1 - x_4 & = & 0 \\ 8x_1 -2x_3 & = & 0 \\ 2x_2 - x_3 - 2x_4 & = & 0 \end{eqnarray*} \]

We then row-reduce the corresponding augmented matrix: \[ \begin{bmatrix} 3 & 0 & 0 & -1 & 0 \\ 8 & 0 & -2 & 0 & 0 \\ 0 & 2 & -1 & -2 & 0 \end{bmatrix} \longrightarrow \begin{bmatrix} 1 & 0 & 0 & -1/3 & 0 \\ 0 & 1 & 0 & -5/3 & 0 \\ 0 & 0 & 1 & -4/3 & 0 \end{bmatrix} \]

This gives the general solution \( x_1 = \frac 1 3 x_4 \), \( x_2 = \frac 5 3 x_4 \), \( x_3 = \frac 4 3 x_4 \), and \( x_4 \) is free. Generally, chemists are looking for the smallest integer coefficients that balance the chemical equation. In this case, the smallest positive value for \( x_4 \) that makes all the variables have integer values is \( x_4 = 3 \). This yields the balanced equation \[ {\color{red} 1}\ {\rm C_3 H_8} + {\color{red} 5}\ {\rm O_2} \rightarrow {\color{red} 4}\ {\rm H_2 O} + {\color{red} 3}\ {\rm C O_2} \]

Remember that these coefficients represent ratios. So, for example, if we had 10 million molecules of propane (\(\rm C_3 H_8\)) before the reaction, then we would have \( (x_3/x_1) \cdot 10 = 40 \) million molecules of water afterwards.

Summary

When using linear algebra to balance a chemical equation, you'll have on equation for each element that appears in the equation. You'll have one variable for each compound that appears in the equation. The system you get will be homogeneous, so there will be infinitely many solution. Find the solution where the variables all equal positive integers that are as small as possible.