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A First Course in Linear Algebra, 2017a

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1.2. Systems of Equations, <strong>Algebra</strong>ic Procedures 33<br />

1.2.5 Balanc<strong>in</strong>g Chemical Reactions<br />

The tools of l<strong>in</strong>ear algebra can also be used <strong>in</strong> the subject area of Chemistry, specifically for balanc<strong>in</strong>g<br />

chemical reactions.<br />

Consider the chemical reaction<br />

SnO 2 + H 2 → Sn + H 2 O<br />

Here the elements <strong>in</strong>volved are t<strong>in</strong> (Sn), oxygen (O), and hydrogen (H). A chemical reaction occurs and<br />

the result is a comb<strong>in</strong>ation of t<strong>in</strong> (Sn) andwater(H 2 O). When consider<strong>in</strong>g chemical reactions, we want<br />

to <strong>in</strong>vestigate how much of each element we began with and how much of each element is <strong>in</strong>volved <strong>in</strong> the<br />

result.<br />

An important theory we will use here is the mass balance theory. It tells us that we cannot create or<br />

delete elements with<strong>in</strong> a chemical reaction. For example, <strong>in</strong> the above expression, we must have the same<br />

number of oxygen, t<strong>in</strong>, and hydrogen on both sides of the reaction. Notice that this is not currently the<br />

case. For example, there are two oxygen atoms on the left and only one on the right. In order to fix this,<br />

we want to f<strong>in</strong>d numbers x,y,z,w such that<br />

xSnO 2 + yH 2 → zSn + wH 2 O<br />

where both sides of the reaction have the same number of atoms of the various elements.<br />

This is a familiar problem. We can solve it by sett<strong>in</strong>g up a system of equations <strong>in</strong> the variables x,y,z,w.<br />

Thus you need<br />

Sn : x = z<br />

O : 2x = w<br />

H : 2y = 2w<br />

We can rewrite these equations as<br />

Sn : x − z = 0<br />

O : 2x − w = 0<br />

H : 2y − 2w = 0<br />

The augmented matrix for this system of equations is given by<br />

⎡<br />

1 0 −1 0<br />

⎤<br />

0<br />

⎣ 2 0 0 −1 0 ⎦<br />

0 2 0 −2 0<br />

The reduced row-echelon form of this matrix is<br />

⎡<br />

1 0 0 − 1 2<br />

0<br />

⎢<br />

⎣<br />

0 1 0 −1 0<br />

0 0 1 − 1 2<br />

0<br />

⎤<br />

⎥<br />

⎦<br />

The solution is given by<br />

x − 1 2 w = 0<br />

y − w = 0<br />

z − 1 2 w = 0

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