Question

Balancing a chemical reaction. Consider the chemical reaction

$\mathbf{a}\mathbf{}{\mathbf{NO}}_{\mathbf{2}}\mathbf{+}\mathbf{b}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\to }\mathbf{c}\mathbf{}{\mathbf{HNO}}_{\mathbf{2}}\mathbf{+}\mathbf{d}\mathbf{}{\mathbf{HNO}}_{\mathbf{3}}$,

where a, b, c, and d are unknown positive integers. The reaction must be balanced; that is, the number of atoms of each element must be the same before and after the reaction. For example, because the number of oxygen atoms must remain the same,

$\mathbf{2}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{=}\mathbf{2}\mathbf{c}\mathbf{+}\mathbf{3}\mathbf{d}$.

While there are many possible values for $\mathit{a}\mathbf{,}\mathit{b}\mathbf{,}\mathit{c}$ and d that balance the reaction, it is customary to use the smallest possible positive integers. Balance this reaction.

Short answer

The smallest possible positive integers are.$a=2,b=c=d=1$,

Step-by-step solution

Write the elements in terms of positive integers.

The elements in terms of integers are:

Oxygen:

$$\begin{gathered} 2a+b=2ca+3d \\ 2a+b-2c-3d=0......\left(1\right) \end{gathered}$$

Nitrogen:

$$\begin{gathered} a=c+d \\ a-c-d=0......\left(2\right) \end{gathered}$$

Hydrogen:

$$\begin{gathered} 2b=c+d \\ 2b-c-d=0......\left(3\right) \end{gathered}$$

Consider the equations.

Re-write the equations in terms of a matrix.

$\left|\begin{array}{c}2a+b-2c-3d=0\\ a-c-d=0\\ 2b-c-d=0\end{array}\right|$

Represent the above system of equations in terms of the matrix form,

$\left[\begin{array}{ccccc}2& 1& -2& -3& 0\\ 1& 0& -1& -1& 0\\ 0& 2& -1& -1& 0\end{array}\right]$

Solve the matrix.

Consider the matrix.

$\left[\begin{array}{ccccc}2& 1& -2& -3& 0\\ 1& 0& -1& -1& 0\\ 0& 2& -1& -1& 0\end{array}\right]$

Using row Echelon form to reduce the matrix.

$\left[\begin{array}{ccccc}2& 1& -2& -3& 0\\ 0& 2& -1& -1& 0\\ 0& 0& -\frac{1}{4}& \frac{1}{4}& 0\end{array}\right]=\left[\begin{array}{ccccc}1& 0& 0& -2& 0\\ 0& 1& 0& -1& 0\\ 0& 0& 1& -1& 0\end{array}\right]$

Consider the equations,

$$\begin{gathered} a-2d=0........\left(1\right) \\ b-c=0........\left(2\right) \\ c-d=0.........\left(3\right) \end{gathered}$$

Computing the equations (1), (2) and (3), we get the values as,$a=2,b=c=d=1$ .

Final answer

$a=2,b=c=d=1$are the smallest possible values.