Chemical equation

Balancing chemical equations

Because no nuclear reactions take place in a chemical reaction, the chemical elements pass through the reaction unchanged. Thus, each side of the chemical equation must represent the same number of atoms of any particular element (or nuclide, if different isotopes are taken into account). The same holds for the total electric charge, as stated by the charge conservation law. An equation adhering to these requirements is said to be balanced.

A chemical equation is balanced by assigning suitable values to the stoichiometric coefficients. Simple equations can be balanced by inspection, that is, by trial and error. Another technique involves solving a system of linear equations.

Balanced equations are usually written with smallest natural-number coefficients. Yet sometimes it may be advantageous to accept a fractional coefficient, if it simplifies the other coefficients. The introductory example can thus be rewritten as

${\displaystyle \text{HCl}+\text{Na}⟶\text{NaCl}+\frac{1}{2}{\text{H}}_2^{}}$

In some circumstances the fractional coefficients are even inevitable. For example, the reaction corresponding to the standard enthalpy of formation must be written such that one molecule of a single product is formed. This will often require that some reactant coefficients be fractional, as is the case with the formation of lithium fluoride:

${\displaystyle \text{Li}(\text{s})+\frac{1}{2}{\text{F}}_2^{}(\text{g})⟶\text{LiF}(\text{s})}$

Inspection method

The method of inspection can be outlined as setting the most complex substance's stoichiometric coefficient to 1 and assigning values to other coefficients step by step such that both sides of the equation end up with the same number of atoms for each element. If any fractional coefficients arise during this process, the presence of fractions may be eliminated (at any time) by multiplying all coefficients by their lowest common denominator.

Example

Balancing of the chemical equation for the complete combustion of methane

${\displaystyle ?{\text{CH}}_4^{}+?{\text{O}}_2^{}⟶?{\text{CO}}_2^{}+?{\text{H}}_2^{}\text{O}}$

is achieved as follows:

1. A coefficient of 1 is placed in front of the most complex formula (CH₄):

   ${\displaystyle 1{\text{CH}}_4^{}+?{\text{O}}_2^{}⟶?{\text{CO}}_2^{}+?{\text{H}}_2^{}\text{O}}$

2. The left-hand side has 1 carbon atom, so 1 molecule of CO₂ will balance it. The left-hand side also has 4 hydrogen atoms, which will be balanced by 2 molecules of H₂O:

   ${\displaystyle 1{\text{CH}}_4^{}+?{\text{O}}_2^{}⟶1{\text{CO}}_2^{}+2{\text{H}}_2^{}\text{O}}$

3. Balancing the 4 oxygen atoms of the right-hand side by 2 molecules of O₂ yields the equation

   ${\displaystyle 1{\text{CH}}_4^{}+2{\text{O}}_2^{}⟶1{\text{CO}}_2^{}+2{\text{H}}_2^{}\text{O}}$

4. The coefficients equal to 1 are omitted, as they do not need to be specified explicitly:

   ${\displaystyle {\text{CH}}_4^{}+2{\text{O}}_2^{}⟶{\text{CO}}_2^{}+2{\text{H}}_2^{}\text{O}}$

5. It is wise to check that the final equation is balanced, i.e. that for each element there is the same number of atoms on the left- and right-hand side: 1 carbon, 4 hydrogen, and 4 oxygen.

System of linear equations

For each chemical element (or nuclide or unchanged moiety or charge) i, its conservation requirement can be expressed by the mathematical equation

${\displaystyle \sum _{j\in \text{reactants}}{a}_{ij}{s}_j=\sum _{j\in \text{products}}{a}_{ij}{s}_j}$

where

a_ij is the number of atoms of element i in a molecule of substance j (per formula in the chemical equation), and

s_j is the stoichiometric coefficient for the substance j.

This results in a homogeneous system of linear equations, which are readily solved using mathematical methods. Such system always has the all-zeros trivial solution, which we are not interested in, but if there are any additional solutions, there will be infinite number of them. Any non-trivial solution will balance the chemical equation. A "preferred" solution is one with whole-number, mostly positive stoichiometric coefficients s_j with greatest common divisor equal to one.

Example

Let us assign variables to stoichiometric coefficients of the chemical equation from the previous section and write the corresponding linear equations:

${\displaystyle {\mathit{\text{s}}}_1^{}{\text{CH}}_4^{}+{\mathit{\text{s}}}_2^{}{\text{O}}_2^{}⟶{\mathit{\text{s}}}_3^{}{\text{CO}}_2^{}+{\mathit{\text{s}}}_4^{}{\text{H}}_2^{}\text{O}}$

${\displaystyle \begin{array}{cccc}\text{C:}& & s_1& =s_3\\ \text{H:}& & 4s_1& =2s_4\\ \text{O:}& & 2s_2& =2s_3+s_4\end{array}}$

All solutions to this system of linear equations are of the following form, where r is any real number:

${\displaystyle \begin{array}{cc}{s}_1& =r\\ s_2& =2r\\ s_3& =r\\ s_4& =2r\end{array}}$

The choice of r = 1 yields the preferred solution,

${\displaystyle \begin{array}{cc}{s}_1& =1\\ s_2& =2\\ s_3& =1\\ s_4& =2\end{array}}$

which corresponds to the balanced chemical equation:

${\displaystyle {\text{CH}}_4^{}+2{\text{O}}_2^{}⟶{\text{CO}}_2^{}+2{\text{H}}_2^{}\text{O}}$

Matrix method

The system of linear equations introduced in the previous section can also be written using an efficient matrix formalism. First, to unify the reactant and product stoichiometric coefficients s_j, let us introduce the quantity

${\displaystyle {\nu }_j=\{\begin{array}{ll}-s_j& \text{for a reactant}\\ +s_j& \text{for a product}\end{array}}$

called stoichiometric number, which simplifies the linear equations to

${\displaystyle \sum _{j=1}^J{a}_{ij}{\nu }_j=0}$

where J is the total number of reactant and product substances (formulas) in the chemical equation.

Placement of the values a_ij at row i and column j of the composition matrix

A =${\displaystyle \left[\begin{array}{cccc}{a}_{1,1}& a_{1,2}& \cdots & a_{1,J}\\ a_{2,1}& a_{2,2}& \cdots & a_{2,J}\\ ⋮& ⋮& \ddots & ⋮\end{array}\right]}$

and arrangement of the stoichiometric numbers into the stoichiometric vector

ν =${\displaystyle \left[\begin{array}{c}{\nu }_1\\ {\nu }_2\\ ⋮\\ {\nu }_J\end{array}\right]}$

allows the system of equations to be expressed as a single matrix equation:

Aν = 0

Like previously, any nonzero stoichiometric vector ν, which solves the matrix equation, will balance the chemical equation.

The set of solutions to the matrix equation is a linear space called the kernel of the matrix A. For this space to contain nonzero vectors ν, i.e. to have a positive dimension J_N, the columns of the composition matrix A must not be linearly independent. The problem of balancing a chemical equation then becomes the problem of determining the J_N-dimensional kernel of the composition matrix. Only for J_N = 1 will there be a unique preferred solution to the balancing problem. For J_N > 1 there will be an infinite number of preferred solutions with J_N of them linearly independent. If J_N = 0, there will be only the unusable trivial solution, the zero vector.

Techniques have been developed to quickly calculate a set of J_N independent solutions to the balancing problem, which are superior to the inspection and algebraic method in that they are determinative and yield all solutions to the balancing problem.

Example

Using the same chemical equation again, write the corresponding matrix equation:

${\displaystyle {\mathit{\text{s}}}_1^{}{\text{CH}}_4^{}+{\mathit{\text{s}}}_2^{}{\text{O}}_2^{}⟶{\mathit{\text{s}}}_3^{}{\text{CO}}_2^{}+{\mathit{\text{s}}}_4^{}{\text{H}}_2^{}\text{O}}$

${\displaystyle \begin{array}{c}\text{C:}\\ \text{H:}\\ \text{O:}\end{array}\left[\begin{array}{cccc}1& 0& 1& 0\\ 4& 0& 0& 2\\ 0& 2& 2& 1\end{array}\right]\left[\begin{array}{c}{\nu }_1\\ {\nu }_2\\ {\nu }_3\\ {\nu }_4\end{array}\right]=\mathbf{0}}$

Its solutions are of the following form, where r is any real number:

${\displaystyle \left[\begin{array}{c}{\nu }_1\\ {\nu }_2\\ {\nu }_3\\ {\nu }_4\end{array}\right]=\left[\begin{array}{c}-s_1\\ -s_2\\ s_3\\ s_4\end{array}\right]=r\left[\begin{array}{c}-1\\ -2\\ 1\\ 2\end{array}\right]}$

The choice of r = 1 and a sign-flip of the first two rows yields the preferred solution to the balancing problem:

${\displaystyle \left[\begin{array}{c}-{\nu }_1\\ -{\nu }_2\\ {\nu }_3\\ {\nu }_4\end{array}\right]=\left[\begin{array}{c}{s}_1\\ s_2\\ s_3\\ s_4\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 1\\ 2\end{array}\right]}$

Ionic equations

An ionic equation is a chemical equation in which electrolytes are written as dissociated ions. Ionic equations are used for single and double displacement reactions that occur in aqueous solutions.

For example, in the following precipitation reaction:

${\displaystyle {\text{CaCl}}_2^{}+2{\text{AgNO}}_3^{}⟶\text{Ca}{({\text{NO}}_3^{})}_2^{}+2\text{AgCl}\downarrow }$

the full ionic equation is:

${\displaystyle {\text{Ca}}^{2+}+2{\text{Cl}}^{-}+2{\text{Ag}}^{+}+2{\text{NO}}_3^{-}⟶{\text{Ca}}^{2+}+2{\text{NO}}_3^{-}+2\text{AgCl}\downarrow }$

or, with all physical states included:

${\displaystyle {\text{Ca}}^{2+}(\text{aq})+2{\text{Cl}}^{-}(\text{aq})+2{\text{Ag}}^{+}(\text{aq})+2{\text{NO}}_3^{-}(\text{aq})⟶{\text{Ca}}^{2+}(\text{aq})+2{\text{NO}}_3^{-}(\text{aq})+2\text{AgCl}\downarrow }$

In this reaction, the Ca²⁺ and the NO₃⁻ ions remain in solution and are not part of the reaction. That is, these ions are identical on both the reactant and product side of the chemical equation. Because such ions do not participate in the reaction, they are called spectator ions. A net ionic equation is the full ionic equation from which the spectator ions have been removed. The net ionic equation of the proceeding reactions is:

${\displaystyle 2{\text{Cl}}^{-}+2{\text{Ag}}^{+}⟶2\text{AgCl}\downarrow }$

or, in reduced balanced form,

${\displaystyle {\text{Ag}}^{+}+{\text{Cl}}^{-}⟶\text{AgCl}\downarrow }$

In a neutralization or acid/base reaction, the net ionic equation will usually be:

${\displaystyle {\text{H}}^{+}(\text{aq})+{\text{OH}}^{-}(\text{aq})⟶{\text{H}}_2^{}\text{O}(\text{l})}$

There are a few acid/base reactions that produce a precipitate in addition to the water molecule shown above. An example is the reaction of barium hydroxide with phosphoric acid, which produces not only water but also the insoluble salt barium phosphate. In this reaction, there are no spectator ions, so the net ionic equation is the same as the full ionic equation.

${\displaystyle 3\text{Ba}{(\text{OH})}_2^{}+2{\text{H}}_3^{}{\text{PO}}_4^{}⟶6{\text{H}}_2^{}\text{O}+{\text{Ba}}_3^{}{({\text{PO}}_4^{})}_2^{}\downarrow }$

${\displaystyle 3{\text{Ba}}^{2+}+6{\text{OH}}^{-}+6{\text{H}}^{+}+\underset{\text{phosphate}}{\underbrace{2{\text{PO}}_4^{3-}}}⟶6{\text{H}}_2^{}\text{O}+\underset{\text{barium}\text{phosphate}}{\overset{}{\underbrace{{\text{Ba}}_3^{}{({\text{PO}}_4^{})}_2^{}\downarrow }}}}$

Double displacement reactions that feature a carbonate reacting with an acid have the net ionic equation:

${\displaystyle 2{\text{H}}^{+}+\underset{\text{carbonate}}{\underbrace{{\text{CO}}_3^{2-}}}⟶{\text{H}}_2^{}\text{O}+{\text{CO}}_2^{}\uparrow }$

If every ion is a "spectator ion" then there was no reaction, and the net ionic equation is null.

Generally, if z_j is the multiple of elementary charge on the j-th molecule, charge neutrality may be written as:

${\displaystyle \sum _{j=1}^J{z}_j{\nu }_j=0}$

where the ν_j are the stoichiometric coefficients described above. The z_j may be incorporated as an additional row in the a_ij matrix described above, and a properly balanced ionic equation will then also obey:

${\displaystyle \sum _{j=1}^J{a}_{ij}{\nu }_j=0}$

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