Skip to main content

Chemical equations and reactions

A chemical reaction is represented by a chemical equation that involves ions, atoms and molecular formulas. Use this resource to will learn about the main components of a chemical equation, and how to write and balance chemical equations.

The main components of a chemical equation are:

  • Reactants – the starting materials of the reaction. Reactants are written on the left side of the equation.
  • Products – the materials obtained at the end of the reaction. Products are written on the right side of the equation.
  • An arrow – drawn between the reactants and products pointing towards the products to indicate a chemical change
  • Conditions required for the reaction to occus—such as pressure and temperature—written above the arrow
  • Physical state of the substances involved – written next to the chemical formulas using brackets and abbreviations
    • solid – (s)
    • liquid – (l)
    • gas – (g)
    • aqueous solution (a substance dissolved in water) – (aq)
  • Coefficients – the numbers written before the chemical formulas to balance the chemical equation.

According to the law of conservation of mass, matter cannot be created nor destroyed. None of the atoms disappear during a chemical reaction. Instead, atoms rearrange to form new chemical compounds. Therefore, the total number of atoms of each type on the reactant side of the arrow must be equal to that on the product side.

Worked example – writing balanced equations for chemical reactions

Write the balanced chemical equation for the reaction between nitrogen gas and hydrogen gas to produce ammonia gas.

Step 1: Write the reactants and products of the chemical equation using correct symbols and formulas. The reactants are written on the left side of the arrow and the products are written on the right. Use a plus (+) sign to separate each reactant or product from others.

\[\ce{N}_{2}+\ce{H}_{2}\ce{\rightarrow}\ce{NH}_{3}\]

Step 2: Balance the chemical equation by writing the correct coefficients in front of each chemical formula. In a balanced chemical equation, an equal number of atoms of each element involved must be present on either side of the arrow. Choose one atom to balance at a time. Let's start with the compound that is composed of the greatest number of atoms.

The compound with the greatest number of atoms is ammonia. The element in ammonia with the highest number of atoms is \(\ce{H}\). The reactant side contains \(3\textrm{ H}\) atoms. On the product side, \(\ce{H}\) contains \(2\textrm{ H}\) atoms.

To balance the number of \(\ce{H}\) atoms, \(6\textrm{ H}\) atoms are needed on either side, as \(6\) is the least common multiple. To achieve \(6\textrm{ H}\) atoms on either side, we need \(2\ce{NH3}\) molecules and \(3\ce{H2}\) molecules.

\[\ce{N}_{2}+3\ce{H}_{2}\rightarrow2\ce{NH}_{3}\]

Step 3: Repeat step 2 with each atom, as required.

The next atom will be nitrogen. The reactant side has \(2\textrm{ N}\) atoms and the product side also has \(2\textrm{ N}\) atoms. Therefore, nitrogen is already balanced.

Step 4: Do a final check to make sure all types of atoms in the equation are balanced.

\[\ce{N}_{2}+3\ce{H}_{2}\rightarrow2\ce{NH}_{3}\]

On the reactant side, there are \(2\ce{N}\) atoms and \(6\ce{H}\) atoms. On the product side, there are \(2\ce{N}\) atoms and \(6\ce{H}\) atoms. The equation is balanced. This is summarised in the table.

Element Reactant side Product side
\(\ce{N}\) \(2\) \(2\)
\(\ce{H}\) \(3\times2=6\) \(2\times3=6\)

Step 5: Check whether the coefficients are in their lowest possible whole numbers.

\[1\ce{N}_{2}+3\ce{H}_{2}\rightarrow2\ce{NH}_{3}\]
The ratio of \(1:3:2\) is the lowest possible in whole numbers. In the final equation, any coefficients of \(1\) do not need to be written. That is, \(1\ce{N2}\) is simply \(\ce{N2}\).

Step 6: Add the states for each compound.

\[\ce{N}_{2}(\textrm{g})+3\ce{H}_{2}(\textrm{g})\rightarrow2\ce{NH}_{3}(\textrm{g})\]

The unbalanced equation for the combustion of liquid ethanol is:
\[\ce{C}_{2}\ce{H}_{6}\ce{O}+\ce{O}_{2}\rightarrow\ce{CO}_{2}+\ce{H}_{2}\ce{O}\]

Step 1: Write the reactants and products of the chemical equation using correct symbols and formulas. The reactants are written on the left side of the arrow and the products are written on the right. Use a plus (+) sign to separate each reactant or product from others.

The equation is given, so you can move on to the next step.

Step 2: Balance the chemical equation by writing the correct coefficients in front of each chemical formula. Start with the compound that is composed of the greatest number of atoms.

The compound with the greatest number of atoms is hydrogen, \(\ce{H2}\). The reactant side contains \(6\textrm{ H}\) atoms and the product side has \(2\textrm{ H}\) atoms.

To balance the number of \(\ce{H}\) atoms, \(6\textrm{ H}\) atoms are needed on either side, as \(6\) is the least common multiple. To achieve \(6\textrm{ H}\) atoms on either side, we need \(1\ce{C2H6O}\) molecules and \(3\ce{H2O}\) molecules.

\[1\ce{C}_{2}\ce{H}_{6}\ce{O}+\ce{O}_{2}\rightarrow\ce{CO}_{2}+\ce{3H}_{2}\ce{O}\]

Step 3: Repeat step 2 with each atom, as required.

  • The next atom will be carbon. The reactant side has \(2\textrm{ C}\) atoms and the product side also has \(1\textrm{ C}\) atom. To balance the number of \(\ce{C}\) atoms, \(2\textrm{ C}\) atoms are need on either side. To achieve this, we need \(2\ce{CO2}\) molecules.
    \[1\ce{C}_{2}\ce{H}_{6}\ce{O}+\ce{O}_{2}\rightarrow\ce{2CO}_{2}+\ce{3H}_{2}\ce{O}\]
  • Finally, balance the oxygen atoms. The reactant side has \(3\textrm{ O}\) atoms, and the product side has \(7\textrm{ O}\) atoms. Placing \(3\) in front of \(\ce{O2}\) on the reactant side gives \(7\textrm{ O}\) atoms on each side.
    \[1\ce{C}_{2}\ce{H}_{6}\ce{O}+\ce{3O}_{2}\rightarrow\ce{2CO}_{2}+\ce{3H}_{2}\ce{O}\]

Step 4: Do a final check to make sure all types of atoms in the equation are balanced.

\[\ce{N}_{2}+3\ce{H}_{2}\rightarrow2\ce{NH}_{3}\]

Element Left side Right side
\(\ce{H}\) \(6\) \(3\times2=6\)
\(\ce{C}\) \(2\) \(2\)
\(\ce{O}\) \(1+\left(3\times2\right)=7\) \(\left(2\times2\right)+3=7\)

Step 5: Check whether the coefficients are in their lowest possible whole numbers.

The ratio of \(1:3:2:3\) is the lowest possible in whole numbers. In the final equation, any coefficients of \(1\) do not need to be written. That is, \(1\ce{C2H6O}\) is simply \(\ce{C2H6O}\).

\[\ce{C}_{2}\ce{H}_{6}\ce{O}+\ce{3O}_{2}\rightarrow\ce{2CO}_{2}+\ce{3H}_{2}\ce{O}\]

Step 6: Add the states for each compound.

\[\ce{C}_{2}\ce{H}_{6}\ce{O}(\textrm{l})+\ce{3O}_{2}(\textrm{g})\rightarrow\ce{2CO}_{2}(\textrm{g})+\ce{3H}_{2}\ce{O}(\textrm{l})\]

Write a balanced chemical equation for the decomposition of liquid hydrogen peroxide in the presence of light to water and oxygen gas.

Step 1: Write the reactants and products of the chemical equation using correct symbols and formulas. The reactants are written on the left side of the arrow and the products are written on the right. Use a plus (+) sign to separate each reactant or product from others.

\[\ce{H}_{2}\ce{O}_{2}\rightarrow\ce{H}_{2}\ce{O}+\ce{O}_{2}\]

Step 2: Balance the chemical equation by writing the correct coefficients in front of each chemical formula. Start with the compound that is composed of the greatest number of atoms.

The compound with the greatest number of atoms is hydrogen peroxide, \(\ce{H2O2}\). The reactant side contains \(2\textrm{ H}\) atoms and the product side has \(2\textrm{ H}\) atoms; therefore, \(\ce{H}\) is already balanced.

\[\ce{H}_{2}\ce{O}_{2}\rightarrow\ce{H}_{2}\ce{O}+\ce{O}_{2}\]

Step 3: Repeat step 2 with each atom, as required.

There are \(2\textrm{ O}\) atoms on the reactant side and \(3\textrm{ O}\) atoms on the product side. The number of oxygen atoms can be balanced by placing a \(\frac{1}{2}\) in front of \(\ce{O2}\). This gives \(2\textrm{ O}\) atoms on each side.

\[\ce{H}_{2}\ce{O}_{2}\rightarrow\ce{H}_{2}\ce{O}+\frac{1}{2}\ce{O}_{2}\]

Step 4: Do a final check to make sure all types of atoms in the equation are balanced.

\[\ce{H}_{2}\ce{O}_{2}\rightarrow\ce{H}_{2}\ce{O}+\frac{1}{2}\ce{O}_{2}\]

Element Left side Right side
\(\ce{H}\) \(2\) \(2\)
\(\ce{O}\) \(2\) \(2\)

Step 5: Check whether the coefficients are in their lowest possible whole numbers.

Except for \(\ce{O2}\), the compounds have whole number coefficients. To make all of the coefficients their smallest whole number, multiply all coefficients by \(2\).
\[\ce{2H}_{2}\ce{O}_{2}\rightarrow\ce{2H}_{2}\ce{O}+\ce{O}_{2}\]

Step 6: Add the states for each compound.

\[\ce{2H}_{2}\ce{O}_{2}(\textrm{l})\rightarrow\ce{2H}_{2}\ce{O}(\textrm{g})+\ce{O}_{2}(\textrm{g})\]