A **chemical equation** is a symbolic representation of a chemical reaction, wherein one set of substances, called the *reactants*, is converted into another set of substances, called the *products*.^{[1]} The reactants and products are shown using their chemical formulas, and an arrow is used to indicate the direction of the reaction. The reactants are usually placed to the left of the arrow, and the products are placed to the right. If the reaction is irreversible, a single arrow is used; if the reaction is reversible, a double arrow (pointing in opposite directions) is used.

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The first chemical equation was diagrammed by Jean Beguin in 1615.

The combustion of methane in oxygen may be shown by the following equation:

- CH
_{4}+ 2 O_{2}→ CO_{2}+ 2 H_{2}O

This equation represents an irreversible reaction in which one molecule of methane reacts with two molecules of oxygen to produce one molecule of carbon dioxide and two molecules of water.

The reaction of sodium metal with oxygen produces sodium oxide, as follows:

- 4Na + O
_{2}→ 2Na_{2}O

In the above equation, four sodium atoms react with one oxygen molecule to produce two molecules of sodium oxide.

The synthesis of ammonia from nitrogen and hydrogen by the Haber process is a reversible reaction:

- N
_{2}(g) + 3H_{2}(g) ↔ 2NH_{3}(g)

The parenthetical "g" after a substance indicates that that substance is in the gaseous form. Likewise, a parenthetical "s" would indicate a solid form, "l" would indicate a liquid form, and "aq" would indicate that the material is in aqueous solution.

When reading a chemical equation, several points need to be considered:

- Each side of the equation represents a mixture of chemicals. The mixture is written as a set of chemical formulas of the atoms and molecules involved in the reaction, separated by
**+**symbols.

- The two sides of the equation are separated by an arrow. If the reaction is irreversible, a right-arrow (→) is used, indicating that the left side represents the reactants (mixture of chemicals before the reaction) and the right side represents the products (mixture obtained after the reaction). For a reversible reaction, a two-way arrow is used.

- The formula of each reactant and product is normally preceded by a scalar number called the
**stoichiometric number**or**stoichiometric coefficient**. (The absence of a scalar number implies that the number is 1.) The stoichiometric numbers indicate the relative quantities of the molecules (or moles) participating in the reaction. For instance, the string 2H_{2}O + 3CH_{4}represents a mixture containing two molecules of H_{2}O for every three molecules of CH_{4}.

- A chemical equation does not imply that all reactants are consumed in a chemical process. For instance, a limiting reactant determines how far a reaction can go.

- In a chemical reaction, the quantity of each element does not change. Thus, each side of the equation must represent the same quantity of any particular element. In other words, the number of atoms of a given element in the products must equal the number of atoms of that element in the reactants. This is known as the "conservation of mass" in a chemical reaction. The process of equalizing these numbers in a chemical equation is known as "balancing the equation."

- Some equations contain the term "ΔH" placed on the right side. This indicates that there is a change of enthalpy (or "heat content") during the reaction. If the reaction releases heat, ΔH has a negative value; if the reaction absorbs heat, ΔH has a positive value.

One may balance a chemical equation by changing the scalar number for each chemical formula in the equation. Also, in case of net ionic reactions, the same charge must be present on both sides of the unbalanced equation.

Simple chemical equations can be balanced by inspection, that is, by trial and error. Generally, it is best to balance the elements in the most complicated molecule first. Hydrogen, oxygen, and elements that occur in the form of single atoms are usually balanced toward the end.

**Example 1:** Na + O_{2} → Na_{2}O

For this equation to be balanced, there must be an equal number of Na atoms and O atoms on the left hand side as there are on the right hand side. Since Na occurs as a single atom on the left whereas O occurs only in molecules on both sides of the equation, it would be better to start by balancing the O atoms.

As it stands now, there are 2 O atoms on the left but only 1 on the right. To balance the O atoms, one places a 2 in front of the O on the right side:

- Na + O
_{2}→ 2Na_{2}O

In the next step, one needs to balance the Na atoms. There is a single Na atom on the left side but 4 Na atoms on the right. To balance the Na atoms, one places a 4 in front of the Na atom on the left side. This process leads to the following equation:

- 4Na + O
_{2}→ 2Na_{2}O

This equation is a balanced equation because there is an equal number of atoms of each element on the left and right hand sides.

**Example 2:** P_{4} + O_{2} → P_{2}O_{5}

In this case, one may start by balancing the P atoms. The left side has 4 P atoms, but the right side has 2 P atoms. To balance the P atoms, one may place a 2 in front of P_{2}O_{5}. This step leads to the following equation:

- P
_{4}+ O_{2}→ 2P_{2}O_{5}

Now the left side has 2 O's and the right side has 10 O's. To fix this unbalanced equation, one may place a 5 in front of the O_{2} on the left side, giving 10 O atoms on both sides of the equation:

- P
_{4}+ 5O_{2}→ 2P_{2}O_{5}

Each element now has equal numbers of atoms on the two sides of the equation. Consequently, the equation is balanced.

**Example 3:** C_{2}H_{5}OH + O_{2} → CO_{2} + H_{2}O

This equation is more complex than the earlier examples and requires more steps. The most complicated molecule here is C_{2}H_{5}OH, so balancing begins by placing the coefficient 2 before the CO_{2} to balance the carbon atoms.

- C
_{2}H_{5}OH + O_{2}→ 2CO_{2}+ H_{2}O

Since C_{2}H_{5}OH contains 6 hydrogen atoms, the hydrogen atoms can be balanced by placing 3 before the H_{2}O:

- C
_{2}H_{5}OH + O_{2}→ 2CO_{2}+ 3H_{2}O

Finally the oxygen atoms must be balanced. Since there are 7 oxygen atoms on the right and only 3 on the left, a 3 is placed before O_{2}, to produce the balanced equation:

- C
_{2}H_{5}OH + 3O_{2}→ 2CO_{2}+ 3H_{2}O

In reactions involving many compounds, equations may be balanced using an algebraic method, based on solving sets of linear equations.

1. Assign variables to each coefficient. (Coefficients represent both the basic unit and mole ratios in balanced equations.):

**a K**_{4}Fe(CN)_{6}+ b H_{2}SO_{4}+ c H_{2}O → d K_{2}SO_{4}+ e FeSO_{4}+ f (NH_{4})_{2}SO_{4}+ g CO

2. There must be the same quantities of each atom on each side of the equation. So, for each element, count its atoms and let both sides be equal.

**K: 4a = 2d****Fe: 1a = 1e****C: 6a = g****N: 3a = f****H: 2b+2c = 8f****S: b = d+e+f****O: 4b+c = 4d+4e+4f+g**

3. Solve the system (Direct substitution is usually the best way.)

**d=2a****e=a****g=6a****f=3a****b=6a****c=6a**

which means that all coefficients depend on a parameter a, just choose a=1 (a number that will make all of them small whole numbers), which gives:

**a=1 b=6 c=6 d=2 e=1 f=3 g=6**

4. And the balanced equation at last:

**K**_{4}Fe(CN)_{6}+ 6 H_{2}SO_{4}+ 6 H_{2}O → 2 K_{2}SO_{4}+ FeSO_{4}+ 3 (NH_{4})_{2}SO_{4}+ 6 CO

To speed up the process, one can combine both methods to get a more practical algorithm:

1. Identify elements which occur in one compound in each member. (This is very usual.)

2. Start with the one among those which has a big index (this will help to keep working with integers), and assign a variable, such as a.

**a K**_{4}Fe(CN)_{6}+ H_{2}SO_{4}+ H_{2}O → K_{2}SO_{4}+ FeSO_{4}+ (NH_{4})_{2}SO_{4}+ CO

3. K_{2}SO_{4} has to be 2a (because of K), and also, FeSO_{4} has to be 1a (because of Fe), CO has to be 6a (because of C) and (NH_{4})_{2}SO_{4} has to be 3a (because of N). This removes the first four equations of the system. It is already known that whatever the coefficients are, those proportions must hold:

**a K**_{4}Fe(CN)_{6}+ H_{2}SO_{4}+ H_{2}O → 2a K_{2}SO_{4}+ a FeSO_{4}+ 3a (NH_{4})_{2}SO_{4}+ 6a CO

4. One can continue by writing the equations now (and having simpler problem to solve) or, in this particular case (although not so particular) one could continue by noticing that adding the Sulfurs yields 6a for H_{2}SO_{4} and finally by adding the hydrogens (or the oxygens) one can find the lasting 6a for H_{2}SO_{4}.

5. Again, having a convenient value for a (in this case 1 will do, but if a results in fractional values in the other coefficients, one would like to cancel the denominators) The result is

**K**_{4}Fe(CN)_{6}+ 6 H_{2}SO_{4}+ 6 H_{2}O → 2 K_{2}SO_{4}+ FeSO_{4}+ 3 (NH_{4})_{2}SO_{4}+ 6 CO

- ↑ Chemical reaction equation IUPAC Gold Book. Retrieved September 24, 2008.

- Chang, Raymond. 2006.
*Chemistry*, 9th ed. New York: McGraw-Hill Science/Engineering/Math. ISBN 0073221031 - Cotton, F. Albert, and Geoffrey Wilkinson. 1980.
*Advanced Inorganic Chemistry*, 4th ed. New York: Wiley. ISBN 0471027758 - McMurry, J., and R.C. Fay. 2004.
*Chemistry*, 4th ed. Upper Saddle River, NJ: Prentice Hall. ISBN 0131402080

All links retrieved February 8, 2017.

- Classic Chembalancer FunBased Learning. (A free, online game to learn how to balance equations by inspection.)
- Stoichiometric equations ScienceSoft.at. (Online calculator to determine the coefficients of a chemical equation.)
- Balancing chemical equation using Equation Balancing and Stoichiometry calculator ChemBuddy. (Works off-line, calculates stoichiometry and limiting reagents, balances charge.)
- Balance Chemical Equation - Online Balancer (Balances equation of any chemical reaction, full or half-cell, in one click.)

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