4.1 Writing and balancing chemical equations (Page 2/14)

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Balancing equations

The chemical equation described in section 4.1 is balanced , meaning that equal numbers of atoms for each element involved in the reaction are represented on the reactant and product sides. This is a requirement the equation must satisfy to be consistent with the law of conservation of matter. It may be confirmed by simply summing the numbers of atoms on either side of the arrow and comparing these sums to ensure they are equal. Note that the number of atoms for a given element is calculated by multiplying the coefficient of any formula containing that element by the element’s subscript in the formula. If an element appears in more than one formula on a given side of the equation, the number of atoms represented in each must be computed and then added together. For example, both product species in the example reaction, CO ₂ and H ₂ O, contain the element oxygen, and so the number of oxygen atoms on the product side of the equation is

(1 {CO}_{2} molecule \times \frac{2 O atoms}{{CO}_{2} molecule}) + (2 H_{2} O molecule \times \frac{1 O atom}{H_{2} O molecule}) = 4 O atoms

The equation for the reaction between methane and oxygen to yield carbon dioxide and water is confirmed to be balanced per this approach, as shown here:

{CH}_{4} + 2 O_{2} ⟶ {CO}_{2} + 2 H_{2} O

Element	Reactants	Products	Balanced?
C	1 $\times$ 1 = 1	1 $\times$ 1 = 1	1 = 1, yes
H	4 $\times$ 1 = 4	2 $\times$ 2 = 4	4 = 4, yes
O	2 $\times$ 2 = 4	(1 $\times$ 2) + (2 $\times$ 1) = 4	4 = 4, yes

A balanced chemical equation often may be derived from a qualitative description of some chemical reaction by a fairly simple approach known as balancing by inspection. Consider as an example the decomposition of water to yield molecular hydrogen and oxygen. This process is represented qualitatively by an unbalanced chemical equation:

H_{2} O ⟶ H_{2} + O_{2} (unbalanced)

Comparing the number of H and O atoms on either side of this equation confirms its imbalance:

Element	Reactants	Products	Balanced?
H	1 $\times$ 2 = 2	1 $\times$ 2 = 2	2 = 2, yes
O	1 $\times$ 1 = 1	1 $\times$ 2 = 2	1 ≠ 2, no

The numbers of H atoms on the reactant and product sides of the equation are equal, but the numbers of O atoms are not. To achieve balance, the coefficients of the equation may be changed as needed. Keep in mind, of course, that the formula subscripts define, in part, the identity of the substance, and so these cannot be changed without altering the qualitative meaning of the equation. For example, changing the reactant formula from H ₂ O to H ₂ O ₂ would yield balance in the number of atoms, but doing so also changes the reactant’s identity (it’s now hydrogen peroxide and not water). The O atom balance may be achieved by changing the coefficient for H ₂ O to 2.

2 H_{2} O ⟶ H_{2} + O_{2} (unbalanced)

Element	Reactants	Products	Balanced?
H	2 $\times$ 2 = 4	1 $\times$ 2 = 2	4 ≠ 2, no
O	2 $\times$ 1 = 2	1 $\times$ 2 = 2	2 = 2, yes

The H atom balance was upset by this change, but it is easily reestablished by changing the coefficient for the H ₂ product to 2.

2 H_{2} O ⟶ 2 H_{2} + O_{2} (balanced)

Element	Reactants	Products	Balanced?
H	2 $\times$ 2 = 4	2 $\times$ 2 = 4	4 = 4, yes
O	2 $\times$ 1 = 2	1 $\times$ 2 = 2	2 = 2, yes

These coefficients yield equal numbers of both H and O atoms on the reactant and product sides, and the balanced equation is, therefore:

2 H_{2} O ⟶ 2 H_{2} + O_{2}

Balancing chemical equations

Write a balanced equation for the reaction of molecular nitrogen (N ₂ ) and oxygen (O ₂ ) to form dinitrogen pentoxide.

Solution

First, write the unbalanced equation.

N_{2} + O_{2} ⟶ N_{2} O_{5} (unbalanced)

Next, count the number of each type of atom present in the unbalanced equation.

Element	Reactants	Products	Balanced?
N	1 $\times$ 2 = 2	1 $\times$ 2 = 2	2 = 2, yes
O	1 $\times$ 2 = 2	1 $\times$ 5 = 5	2 ≠ 5, no

Though nitrogen is balanced, changes in coefficients are needed to balance the number of oxygen atoms. To balance the number of oxygen atoms, a reasonable first attempt would be to change the coefficients for the O ₂ and N ₂ O ₅ to integers that will yield 10 O atoms (the least common multiple for the O atom subscripts in these two formulas).

N_{2} + 5 O_{2} ⟶ 2 N_{2} O_{5} (unbalanced)

Element	Reactants	Products	Balanced?
N	1 $\times$ 2 = 2	2 $\times$ 2 = 4	2 ≠ 4, no
O	5 $\times$ 2 = 10	2 $\times$ 5 = 10	10 = 10, yes

The N atom balance has been upset by this change; it is restored by changing the coefficient for the reactant N ₂ to 2.

2 N_{2} + 5 O_{2} ⟶ 2 N_{2} O_{5}

Element	Reactants	Products	Balanced?
N	2 $\times$ 2 = 4	2 $\times$ 2 = 4	4 = 4, yes
O	5 $\times$ 2 = 10	2 $\times$ 5 = 10	10 = 10, yes

The numbers of N and O atoms on either side of the equation are now equal, and so the equation is balanced.

Check your learning

Write a balanced equation for the decomposition of ammonium nitrate to form molecular nitrogen, molecular oxygen, and water. (Hint: Balance oxygen last, since it is present in more than one molecule on the right side of the equation.)

Answer:

2 {NH}_{4} {NO}_{3} ⟶ 2 N_{2} + O_{2} + 4 H_{2} O

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Source: OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9

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