Chemical reactions are represented in a concise way by chemical equations. For example, when hydrogen, H2, burns, it reacts with oxygen, O2, in the air to form water, H2O. We write the chemical equation for this reaction as follows:
We read the + sign as "reacts with" and the arrow as "produces." The chemical formulas on the left of the arrow represent the starting substances, called reactants. The substances produced in the reaction, called products, are shown to the right of the arrow. The numbers in front of the formulas are coefficients. (As in algebraic equations, the numeral 1 is usually not written.)
Because atoms are neither created nor destroyed in any reaction, a chemical equation must have an equal number of atoms of each element on each side of the arrow. When this condition is met, the equation is said to be balanced. For example, on the right side of Equation 3.1, there are two molecules of H2O, each containing two atoms of hydrogen and one atom of oxygen. Thus, 2H2O (read "two molecules of H2O") contains 2 2 = 4 H atoms and 2 1 = 2 O atoms as seen in the art shown below. Because there are also 4 H atoms and 2 O atoms on the left side of the equation, the equation is balanced.
Before we can write the chemical equation for a reaction, we must determine by experiment those substances that are reactants and those that are products. Once we know the chemical formulas of the reactants and products in a reaction, we can write the unbalanced chemical equation. We then balance the equation by determining the coefficients that provide equal numbers of each type of atom on each side of the equation. Generally, whole-number coefficients are used.
In balancing equations, it is important to understand the difference between a coefficient in front of a formula and a subscript in a formula. Refer to Figure 3.2 . Notice that changing a subscript in a formula—from H2O to H2O2, for example—changes the identity of the chemical. The substance H2O2, hydrogen peroxide, is quite different from water. Subscripts should never be changed in balancing an equation. In contrast, placing a coefficient in front of a formula changes only the amount and not the identity of the substance; 2H2O means two molecules of water, 3H2O means three molecules of water, and so forth.
Figure 3.2 Illustration of the difference between a subscript in a chemical formula and a coefficient in front of the formula. Notice that the number of atoms of each type (listed under composition) is obtained by multiplying the coefficient and the subscript associated with each element in the formula.
To illustrate the process of balancing equations, consider the reaction that occurs when methane, CH4, the principal component of natural gas, burns in air to produce carbon dioxide gas, CO2, and water vapor, H2O. Both of these products contain oxygen atoms that come from O2 in the air. We say that combustion in air is "supported by oxygen," meaning that oxygen is a reactant. The unbalanced equation is
It is usually best to balance first those elements that occur in the fewest chemical formulas on each side of the equation. In our example both C and H appear in only one reactant and, separately, in one product each, so we begin by focusing attention on CH4. Let's consider first carbon and then hydrogen.
Notice that one molecule of CH4 contains the same number of C atoms (one) as does one molecule of CO2. Therefore, the coefficients for these substances must be the same, and we choose them both to be 1 as we start the balancing process. However, the reactant CH4 contains more H atoms (four) than does the product H2O (two). If we place a coefficient 2 in front of H2O, there will be four H atoms on each side of the equation:
At this stage the products have more total O atoms (four—two from CO2 and two from 2H2O) than the reactants (two). If we place a coefficient 2 in front of O2, we complete the balancing by making the number of O atoms equal on both sides of the equation:
The molecular view of the balanced equation is shown in Figure 3.3 . For most purposes a balanced equation should contain the smallest possible whole-number coefficients, as in this example.
Figure 3.3 Balanced chemical equation for the combustion of CH4. The drawings of the molecules involved call attention to the conservation of atoms through the reaction.
The approach we have taken to balancing Equation 3.4 is largely trial and error. We balance each kind of atom in succession, adjusting coefficients as necessary. This approach works for most chemical equations.
The physical state of each chemical in a chemical equation is often indicated parenthetically. We use the symbols (g), (l), (s), and (aq) for gas, liquid, solid, and aqueous (water) solution, respectively. Thus, the balanced equation above can be written
Sometimes the conditions under which the reaction proceeds appear above or below the arrow between the two sides of the equation. For example, the temperature or pressure at which the reaction occurs could be indicated. The symbol is often placed above the arrow to indicate the addition of heat.
Balance the following equation:
SOLUTION We begin by counting the atoms of each kind on both sides of the arrow. The Na and O atoms are balanced (one Na and one O on each side), but there are two H atoms on the left and three H atoms on the right. To increase the number of H atoms on the left, we place a coefficient 2 in front of H2O:
This choice is a trial beginning, but it sets us on the correct path. Now that we have 2H2O, we must regain the balance in O atoms. We can do so by moving to the other side of the equation and putting a coefficient 2 in front of NaOH:
This brings the H atoms into balance, but it requires that we move back to the left and put a coefficient 2 in front of Na to rebalance the Na atoms:
Finally, we check the number of atoms of each element and find that we have two Na atoms, four H atoms, and two O atoms on each side of the equation. The equation is balanced.
Balance the following equations by providing the missing coefficients:
(a) __C2H4 + __O2 __CO2 + __H2O
(b) __Al + __HCl __AlCl3 + __H2
Answers: (a) 1, 3, 2, 2; (b) 2, 6, 2, 3