3.1 Chemical Equations Chemical reactions are represented in a concise way by chemical equations. When hydrogen () burns, for example, it reacts with oxygen () in the air to form water () (Figure 3.2). We write the chemical equation for this reaction as follows: [3.1] Figure 3.2 Combustion of hydrogen gas. The gas is bubbled through a soap solution forming hydrogen-filled bubbles. As the bubbles float upwards, they are ignited by a candle on a long pole. The orange flame is due to the reaction of the hydrogen with oxygen in the air and results in the formation of water vapor. 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 chemical formulas on the right of the arrow represent substances produced in the reaction, called products. 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. On the right side of Equation 3.1, for example, there are two molecules of each composed of two atoms of hydrogen and one atom of oxygen. Thus, (read "two molecules of water") contains H atoms and O atoms as seen in the art shown in the margin. Because there are also 4 H atoms and 2 O atoms on the left side of the equation, the equation is balanced. Once we know the formulas of the reactants and products in a reaction, we can write the unbalanced 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. For most purposes a balanced equation should contain the smallest possible whole-number coefficients. 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 3.3. Notice that changing a subscript in a formula—from to for example—changes the identity of the chemical. The substance 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. Thus, means two molecules of water, means three molecules of water, and so forth. Figure 3.3 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 (), the principal component of natural gas, burns in air to produce carbon dioxide gas () and water vapor (). Both of these products contain oxygen atoms that come from in the air. We say that combustion in air is "supported by oxygen," meaning that oxygen is a reactant. The unbalanced equation is [3.2] 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 Let's consider first carbon and then hydrogen. One molecule of contains the same number of C atoms (one) as one molecule of The coefficients for these substances must be the same, therefore, and we choose them both to be 1 as we start the balancing process. However, the reactant contains more H atoms (four) than the product (two). If we place a coefficient 2 in front of there will be four H atoms on each side of the equation: [3.3] At this stage the products have more total O atoms (four—two from and two from ) than the reactants (two). If we place a coefficient 2 in front of we complete the balancing by making the number of O atoms equal on both sides of the equation: [3.4] The molecular view of the balanced equation is shown in Figure 3.4. Figure 3.4 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. Additional information is often added to the formulas in balanced equations to indicate the physical state of each reactant and product. We use the symbols (g), (l), (s), and (aq) for gas, liquid, solid, and aqueous (water) solution, respectively. Thus, Equation 3.4 can be written [3.5] Sometimes the conditions (such as temperature or pressure) under which the reaction proceeds appear above or below the reaction arrow. The symbol is often placed above the arrow to indicate the addition of heat. SAMPLE EXERCISE 3.1 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 This choice is a trial beginning, but it sets us on the correct path. Now that we have 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. PRACTICE EXERCISE Balance the following equations by providing the missing coefficients: (a) (b) (c) Answers: (a) 4, 3, 2; (b) 1, 3, 2, 2; (c) 2, 6, 2, 3 SAMPLE EXERCISE 3.2 (CONCEPTUAL) The following diagrams represent a chemical reaction in which the red spheres are oxygen atoms and the blue spheres are nitrogen atoms. (a) Write the chemical formulas for the reactants and products. (b) Write a balanced equation for the reaction. (c) Is the diagram consistent with the law of conservation of mass? Solution (a) The left box, which represents the reactants, contains two kinds of molecules, those composed of two oxygen atoms () and those composed of one nitrogen atom and one oxygen atom (NO). The right box, which represents the products, contains only molecules composed of one nitrogen atom and two oxygen atoms (). (b) The unbalanced chemical equation is In this equation, there are three O atoms on the left side of the arrow and two O atoms on the right side. We can increase the number of O atoms by placing a coefficient 2 on the product side: Now there are two N atoms and four O atoms on the right. Placing a coefficient 2 in front of NO brings both the N atoms and O atoms into balance: (c) The left box (reactants) contains four molecules and eight NO molecules. Thus, the molecular ratio is one for each two NO as required by the balanced equation. The right box (products) contains eight molecules. The number of molecules on the right equals the number of NO molecules on the left as the balanced equation requires. Counting the atoms, we find eight N atoms in the eight NO molecules in the box on the left. There are also O atoms in the molecules and eight O atoms in the NO molecules giving a total of 16 O atoms. In the box on the right, we find eight N atoms and O atoms in the eight molecules. Because there are equal numbers of both N and O atoms in the two boxes, the drawing is consistent with the law of conservation of mass. PRACTICE EXERCISE In order to be consistent with the law of conservation of mass, how many molecules should be shown in the right box of the following diagram: Answer: Six molecules