1 L The use of this conversion factor is illustrated in Figure 3.6, and the quantities of
4. Choose coefficients for the other substances so that the numbers of atoms of each ele-
O
N F
H
N2 O2 F2
Br2
I2 Cl2 H2
Cl Br I
FIGURE 3.18 The diatomic elements are hydrogen, nitrogen, oxygen, and the group 17 halogens. The halogen astatine (At) at the bottom of group 17 is the rarest terrestrial element, and virtually none of its bulk physical properties are known.
3. 3 Writing Balanced Chemical Equations 99 We can balance the H atoms by placing a coefficient of 4 in front of H2, which
finishes the balancing process:
CH4(g) 1 __2 H2O(g) S CO2(g) 1 __4 H2(g)
+ +
element reactant Side product Side Balanced?
C 1 1 ✔
O 2 3 1 5 2 2 ✔
H 4 1 (2 3 2) 5 8 4 3 2 5 8 ✔
SAmpLe eXerCiSe 3.9 Writing a Balanced chemical equation Lo2 Write a balanced chemical equation for the gas-phase reaction between SO2 and O2 that forms SO3. This reaction, one of the causes of acid rain, occurs when fuels containing sulfur are burned. Acid rain is harmful to aquatic life and damages terrestrial plants.
Collect, Organize, and Analyze We are given the chemical formulas and physical states of the reactants and products. We need to write a balanced chemical equation such that we have the same number of atoms of each element on both sides of the equation. To balance the equation, we assign coefficients as needed to make the numbers of atoms of each element on the reactant (left) and product (right) sides of the reaction arrow equal.
Solve
1. Write a reaction expression based on single molecules of the reactants and product:
SO2(g) 1 O2(g) S SO3(g) 2. A check of whether the expression is balanced:
element reactant Side product Side Balanced?
S 1 1 ✔
O 2 1 2 5 4 3 ✗
discloses that it is not.
3. Sulfur is balanced, so we move on to oxygen.
4. Balancing the O atoms presents a challenge because increasing their number on the product side by increasing the SO3 coefficient from 1 to 2:
SO2(g) 1 O2(g) S __2 SO3(g)
means that there are now two more O atoms and one more S atom on the product side than on the reactant side. Fortunately, both of these imbalances can be solved simultaneously by increasing the SO2 coefficient from 1 to 2:
__2 SO2(g) 1 O2(g) S __2 SO3(g)
coNcePt teSt
Why can’t we write a balanced equation for the reaction between SO2 and O2 in Sample Exercise 3.9 this way?
SO2(g) 1 O(g) S SO3(g)
(Answers to Concept Tests are in the back of the book.)
Note that a chemical equation is analogous to a mathematical equation. The equation in Sample Exercise 3.9 is still balanced if we multiply all the coefficients by the same number, such as 2:
4 SO2(g) 1 2 O2(g) S 4 SO3(g)
It is also still balanced if we multiply all the coefficients by a fraction, such as 12. However, it is conventional to report a balanced equation with the smallest whole-number coefficients.
SAmpLe eXerCiSe 3.10 Balancing chemical equations using Fractional coefficients
Lo2
Write a balanced chemical equation for the gas-phase reaction in which dinitrogen pentoxide is formed from nitrogen and oxygen.
Collect, Organize, and Analyze We are asked to write a balanced chemical equation given the names and physical states of the reactants and products. Nitrogen and oxygen A check of the atom inventory on both sides confirms that the equation is now
balanced:
element reactant Side product Side Balanced?
S 2 3 1 5 2 2 3 1 5 2 ✔
O (2 3 2) 1 2 5 6 2 3 3 5 6 ✔
Think About It Instead of increasing the number of O atoms on the product side to bring the equation into balance, we might have tried decreasing their number on the reactant side by reducing the coefficient in front of O2 from 1 to 12:
SO2(g) 1 12O2(g) S SO3(g)
This fix has the advantage of bringing the number of O atoms into balance without imbalancing the number of S atoms. It has the disadvantage of creating a fractional coefficient, but this problem can be solved by multiplying all of the coefficients by 2, which gives us the same balanced equation we obtained in the Solve section:
2 SO2(g) 1 O2(g) S 2 SO3(g)
We will use this strategy again in Sample Exercise 3.10 and in writing chemical equations for some of the combustion reactions in Section 3.4.
d Practice Exercise Balance the chemical equations for (a) the reaction between elemental phosphorus, P4(s), and oxygen to make P4O10(s) and (b) the reaction of water with P4O10(s) to produce phosphoric acid, H3PO4(aq).
(Answers to Practice Exercises are in the back of the book.)
3.4 Combustion Reactions 101
3.4 Combustion Reactions
Both methane and carbon monoxide are still present in our atmosphere, although at much lower concentrations than on early Earth. Atmospheric methane (natural gas) can be traced to several sources, including wetlands, cattle ranching, rice production, and oil drilling and fracking operations. Carbon dioxide in the atmo- sphere arises from burning carbon-containing compounds. Natural sources of CO2 include volcanic activity and forest fires. In our industrialized world, the greatest use of methane and petroleum-based fuels is energy production.
C nneCtion Molecules like methane (CH4) are members of a class of organic compounds known as hydrocarbons because they are composed of only hydrogen and carbon, as we saw in Chapter 2.
gases exist in nature as diatomic molecules, so their molecular formulas are N2 and O2. The formula for dinitrogen pentoxide is N2O5. To write a balanced chemical equation, we must assign coefficients as needed to make the numbers of atoms of each element on the reactant (left) and product (right) sides of the reaction arrow equal.
Solve
1. Write an equation using correct formulas:
N2(g) 1 O2(g) S N2O5(g) 2. The equation is not balanced as written:
element reactant Side product Side Balanced?
N 2 2 ✔
O 2 5 ✗
3. Two N atoms appear on each side of the equation. Oxygen is the only other element to balance in the equation. With 2 O atoms on the reactant side and 5 O atoms on the product side, there is no whole-number coefficient that we can place in front of the O2 to make the number of O atoms equal on both sides of the equation. A coefficient of 52 in front of O2 would balance the number of O atoms, but we need to use whole numbers. Because 52 times 2 is the whole number 5, we can place a coefficient of 5 in front of O2 and a coefficient of 2 in front of N2O5:
__ N2(g) 1 __5 O2(g) S __2 N2O5(g)
However, the number of nitrogen atoms is no longer equal on both sides of the equation:
element reactant Side product Side Balanced?
N 2 2 3 2 5 4 ✗
O 5 3 2 5 10 2 3 5 5 10 ✔
To balance the number of nitrogen atoms, we add a coefficient of 2 in front of N2: __2 N2(g) 1 __5 O2(g) S __2 N2O5(g)
The equation is now balanced.
Think About It In balancing a chemical equation, always check the equation after you are done to make sure that an element balanced in a previous step is still balanced at the end.
d Practice Exercise Balance the chemical equation for the reaction between carbon monoxide, CO(g), and oxygen to form carbon dioxide, CO2(g).
(Answers to Practice Exercises are in the back of the book.)
Natural gas and other hydrocarbon fuels are burned to heat buildings and to warm water. Burning CH4 is an example of an important class of chemical reac- tions known as combustion reactions, or simply combustion (Figure 3.19). Com- bustion refers to the reaction of a fuel with oxygen, such as the burning of a substance in air, which contains 21% oxygen.
When the combustion of a hydrocarbon is complete, the only products are carbon dioxide and water. In the presence of O2, the more stable form of carbon is CO2, not CO, so carbon monoxide also reacts with oxygen to form carbon dioxide. In the combustion reaction between methane and oxygen, the carbon becomes carbon dioxide when a sufficient supply of oxygen is available. The hydrogen in the fuel combines with oxygen to form water vapor. Let’s see how to balance the equation describing the combustion of methane.