Chemistry Essentials for Dummies
Chapter 7. Chemical Reactions
Balancing Chemical Equations
If you carry out a chemical reaction and carefully sum up the masses of all the reactants, and then you compare the sum to the sum of the masses of all the products, you see that they’re the same. In fact, a law in chemistry, the law of conservation of mass, states, “In an ordinary chemical reaction, matter is neither created nor destroyed.” This means that you neither gain nor lose any atoms during the reaction. They may be combined differently, but they’re still there.
A chemical equation represents the reaction, and that chemical equation needs to obey the law of conservation of mass. You use that chemical equation to calculate how much of each element you need and how much of each element will be produced. You need to have the same number of each kind of element on both sides of the equation. The equation should balance.
Before you start balancing an equation, you need to know the reactants and the products for that reaction. You can’t change the compounds, and you can’t change the subscripts, because that would change the compounds. So the only thing you can do to balance the equation is put in coefficients, whole numbers in front of the compounds or elements in the equation.
Coefficients tell you how many atoms or molecules you have. For example, if you write 2H2O, it means you have two water molecules:
Each water molecule is composed of two hydrogen atoms and one oxygen atom. So with 2H2O, you have a total of four hydrogen atoms and two oxygen atoms.
In this section, I show you how to balance equations using a method called balancing by inspection (or as I call it, “fiddling with coefficients”). You take each atom in turn and balance it by inserting appropriate coefficients on one side or the other. You can balance most simple reactions in this fashion, but one class of reactions is so complex that this method doesn’t work well for them: redox reactions. I show you a special method for balancing those equations in Chapter 8.
Balancing the Haber process
My favorite reaction is the Haber process, a method for preparing ammonia (NH3) by reacting nitrogen gas with hydrogen gas:
This equation shows you what happens in the reaction, but it doesn’t show you how much of each element you need to produce the ammonia. To find out how much of each element you need, you have to balance the equation — make sure that the number of atoms on the left side of the equation equals the number of atoms on the right. You can’t change the subscripts, so you have to put in some coefficients.
TIP. In most cases, waiting until the end to balance hydrogen atoms and oxygen atoms is a good idea; balance the other atoms first.
So in this example, you need to balance the nitrogen atoms first. You have two nitrogen atoms on the left side of the arrow (reactant side) and only one nitrogen atom on the right side (product side). To balance the nitrogen atoms, use a coefficient of 2 in front of the ammonia on the right. Now you have two nitrogen atoms on the left and two nitrogen atoms on the right:
Next, tackle the hydrogen atoms. You have two hydrogen atoms on the left and six hydrogen atoms on the right (two NH3 molecules, each with three hydrogen atoms, for a total of six hydrogen atoms). So put a 3 in front of the H2 on the left, giving you the following:
That should do it. Do a check to be sure: You have two nitrogen atoms on the left and two nitrogen atoms on the right. You have six hydrogen atoms on the left (3 x 2 = 6) and six hydrogen atoms on the right (2 x 3 = 6). The equation is balanced. You can read the equation this way: one nitrogen molecule reacts with three hydrogen molecules to yield two ammonia molecules.
Here’s a tidbit for you: This equation would also balance with coefficients of 2, 6, and 4 instead of 1, 3, and 2. In fact, any multiple of 1, 3, and 2 would balance the equation, but chemists have agreed always to show the lowest whole-number ratio (see the discussion of empirical formulas in Chapter 6 for details).
Balancing the burning of butane
Take a look at an equation showing the burning of butane, a hydrocarbon, with excess oxygen available. (This is the reaction that takes place when you light a butane lighter.) The unbalanced reaction is
Because waiting until the end to balance hydrogen atoms and oxygen atoms is always a good idea, balance the carbon atoms first. You have four carbon atoms on the left and one carbon atom on the right, so add a coefficient of 4 in front of the carbon dioxide:
Balance the hydrogen atoms next. You have ten hydrogen atoms on the left and two hydrogen atoms on the right, so use a coefficient of 5 in front of the water on the right:
Now work on balancing the oxygen atoms. You have two oxygen atoms on the left and a total of thirteen oxygen atoms on the right [(4 x 2) + (5 x 1) = 13]. What can you multiply 2 by to equal 13? How about 6.5?
But you’re not done. You want the lowest whole-number ratio of coefficients. Multiply the entire equation by 2 to generate whole numbers:
Multiply every coefficient by 2 (don’t touch the subscripts!) to get
If you check the atom count on both sides of the equation, you find that the equation is balanced, and the coefficients are in the lowest whole-number ratio.
TIP. After balancing an equation, make sure that the same number of each atom is on both sides and that the coefficients are in the lowest whole-number ratio.