Advanced Placement Chemistry

Lesson Plan #5

1 class period

Chemical Reactions and Patterns of Reactivity

• 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: 2H₂ + O₂ ß 2H₂O
• 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.
• In balancing equations, it is important to understand the difference between a coefficient in front of a formula and a subscript in a formula. Notice that changing a subscript in a formula - from H₂O to H₂O₂, for example - changes the identity of the chemical. 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.

Sample Exercise 3.1

Balance the following equation:

Na(s) + H₂O(l) ß NaOH(aq) + H₂(g)

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 replace a coefficient 2 in front of H₂O:

Na(s) + 2H₂O(l) ß NaOH(aq) + H₂(g)

This choice is a trial beginning, but it set 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:

Na(s) + 2H₂O(l) ß 2NaOH(aq) + H₂(g)

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:

2Na(s) + 2H₂O(l) ß 2NaOH(aq) + H₂(g)

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:

1. __C₂H₄ + __O₂ ß __CO₂ + __H₂O
2. __Al + __HCl ß __AlCl₃ + __H₂

answers: (a) 1,3,2,2; (b) 2,6,2,3

Using the Periodic Table

We may identify reactants and products experimentally by their properties. However, we can often predict what will happen in a reaction if we have seen a similar reaction before. Naturally, recognizing a general pattern of reactivity for a class of substances gives you a broader understanding than merely memorizing a large number of unrelated reactions. The periodic table can be a powerful ally in this regard.

For example, knowing that sodium, Na, reacts with water, H₂O, to form NaOH and H₂, we can predict what happens when potassium, K, is placed in water. Both sodium and potassium are in the same group of the periodic table (the alkali metal group - group 1A). We expect them to behave similarly, producing the same kinds of products. Indeed, this prediction is correct:

2K(s) + 2H₂O(l) ß 2KOH(aq) + H₂(g)

In fact, all alkali metals react with water to form their hydroxide compounds and hydrogen. If we let M represent any alkali metal, we can write the general reaction as follows:

2M(s) + 2H₂O(l) ß 2MOH(aq) + H₂(g)

Alkali metal + water ß Metal hydroxide + hydrogen

Combustion reactions are rapid reactions that produce a flame. Most of the combustion reactions we observe involve O₂ from air as a reactant.

When hydrocarbons are combusted, they react with O₂ to form CO₂ and H₂O*.(*When there is an insufficient quantity of O₂ present, carbon monoxide, CO, will be produced. Even more severe restriction of O₂ will cause the production of fine particles of carbon that we call soot. Complete combustion produces CO₂. Unless specifically stated to the contrary, we will take combustion to mean complete combustion.)

Combustion of compounds containing oxygen atoms as well as carbon and hydrogen atoms (for example, CH₃OH and C₆H₁₂O₆) also produce CO₂ and H₂O. The simple rule that hydrocarbons and related compounds form CO₂ and H₂O when they burn in air summarizes the behavior of about 3 million compounds.

Sample exercise 3.2

Write the balanced chemical equation for the reaction that occurs when methanol, CH₃OH(l), is burned in air.

Solution We first recall that when any compound containing C, H, and O is combusted, it reacts with the O₂(g) in air to produce CO₂(g) and H₂O(l). thus, the unbalanced equation is

CH₃OH(l) + O₂(g) ß CO₂(g) + H₂O(l)

Because CH₃OH has only one C atom, we can start balancing the equation using the coefficient 1 for CO₂. Because CH₃OH has four H atoms, we place a coefficience 2 in front of H₂O to balance the H atoms:

This gives four O atoms among the products and three among the reactants (one in CH₃OH and two in O₂). We can use the fractional coefficient 3/2 in front of O₂ to provide four O atoms among the reactants (there are 3/2 X 2 = 3 O atoms in 3/2O₂):

CH₃OH(l) + 3/2O₂(g) ß CO₂(g) + 2H₂O(l)

Although the equation is now balanced, it is not in its most conventional form because it contains a fractional coefficient. If we multiply each side of the equation by 2, we will remove the fraction and achieve the following balanced equation:

2CH₃OH(l) + 3O₂(g) ß 2CO₂(g) + 4H₂O(l)

Practice Exercise

Write the balanced chemical equation for the reaction that occurs when ethanol, C₂H₅OH(l), is burned in air.

Answer: C₂H₅OH(l) + 3O₂(g) ß 2CO₂(g) + 3H₂O(l)

• In combination reactions two or more substances react to form one product.
• In a decomposition reaction one substance undergoes a reaction to produce two or more other substances.