One common way of classifying chemical reactions is to separate them into four categories: combination, decomposition, displacement, and double displacement. We give several examples of each type for two reasons: (1) to ensure that you understand the scope of each category and (2) to help you gain experience in interpreting and balancing equations. You may want to refer back to Section 3.4A. for a list of the rules and steps to follow in balancing equations.

TABLE 8.1 Parts of a chemical equation

Reactants	The substances that combine in the reaction. Formulas must be correct.
Products	The substances that are formed by the reaction. Formulas must be correct.
H	The enthalpy (heat energy) change accompanying the reaction. Energy is released if H < 0; energy is absorbed if H > 0
Arrows
	Found between reactants and products, means "reacts to form."
	Means the equation is not balanced.
	Placed after the formula of a product that is a gas.
	Placed after the formula of a product that is an insoluble solid - that is, a precipate.
Physical state	Indicates the physical state of the substance whose formula it follows.
	(g) Indicates that the substance is a gas
	(l) Indicates that the substance is a liquid
	(s) Indicates that the substance is a solid
	(aq) Means that the substance is in aqueous (water) solution
Coefficients	The numbers placed in front of the formulas to balance the equation.
Conditions	Words or symbols placed over the arrow () to indicate conditions used to make the reaction occur.
	Heat is added hv Light is added elec Electrical energy is added

A. Combination Reactions
In a two substances combine to form a single compound. Two examples are the reaction of solid magnesium with gaseous oxygen to form magnesium oxide, a solid:

2 Mg(s) + O₂(g) 2 MgO(s)

and the reaction of a hydrogen gas with chlorine gas to form gaseous hydrogen chloride:

H₂(g) + Cl₂(g) 2 HCl(g)

FIGURE 8.2 The brilliant white light associated with some fireworks is due to the release of energy when magnesium reacts with oxygen.

Figure 8.2 illustrates an example of a combination reaction.

Other combination reactions have compounds as reactants. The reaction of gaseous carbon dioxide with solid calcium oxide to form solid calcium carbonate is an example of such a reaction.

CaO(s) + CO₂(g) CaCO₃(s)

Example Write balanced equations for the following combination reactions: a. When solid phosphorus, P4, is burned in chlorine gas, solid phosphorus trichloride is formed. b. When gaseous dinitrogen pentoxide is bubbled through a solution of water, nitric acid is formed. Solution a. Write the reactants, an arrow, and then the products, with the physical state of each reactant or product shown after its formula. Write conditions for the reaction over the arrow. P4(s) + Cl2(g) PCl3(s) For atoms of phosphorus will yield four moleucles of phosphorus trichloride, which will require 12 atoms (six molecules) of chlorine. Putting in thesecoefficients gives the balanced equation P4(s) + 6 Cl2(g) 4 PCl3(s) b. Write the reactants, an arrow, and the products. N2O5(g) + H2O(l) HNO3(aq) Include the physical states and conditions of the reaction as given in its statement. Starting with nitrogen (it is always wise to leave hydrogen and oxygen to the last), two atoms of nitrogen in one molecule of N2O5 will form two molecules of nitric acid. Each molecule of nitric acid contains one hydrogen atom, thus two molecules will require two hydrogen atoms or one molecule of water. We then have six atoms of oxygen in the reactants - the same number of oxygen atoms required by two molecules of nitric acid. The equation is now balanced: N2O5(g) + H2(l) 2 HNO3(aq)

B. Decomposition Reactions
In a decomposition reaction, a compound is decomposed to its component elements or to other compounds. Although some compounds decompose spontaneously, usually light or heat is needed to initiate the decomposition. Following are three examples of decomposition - one induced by light, one induced chemically by a catalyst, and the third caused by heat. The antiseptic hydrogen peroxide is sold in opaque brown bottles because hydrogen peroxide decomposes in light (Figure 8.3). The equation for this decomposition is:

2 H2O2(aq)

hv

2 H2O(l) + O2(g)

Oxygen can be prepared by heating solid potassium chlorate in the presence of manganese dioxide, a catalyst. A catalyst is a chemical that, when added to a reaction mixture, hastens the reaction but can be recovered unchanged after the reaction is complete.

2 KClO3(s)

MnO2

2 KCl(s) + 3 O2(g)

FIGURE 8.3 Light hastens the decomposition of hydrogen peroxide. The dark bottle in which hydrogen peroxide is usually stored keeps out the light, thus protecting the hydrogen peroxide from decomposition.

When slaked lime, Ca(OH)₂(s), is heated, lime (CaO) and water vapor are produced:

Ca(OH)₂(s) CaO(s) + H₂O(g)

Example Write balanced equations for the following decomposition reactions: a. Solid amonium carbonate decomposes at room temperature to ammonia, carbon dioxide, and water. (Because of the ease of decomposition and the penetrating odor of ammonia, ammonium carbonate can be used as smelling salts.) b. On heating, lead(II) nitrate crystals decompose to yield a solid lead(II) oxide and the gases oxygen and nitrogen dioxide. Solution a. The unbalanced equation for the decomposition of ammonium carbonate is: (NH4)2CO3(s) NH3(g) + CO2(g) + H2O(l) Inspection of this equation indicates that two nitrogen atoms, therefore two ammonia molecules, are needed on the right. With this change, all other atoms are balanced: (NH4)2CO3(s)) 2 NH3(g) + CO2(g) + H2O(l) b. Writing the formulas for the reactant and the products in the form of an equation gives:. Pb(NO3)2(s) PbO(s) + O2(g) + NO2(g) Balance the elements in the order Pb, N, O, to leave oxygen for the last, which is in general a good practice. The lead is balanced as it stands, one atom on each side. There are two nitrogen atoms on the left, therefore we need 2 NO2 as a product. The oxygen is unbalanced, with six atoms in the reactants and five in the products. Try 2 Pb(NO3)2, which will give 2 PbO, 4 NO2, and two atoms of oxygen, making one molecule of O2. Now the equation is balanced. 2 Pb(NO3)2(s) 2 PbO(s) + 4 NO2(g) + O2(g)

C. Displacement Reactions
In displacement reactions, an uncombined element reacts with a compound and displaces an element from that compound. For example, bromine is found in seawater as sodium bromide. When chlorine is bubbled through seawater, bromine gas is released and a solution of sodium chloride is formed:

2 NaBr(aq) + Cl₂(g) 2NaCl(aq) + Br₂(g)

As another example, when an iron nail is dropped into a solution of copper(II) sulfate, iron(II) sulfate is formed in solution and metallic copper is deposited:

CuSO₄(aq) + Fe(s) FeSO₄(aq) + Cu(s)

Another displacement reaction, the reaction of metallic copper with silver nitrate, is shown in Figure 8.4.

FIGURE 8.4 A displacement reaction. In the tube on the left a copper wire has just been placed in a solution of silver nitrate. In the tube on the right the reaction is almost complete, and a great deal of silver has been deposited. The copper has displaced silver. The equation for this reaction is 2 AgNO3 + Cu Cu(NO3)2 + 2 Ag.

Example Write balanced equations for the following displacement reactions:: a. When a piece of aluminum is dropped into hydrochloric acid hydrogen is released as a gas and a solution of aluminum chloride is formed. b. When chlorine is bubbled through a solution of sodium iodide crystals of iodine appear in a solution of sodium chloride. Solution a. The unbalanced equation is: Al(s) + HCl(aq) AlCl3(aq) + H2(g) Aluminum is balanced. To balance the chlorine, we need 3 HCl on the reactant side, which will give 1 1/2 molecules of hydrogen. Al(s) + 3 HCl(aq) AlCl3(aq) + 1 1/2 H2(g) We need whole molecules of hydrogen. To get a whole number coefficient without unbalancing the other elements, we can multiply the whole equation by 2 to get: 2 Al(s) + 6 HCl(aq) 2 AlCl3(aq) + 3 H2(g) b. The unbalanced equation is: Cl2(g) + NaI(aq) NaCl(aq) + I2(s) to balance the chlorine, we must form 2 NaCl. Two NaCl require two sodium atoms, or 2 NaI, which gives two iodine atoms, as needed. The balanced equation then is: Cl2(g) + 2NaI(aq) 2 NaCl(aq) + I2(s)

Much more information is needed before you can predict whether or not a proposed displacement will take place. That information is given in Chapter 14 (Oxidation-Reduction). However, the displacement reactions we have discussed here will occur.

D. Double-Displacement Reactions
In double displacement, sometimes called metathesis or ion exchange, two ionic compounds react to form two different compounds. These reactions fall into a pattern that can be expressed as:

AB + CD CB + AD

in which A and C are cations, B and D are anions. These reactions are often called "exchanging-partner" reactions because the cations A and C exchange the anions with which they are associated.

Double-displacement reactions fall into two categories: (1) those in which an acid reacts with a base to form a salt and water, which are known as neutralization reactions, and (2) those in which one of the products is insoluble, which are usually precipitation reactions, although occasionally the insoluble product is a gas.

1. Reaction of an acid with a base: Neutralization reactions
In neutralization reactions, an acid reacts with a base to form a salt and water. Recall from Section 5.7D that an acid is a compound that liberates hydrogen ions in solution and a base (we will center here on hydroxides, a subgroup of bases) is a compound that liberates hydroxide ions in solution. A salt is defined as an ionic compound in which the cation is not hydrogen and the anion is not hydroxide. These reactions are called neutralization reactions because the base neutralizes the acid. Some examples are:

1. The reaction of sodium hydroxide with hydrochloric acid to form sodium chloride and water:

   NaOH(aq) + HCl(aq) NaCl(aq) + H₂O(l)

   Note that the salt formed, sodium chloride, combines the cation of the base, Na⁺, with the anion of the acid, Cl^-. The formula of the salt is the neutral combination of these ions, here a 1:1 combination in sodium chloride, NaCl.

2. The reaction of magnesium hydroxide with phosphoric acid to form magnesium phosphate and water:

   3 Mg(OH)₂(aq) + 2 H₃PO₄(aq) Mg₃(PO₄)₂(aq) + 6 H₂O(l)

   Here again the salt formed, magnesium phosphate, Mg₃(PO₄)₂, is a neutral combination of the cation of the base, Mg²⁺, with the anion of the acid, PO₄^3- .

A polyprotic acid is one whose molecules ionize to yield more than one hydrogen ion. Sulfuric acid, H₂SO₄, phosphoric acid, H₃PO₄, and carbonic acid, H₂CO₃, are examples of polyprotic acids. When a polyprotic acid is one of the reactants, neutralization may be incomplete and an acid salt may form. An example of this reaction is:

NaOH(aq) + H₂SO₄(aq) NaHSO₄(aq) + H₂O(l)

In this reaction, only one of the hydrogens of the diprotic acid reacted; the product is an acid salt, sodium hydrogen sulfate. The addition of more sodium hydroxide neutralizes the second hydrogen of this diprotic acid:

NaHSO₄(aq) +NaOH(aq) Na₂SO₄(aq) + H₂O(l)

 

Example Write balanced equations for the following neutralization reactions: a. the complete reaction of sulfuric acid with calcium hydroxide b. the complete reaction of magnesium hydroxide with hydrochloric acid c. the reaction of sodium hydroxide with carbonic acid to form an acid salt. Solutions a. The reactants are H2SO4 and Ca(OH)2. The products will show Ca2+ with SO42- instead of with OH- and H+ with OH- (HOH is the same as H2O). Write these facts in the form of an equation: Ca(OH)2(aq) + H2SO4(aq) CaSO4(aq) + H2O(l) To balance the equation, note that there are two H+ and two OH-. They will combine to give 2 H2O and the balanced equation: Ca(OH)2(aq) + H2SO4(aq) CaSO4(aq) + 2 H2O(l)) b. The formulas of the reactants are Mg(OH)2 and HCl. The products will show Mg2+ and Cl- instead of OH- and H+ with OH- instead of Cl-. We write the equation as: Mg(OH)2(aq) + HCl(aq) MgCl2(aq) + H2O(s) We need two chloride ions to combine with the Mg2+, so we write 2 HCl. We now have 2 H+ and 2 OH- to combine with them to form 2 H2O: Mg(OH)2(aq) + 2 HCl(aq) MgCl2(aq) + 2 H2O(l) c. The reactants are NaOH and H2CO3. If an acid salt is to be formed, only one of the H+ in carbonic acid will be replaced. The cations changing partners are Na+ and H+. The anions are HCO3- and OH-. Writing the equation we get NaOH(aq) + H2CO3(aq) NaHCO3(aq) + H2O(l)

2. Double-displacement reactions that form insoluble ionic products
Precipitation reactions, the second group of double-displacement reactions, result in the formation of insoluble ionic compounds. Ionic compounds differ enormously in the extent to which they dissolve in water, or their solubility. Table 8.2 illustrates this point by listing the solubilities of several ionic compounds in cold water. Notice that several, such as barium iodide and silver(I) nitrate, are very soluble in water, whereas others, such as lead(II) chloride, are only slightly soluble. Others, such as silver(I) chloride, are virtually insoluble. Generally, if more than 0.1 g of an ionic solid dissolves in 100 mL (0.1 L) of water, the compound is said to be soluble. Less than 0.1 g calcium carbonate, barium sulfate, and silver(I) chloride dissolve in 100 mL water. Therefore, they are classified as insoluble compounds.

Example Write the formulas of the following salts and predict whether each is soluble in water. a. lead(II) nitrate    b. iron(II) chloride   c. ammonium sulfide d. barium sulfate Solutions     Formula Solubility Reason lead(II) nitrate Pb(NO3)2 soluble It is a nitrate iron(II) chloride FeCl2 soluble it is a chloride, but not one of the listed exceptions. ammonium sulfide (NH4)2S soluble It is an ammonium salt. barium sulfate BaSO4 insoluble It is listed as an insoluble sulfate.

In precipitation reactions, solutions of two ionic compounds are combined. If two of the ions in the resulting mixture combine to form an insoluble compound or precipitate, a reaction occurs. (Figure 8.5 shows the formation of a precipitate.) If no insoluble product is produced, no reaction occurs.

FIGURE 8.5 The formation of a precipitate. When a clear colorless solution of lead nitrate (Pb(NO3)2) is added to a clear colorless solution of sodium iodide (NaI), a yellow precipitate of lead iodide (PbI2) appears. The equation for this reaction is given in Example 8.7a.

For example, if a solution of barium iodide is added to a solution of ammonium nitrate, no reaction takes place because the predicted products barium nitrate and ammonium iodide are both soluble.

BaI₂(aq) + 2 NH₄NO₃(aq) Ba(NO₃)₂(aq) + NH₄I(aq)

A reaction would occur if a solution of barium iodide were added to a solution of silver nitrate, because one of the products, silver iodide, is insoluble.

BaI₂(aq) + 2 AgNO₃(aq) Ba(NO₃)₂(aq) + 2 AgI()

In the equations for these reactions, the physical state of the insoluble product, the precipitate, is indicated either by (s) or by a downward-pointing arrow after its formula; the soluble components of the reaction are shown as (aq).

Example Write the balanced equatio for the following reactions. Indicate with a down-ward-pointing arrow any precipitate formed; name the precipitate. a. Solutions of lead(II) nitrate and sodium iodide react to form a yellow precipitate. b. the reaction between a solution of copper(II) nitrate and a solution of potassium sulfide yields a heay black precipitate.   Solutions a. The formulas for the reactants are Pb(NO3)2 and NaI. The formulas of the products of a reaction between these two compounds would have an interchange of anions, yielding PbI2 and NaNO3. Arranging these formlas in an unbalanced equation, we get: Pb(NO3)2(aq) + NaI(aq) PbI2 + NaNO3 Balancing this equation requires two iodide ions and therefore 2 NaI. Two sodium nitrate are formed: Pb(NO3)2(aq) + 2 NaI(aq) PbI2() + 2 NaNO3(aq) Because all sodium salts are soluble, the precipitate must be lead(II) iodide; we place an arrow after that formula. b. The formulas of the reactants are Cu(NO3)2 and K2S. The formulas of the products are CuS and KNO3. From Table 8.3 we know that potassium nitrate is soluble, so the precipitate must be CuS, copper(II) sulfide. The unbalanced equation is: Cu(NO3)2(aq) + K2S(aq) CuS{) + KNO3(aq) Balancing this equation requires two potassium nitrate. The balanced equation is: Cu(NO3)2(aq) + K2S(aq) CuS() + 2 KNO3(aq) All nitrates are soluble, so the precipitate is copper(II) sulfide. A downward-pointing arrow is put after its formula.

As mentioned earlier, occasionally the insoluble product is a gas, as in the following examples:

1. Hydrogen chloride is released as a gas when concentrated sulfuric acid is added to solid sodium chloride. Although hydrogen chloride is very soluble in water, it is quite insoluble in concentrated sulfuric acid. The acid salt sodium hydrogen sulfate is the second product.

   NaCl(s) + H₂SO₄(l) HCl(g) + NaHSO₄(s)

2. Acetic acid may be released as a gas in a reaction similar to that in the first example. The equation for the reaction of concentrated hydrochloric acid with sodium acetate is:

   NaC₂H₃O₂(s) + HCl(aq) HC₂H₃O₂(g) + NaCl(s)

3. A carbonate reacts with an acid to form carbonic acid, which immediately decomposes to gaseous carbon dioxide and water.

   Na₂CO₃(s) + 2 HCl(aq) 2NaCl(aq) + CO₂(g) + H₂O(l)