Chemical Reactions

5.1 Introduction to Chemical Reactions

A. General Features of Physical and Chemical Changes

• A physical change alters the physical state of a substance without changing its composition.
• A chemical change (a chemical reaction) converts one substance into another.
• Chemical reactions involve:
  • Breaking bonds in the reactants (starting materials).
  • Forming new bonds in the products.

5.1 Introduction to Chemical Reactions

B. Writing Chemical Equations

• A chemical equation uses chemical formulas and other symbols to show the reactants (starting materials) and products in a reaction.
• The reactants are written on the left.
• The products are written on the right.
• Coefficients show the number of molecules of a given element or compound that react or are formed.
• The law of conservation of mass states that atoms cannot be created or destroyed in a chemical reaction.
• Coefficients are used to balance an equation.
• A balanced equation has the same number of atoms of each element on both sides of the equation.

5.1 Introduction to Chemical Reactions

B. Writing Chemical Equations

• Symbols Used in Chemical Equations
  • $\rightarrow$ Reaction arrow
  • $\Delta$ Heat
  • $(s)$ Solid
  • $(l)$ Liquid
  • $(g)$ Gas
  • $(aq)$ Aqueous solution

5.2 Balancing Chemical Equations

HOW TO Balance a Chemical Equation

Balancing chemical equations involves adjusting coefficients to ensure the number of atoms of each element is the same on both sides of the equation. Subscripts in chemical formulas should never be changed during balancing, as this alters the compound's identity.

• Example: Balancing the reaction of propane (C3H8) with oxygen (O2) to form carbon dioxide (CO2) and water (H2O).
• Step 1: Write the unbalanced equation with correct chemical formulas:
  $C<em>3H</em>8 + O<em>2 \rightarrow CO</em>2 + H_2O$
• Step 2: Balance the equation with coefficients, balancing one element at a time:
  • Balance Carbons (C):
    $C<em>3H</em>8 + O<em>2 \rightarrow 3CO</em>2 + H_2O$
  • Balance Hydrogens (H):
    $C<em>3H</em>8 + O<em>2 \rightarrow 3CO</em>2 + 4H_2O$
  • Balance Oxygens (O):
    $C<em>3H</em>8 + 5O<em>2 \rightarrow 3CO</em>2 + 4H_2O$
• Step 3: Check that the smallest set of whole numbers is used.
• Atoms in the reactants:
  • 3 Carbons
  • 8 Hydrogens
  • 10 Oxygens (5 × 2 O’s)
• Atoms in the products:
  • 3 Carbons
  • 8 Hydrogens
  • 10 Oxygens ((3 × 2 O’s) + (4 × 1 O))

Activity: Balancing Equations

• Balance the following equations:
  1. Potassium chlorate → potassium chloride + oxygen
  2. Aluminum acetate, $Al(C<em>2H</em>3O<em>2)</em>3$, reacts with potassium sulfate to form potassium acetate and aluminum sulfate
  3. Hexane ($C<em>6H</em>{14}$) reacts with oxygen gas to form carbon dioxide and water
  4. Zinc reacts with hydrochloric acid to form zinc chloride and hydrogen gas

Activity Solutions: Balancing Equations

1. Potassium chlorate → potassium chloride + oxygen
   • Translate chemical names into chemical formulas:
     $KClO<em>3(s) \rightarrow KCl(aq) + O</em>2(g)$
   • Balance the chemical equation:
     $2KClO<em>3(s) \rightarrow 2KCl(aq) + 3O</em>2(g)$
2. Aluminum acetate reacts with potassium sulfate to form potassium acetate and aluminum sulfate
   • Aluminum acetate + potassium sulfate → potassium acetate + aluminum sulfate
     $Al(C<em>2H</em>3O<em>2)</em>3 + K<em>2SO</em>4 \rightarrow KC<em>2H</em>3O<em>2 + Al</em>2(SO<em>4)</em>3$
   • $Al(C<em>2H</em>3O<em>2)</em>3(aq) + K<em>2SO</em>4(aq) \rightarrow KC<em>2H</em>3O<em>2(aq) + Al</em>2(SO<em>4)</em>3(s)$
   • Balanced:
     $2Al(C<em>2H</em>3O<em>2)</em>3(aq) + 3K<em>2SO</em>4(aq) \rightarrow 6KC<em>2H</em>3O<em>2(aq) + Al</em>2(SO<em>4)</em>3(s)$
3. Hexane ($C<em>6H</em>{14}$) reacts with oxygen gas to form carbon dioxide and water
   • Hexane was given. Oxygen gas is diatomic. Carbon dioxide has 1 carbon atom and 2 oxygen atoms (using the Greek prefixes), and water has 2 hydrogen atoms and 1 oxygen atom.
     $C<em>6H</em>{14}(l) + O<em>2(g) \rightarrow CO</em>2(g) + H_2O(g)$
   • Balanced:
     $C<em>6H</em>{14}(l) + 9.5O<em>2(g) \rightarrow 6CO</em>2(g) + 7H_2O(g)$
   • Fractional coefficients are not acceptable, so we multiply all coefficients by 2:
     $2C<em>6H</em>{14}(l) + 19O<em>2(g) \rightarrow 12CO</em>2(g) + 14H_2O(g)$
4. Zinc reacts with hydrochloric acid to form zinc chloride and hydrogen gas
   • Zinc + Hydrochloric acid → Zinc chloride + hydrogen
     $Zn(s) + HCl(aq) \rightarrow ZnCl<em>2(aq) + H</em>2(g)$
   • Balanced:
     $Zn(s) + 2HCl(aq) \rightarrow ZnCl<em>2(aq) + H</em>2(g)$

5.3 Types of Reactions

The majority of chemical reactions fall into 6 categories: combination, decomposition, single replacement, double replacement, oxidation and reduction (excluded in this section), and acid-base (covered in Chapter 9).

A. Combination and Decomposition

• A combination reaction is the joining of two or more reactants to form a single product.
• Examples of Combination Reactions:
  1. $N<em>2 + 3H</em>2 \rightarrow 2NH_3$
  2. $Ca + Br<em>2 \rightarrow CaBr</em>2$
  3. $H<em>2C=CH</em>2 + Cl<em>2 \rightarrow ClCH</em>2CH_2Cl$
• A decomposition reaction is the conversion of a single reactant to two or more products.
• Examples of Decomposition Reactions:
  1. $2NH<em>3 \rightarrow N</em>2 + 3H_2$
  2. $2KClO<em>3 \rightarrow 2KCl + 3O</em>2$
  3. $CH<em>3CH</em>2Cl \rightarrow H<em>2C=CH</em>2 + HCl$

B. Replacement Reactions

• A single replacement reaction is a reaction in which one element replaces another element in a compound to form a different compound and element as products.
• A double replacement reaction is a reaction in which two compounds exchange “parts”–atoms or ions—to form two new compounds.

B. Replacement Reactions

• Examples of Single and Double Replacement Reactions
  • Single replacement: $2NaCl + Br<em>2 \rightarrow 2NaBr + Cl</em>2$ (The element Br replaces Cl in the compound NaCl.)
  • Single replacement: $Fe + CuSO<em>4 \rightarrow FeSO</em>4 + Cu$ (The element Fe replaces Cu in the compound CuSO4.)
  • Double replacement: $AgNO<em>3 + NaCl \rightarrow AgCl + NaNO</em>3$ (The ions Ag+ and Na+ exchange.)

Types of Reactions - Overview

5.5 The Mole and Avogadro’s Number

• A mole is a quantity that contains $6.02 \times 10^{23}$ items.

  • 1 mole of C = $6.02 \times 10^{23}$ C atoms
  • 1 mole of $H<em>2O$ = $6.02 \times 10^{23} H</em>2O$ molecules
  • 1 mole of Vitamin C = $6.02 \times 10^{23}$ Vitamin C molecules

• The number $6.02 \times 10^{23}$ is Avogadro’s number.

• It can be used as a conversion factor to relate the number of moles of a substance to the number of atoms or molecules:

  • $\frac{6.02 \times 10^{23} atoms}{1 mol}$ or $\frac{1 mol}{6.02 \times 10^{23} atoms}$
  • $\frac{6.02 \times 10^{23} molecules}{1 mol}$ or $\frac{1 mol}{6.02 \times 10^{23} molecules}$

• Sample Problem 5.9

  • How many molecules are contained in 5.0 moles of carbon dioxide ($CO_2$)?
  • Step [1] Identify the original quantity and the desired quantity.
    • Original quantity: 5.0 mol of $CO_2$
    • Desired quantity: ? number of molecules of $CO_2$
  • Step [2] Write out the conversion factors.
    • $\frac{6.02 \times 10^{23} molecules}{1 mol}$
    • Choose the right side to cancel mol.
  • Step [3] Set up and solve the problem.
    • Unwanted unit cancels. (Problem solution would be here)

5.6 Mass to Mole Conversions

• The formula weight is the sum of the atomic weights of all the atoms in a compound, reported in atomic mass units (amu).

HOW TO Calculate the Formula Weight of a Compound

• Example: Calculate the formula weight for $FeSO_4$.
• Step [1] Write the correct formula and determine the number of atoms of each element from the subscripts.
  • $FeSO_4$ contains 1 Fe atom, 1 S atom, and 4 O atoms.
• Step [2] Multiply the number of atoms of each element by the atomic weight and add the results.
  • 1 Fe atom × 55.85 amu = 55.85 amu
  • 1 S atom × 32.07 amu = 32.07 amu
  • 4 O atoms × 16.00 amu = 64.00 amu
  • Formula weight of $FeSO_4$ = 151.92 amu

A. Molar Mass

• The molar mass is the mass of one mole of any substance, reported in grams per mole (g/mol).
• The value of the molar mass of a compound in grams equals the value of its formula weight in amu.

B. Relating Grams to Moles

• The molar mass relates the number of moles to the number of grams of a substance.
• In this way, molar mass can be used as a conversion factor.
• The molar mass of $H_2O$ is 18.02 g/mol, so the conversion factor can be written as:
  • $\frac{1 mol}{18.02 g H<em>2O}$ or $\frac{18.02 g H</em>2O}{1 mol}$

B. Relating Grams to Moles

• Sample Problem 5.13
  • How many moles are present in 100. g of aspirin ($C<em>9H</em>8O_4$, molar mass 180.2 g/mol)?
  • Step [1] Identify the original quantity and the desired quantity.
    • Original quantity: 100. g of aspirin
    • Desired quantity: ? mol of aspirin
  • Step [2] Write out the conversion factors.
    • The conversion factor is the molar mass, and it can be written in two ways. Choose the one that places the unwanted unit, grams, in the denominator so that the units cancel:
      • $\frac{1 mol}{180.02 g aspirin}$ or $\frac{180.02 g aspirin}{1 mol}$
    • Choose the right one to cancel g aspirin.
  • Step [3] Set up and solve the problem.
    • Unwanted unit cancels.

C. Relating Grams to Number of Atoms or Molecules

• We can also use the molar mass to show the relationship between grams and number of molecules (or atoms).
  • $\frac{180.02 g}{1 mol} aspirin = \frac{180.02 g}{6.02 \times 10^{23} molecules} aspirin$
  • $\frac{1 mol}{6.02 \times 10^{23} molecules}$

C. Relating Grams to Number of Atoms or Molecules

• Sample Problem 5.14
  • How many molecules are in a 325-mg tablet of aspirin ($C<em>9H</em>8O_4$, molar mass 180.2 g/mol)?
  • Step [1] Identify the original and desired quantities.
    • Original quantity: 325 mg aspirin
    • Desired quantity: ? molecules aspirin

C. Relating Grams to Number of Atoms or Molecules

• Step [2] Write out the conversion factors.
  • To convert mg to g:
    • $\frac{1 g}{1000 mg}$ or $\frac{1000 mg}{1 g}$
    • Choose the right one to cancel mg.
  • Then, to convert g to number of moles:
    • $\frac{6.02 \times 10^{23} molecules}{180.02 g aspirin}$ or $\frac{180.02 g aspirin}{6.02 \times 10^{23} molecules}$
    • Choose the right one to cancel g aspirin.
• Step [3] Set up and solve the problem.

5.7 Mole Calculations in Chemical Equations

• A balanced chemical equation also tells us:

  • The number of moles of each reactant that combine
  • The number of moles of each product formed

• $N<em>2(g) + O</em>2(g) \rightarrow 2NO(g)$

  • 1 mole of N2
  • 1 molecule N2
  • 1 mole of O2
  • 1 molecule O2
  • 2 moles of NO
  • 2 molecules NO
  • (The coefficient “1” has been written for emphasis.)

• Coefficients are used to form mole ratios, which can serve as conversion factors.

• $N<em>2(g) + O</em>2(g) \rightarrow 2NO(g)$

  • $\frac{1 mol N<em>2}{1 mol O</em>2}$
  • $\frac{2 mol NO}{1 mol N_2}$
  • $\frac{2 mol NO}{1 mol O_2}$

• Use the mole ratios from the coefficients in the balanced equation to convert moles of one compound (A) into moles of another compound (B).

• Sample Problem 5.15

  • Using the balanced chemical equation, how many moles of $CO$ are produced from 3.5 moles of $C<em>2H</em>6$?
  • $2C<em>2H</em>6(g) + 5O<em>2(g) \rightarrow 4CO(g) + 6H</em>2O(g)$
  • Step [1] Identify the original and desired quantities.
    • Original quantity: 3.5 mol $C<em>2H</em>6$
    • Desired quantity: ? mol $CO$

• $2C<em>2H</em>6(g) + 5O<em>2(g) \rightarrow 4CO(g) + 6H</em>2O(g)$

  • Step [2] Write out the conversion factors.
    • $\frac{4 mol CO}{2 mol C<em>2H</em>6}$ or $\frac{2 mol C<em>2H</em>6}{4 mol CO}$
    • Choose the right one to cancel mol $C<em>2H</em>6$.
  • Step [3] Set up and solve the problem.

5.8 Mass Calculations in Chemical Equations

HOW TO Convert Moles of Reactant to Grams of Product

• Example

  • Using the balanced equation, how many grams of $O<em>3$ (theoretical yield of $O</em>3$) are formed from 9.0 mol of $O_2$?
  • $3O<em>2(g) \xrightarrow{sunlight} 2O</em>3(g)$
  • Step [1] Convert the number of moles of reactant to the number of moles of product using a mole–mole conversion factor.
    • Mole–mole conversion factor
      • $\frac{2 mol O<em>3}{3 mol O</em>2}$ or $\frac{3 mol O<em>2}{2 mol O</em>3}$
      • Choose the right one to cancel mol $O_2$ in Step [1].
    • Multiply the number of moles of starting material by the conversion factor to give the number of moles of product.

• Step [2] Convert the number of moles of product to the number of grams of product using the product’s molar mass.

  • $MM O_3 = 3 \times 16.0 = 48.0 \frac{g}{mol}$
  • Molar mass conversion factor
    • $\frac{48.0 g O<em>3}{1 mol O</em>3}$ or $\frac{1 mol O<em>3}{48.0 g O</em>3}$
  • Choose the right one to cancel mol $O_3$ in Step [2].
    • Multiply the number of moles of product from Step [1] by the conversion factor to give the number of grams of product.

HOW TO Convert Grams of Reactant to Grams of Product

• Example

  • Ethanol ($C<em>2H</em>6O$, molar mass 46.1 g/mol) is synthesized by reacting ethylene ($C<em>2H</em>4$, molar mass 28.1 g/mol) with water.
  • How many grams of ethanol (theoretical yield of ethanol) are formed from 14 g of ethylene and excess water?
  • $C<em>2H</em>4 + H<em>2O \rightarrow C</em>2H_6O (g)$

• $C<em>2H</em>4 + H<em>2O \rightarrow C</em>2H_6O$

• Grams of reactant -> molar mass conversion factor -> mole–mole conversion factor -> molar mass conversion factor

5.9 Percent Yield

• The theoretical yield is the amount of product expected from a given amount of reactant based on the coefficients in the balanced chemical equation.

• Usually, however, the amount of product formed is less than the maximum amount of product predicted.

• The actual yield is the amount of product isolated from a reaction.

• Sample Problem 5.18

  • If the reaction of ethylene with water to form ethanol has a calculated theoretical yield of 23 g of ethanol, what is the percent yield if only 15 g of ethanol are actually formed?
  • $\% Yield = \frac{Actual Yield}{Theoretical Yield} \times 100\%$
  • $\% Yield = \frac{15 g}{23 g} \times 100\% = 65\%$