Chemical Reactions. A physical change alters the physical state of a substance without changing its composition ◦ Examples Boiling melting Freezing

Chapter 5Chemical Reactions

A physical change alters the physical state of a substance without changing its composition◦ Examples

Boiling melting Freezing Vaporization Condensation Sublimation Breaking a bottle Crushing a can

Chemical vs physical changes

A chemical change (a chemical reaction) converts one substance into another◦ Breaking bonds in the reactants (starting

materials)◦ Forming new bonds in the products

Chemical change

Chemical reactions

aA (physical state) + bB (state) cC (state) + dD (state)

A, B = reactants C, D = products

a, b, c, d = coefficients to indicate molar ratios of reactants and products

Chemical reactions

CH4 and O2 CO2 and H2O

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)

Balancing chemical reactions

2 molecules of C4H10 13 molecules of O2

10 molecules of C4H108 molecules of

CO2

Balancing Chemical Equations:Unbalanced equation:

C4H10 + O2 CO2 + H2O

Balanced equation:

2C4H10 + 13O2 8CO2 + 10H2O

H2 + O2 → H20

Reactants are on the left ◦ Things that are used◦ H2 + O2

Product(s) are on the right◦ Things that are made◦ H20

This equation is not yet balanced

Parts of an Equation

Do the number of atoms of each element on either side of the arrow balance?◦ Compounds consist of more than one element◦ Examples: NaCl, H2SO4, H20◦ Look at numbers of each atoms within the

compounds

Evaluating the Equation

Correct any imbalances with a coefficient◦ Coefficient = the large number to the left of a substance in the equation

◦Don’t change subscripts This will change what the molecule is

Changing the Equation

Example: to balance an equation you need two atoms of oxygen from water (H2O)◦ If you change the subscript you change the

compound

◦H2O2 does provide two atoms of oxygen but

H2O (water) ≠ H2O2 (hydrogen peroxide)

◦2 H2O provides 2 atoms of oxygen and keeps the compound as water

◦ It also gives you 4 atoms of hydrogen that you should then make sure is balanced

Changing the Equation

Relax and calmly go through what is on either side of the equation

There are many different ways to start balancing the equation

Once you start the remaining coefficients should fall into place

Where to start?◦ Compounds (NaCl, H2SO4, H20)

See what elements they have in common

◦ Molecular elements (O2, H2) More than one atom present

◦ Monoatomic elements (Ca, Cl) Only one atom present

Tips

Example: Formation of Water

H2 + O2 → H20

List out what is on each side Oxygen does not balance

Left Right

2 hydrogen 2 hydrogen

2 oxygen 1 oxygen


H2 + O2 → 2H20

Multiply right side by 2 to balance oxygen Now hydrogen does not balance

Left Right


2 oxygen 2 oxygen


2H2 + O2 → 2H20

Multiply hydrogen by 2 to balance hydrogen Equation is now balanced

Left Right


2 oxygen 2 oxygen

List out what is on each side Carbon does not balance


C6H12O6 + O2 → CO2 + H2O Left Right

6 carbon 1 carbon


8 oxygen 3 oxygen

Look at compounds first Multiply CO2 on right by 6 to balance C Hydrogen does not balance


C6H12O6 + O2 → 6CO2 + H2O Left Right

6 carbon 6 carbon


8 oxygen 13 oxygen

Multiply water on right side by 6 to balance hydrogen

Oxygen does not balance


C6H12O6 + O2 → 6CO2 + 6H2O Left Right

6 carbon 6 carbon


8 oxygen 18 oxygen

Multiply oxygen on left side by 6 to balance oxygen

Equation is now balanced


C6H12O6 + 6O2 → 6CO2 + 6H2O Left Right

3 carbon 6 carbon


2 oxygen 18 oxygen

Propane + oxygen → carbon dioxide + water

C3H8 + O2 → CO2 + H2O

Example: Combustion of Propane

List out what is on each side Carbon does not balance

Left Right

3 carbon 1 carbon


2 oxygen 3 oxygen


C3H8 + O2 → 3CO2 + H2O


Look at compounds first Multiply CO2 on right side by 3 to balance carbon Hydrogen does not balance

Left Right

3 carbon 3 carbon


2 oxygen 7 oxygen


C3H8 + O2 → 3CO2 + 4H2O


Multiply water on right side by 4 to balance hydrogen

Oxygen does not balance

Left Right

3 carbon 3 carbon


2 oxygen 10 oxygen


C3H8 + 5O2 → 3CO2 + 4H2O


Multiply oxygen on left side by 5 to balance oxygen


Left Right

3 carbon 3 carbon


10 oxygen 10 oxygen

Sodium azide → Sodium + nitrogen

NaN3 → Na + N2

Example: Air Bag Inflation

Identify what you have Nitrogen does not balance How do we balance nitrogen when left side has 3 and the

right has 2?o find the lowest common multiple for botho In this case 6o Multiply each side to make 6 atoms

Left Right

1 sodium 1 sodium

3 nitrogen 2 nitrogen


2NaN3 → Na + N2


Multiply sodium azide on left side by 2 to get 6 nitrogen atoms

Still need to make 6 atoms of nitrogen on right side

Left Right

2 sodium 1 sodium



2NaN3 → Na + 3N2


Multiply nitrogen by 3 on left side to get 6 nitrogen atoms

Sodium is not balanced

Left Right

2 sodium 1 sodium



2NaN3 → 2Na + 3N2


Multiply sodium on right side by 2 to balance sodium


Left Right

2 sodium 2 sodium


A mole is a quantity that contains 6.02 X 1023 items (usu. atoms, molecules or ions)◦ An amount of a substance whose weight, in grams

is numerically equal to what its molecular weight was in amu

◦ Just like a dozen is a quantity that contains 12 items◦ 1 mole of C atoms = 6.02 x 1023 C atoms◦ 1 mole of CO2 molecules = 6.02 x 1023 CO2

molecules◦ 1 mole of H2O molecules = 6.02 x 1023 H2O

molecules The number 6.02 X 1023 is Avogadro’s number

Avogadro’s Number and the Mole

1 mol6.02 x 1023 atoms

1 mol6.02 x 1023 molecules

How many molecules are in 2.5 moles of penicillin

2.5 moles penicillin X 6.02 x 1023 molecules = 1 mole

1.5 X 1024 molecules

Example

The formula weight is the sum of the atomic weights of all the atoms in a compound, reported in atomic mass units

The molar mass is the mass of one mole of any substance, reported in grams◦ The value of the molar mass of a compound in

grams equals the value of its formula weight in amu.

Molar mass

The sum of the atomic weights of all the atoms in a compound, reported in atomic mass units◦ This may be called molecular weight for covalent

compounds Example

◦ H2O Contains 2 Hydrogen atoms and 1 Oxygen atom Hydrogen weighs 1.01 amu (1.01 g H/mole) Oxygen weighs 16.0 amu (16.0 g O/mole) Formula/molecular weight = 2(1.01 amu) +16.0 amu =

18.0 amu Looking only at one molecule

Molar mass = 2(1.01 g/mol) +16.0 g/mol = 18.0 g/mol Looking at one mole of the substance

Formula weight

Stoichiometry is the study of the quantitative relationships that exist between substances involved in a chemical reaction

Mole ratios within molecules:

Stoichiometry: mole ratios

AxBy

Mole ratio of A:B = x:y

Example: H2O2 ⇒ H:O = 2:2 = 1:1

Mole ratios between molecules A balanced equation tells us the number of

moles of each reactant that combine and number of moles of each product formed

Stoichiometry

aA + bB cC + dD

Mole ratio of A:B:C:D = a:b:c:d

Example: 2H2 + O2 → 2H20

H2:O2:H20 = 2:1:2

The coefficients in the balanced chemical equation can represent the ratio of molecules of the substances that are consumed or produced

The coefficients in the balanced chemical equation can represent the ratio of moles of the substances that are consumed or produced

2H2 + O2 → 2H20

Stoichiometry

There are 4 basic types of stoichiometry problems: ◦ Moles to moles◦ Moles to grams◦ Grams to moles◦ Grams to grams

However, all stoichiometry problems are really very similar, and the same general approach can be used to solve any of them◦ So really only one type of problem

Stoichiometry

Grams to moles to moles to grams

How to visualize problems

mol mol

g g

N2 +3H2→2NH3

How many moles of H2 are required to produce 3.89 mol of NH3?

◦ Equation says H2:NH3 is 3:2

◦ 3.89 mol NH3 * 2 mol H2 = 5.84 moles H2

3 mol NH3

Moles to moles example

mol mol

g g

N2 +3H2→2NH3

How many grams of NH3 are produced from 3.44 mol of N2?◦ Use mole ratio between NH3 and N2

◦ Then use molar mass to convert moles to g

◦ 3.44 mol N2 * 2 mol NH3 * 17 g NH3 =117 g NH3

1 mol N2 1 mol NH3

Moles to grams

mol mol

g g

Grams to moles

N2 +3H2→2NH3

How many moles H2 react with 6.77 g of N2?

Convert grams to moles using molar mass Use molar ratio between N2 and H2

◦ 6.77 g N2 * 1 mol N2 * 3 mol H2 = 0.725 mol H2

28.0 g N2 1 mol N2

mol mol

g g

Grams to grams

N2 +3H2→2NH3

How many grams of NH3 are produced from 8.23 g of H2?

Use molar mass to convert g to moles Use molar ratio to convert between moles Use molar mass to convert moles to g

8.23 g H2 * 1 mol H2 * 2 mol NH3 * 17.0 g NH3= 46.2 g NH3

2.02 g H2 3 mol H2 mol NH3

mol mol

g g

Once you have converted things into moles you can use molar ratios in balanced equations to convert between different elements, compounds, and molecules

Problems may ask for g, molecules, atoms, or many other units

For most problems

mol mol

unit given unit requested

If I gave you a value in g how would you convert it to mol?◦ Use formula weight (mass)

Calculate using the periodic table

◦ Example: 53.21 g C6H12O6

Formula weight/molecular mass:

6*(12.01 g/mol) + 12*(1.01 g/mol) +6*(16.00 g/mol) = 180.18 g/mol

◦ Use value to calculate moles (may have to invert value)

53.21 g C6H12O6 * 1 mol C6H12O6 = 0.2953 mol C6H12O6

180.18 g C6H12O6

Stoichiometry

If I give you a value in mol how would you compare it to mol of another substance?◦ Use a balanced equation

Example 5 mol CO2 to mol O2

◦ 6CO2 + 6H2O C6H12O6 + 6O2

◦ Use the stoichiometric coefficients to convert between any of the substances in this equation

6CO2 6CO2 6CO2

6H2O C6H12O6 6O2

◦ In our case only interested in the relationship with O2

5 mol CO2 * 6CO2 = 5 mol O2

6O2

Stoichiometry

All questions will build on the examples on the previous two slides◦ May ask you to go g mol mol g◦ May ask you to convert mol to number of atoms

or number of molecules Use Avagadro’s number 6.02 X 1023

◦ May ask you to use some other value in the future, but it will be something that you can relate to g or to mol

Stoichiometry

INGREDIENTS:◦ 3 cups all-purpose flour◦ 1 teaspoon salt◦ 1 cup shortening◦ 1/2 cup cold water◦ 2 cups pumpkin◦ 2 eggs, beaten◦ 3/4 cup packed brown sugar◦ 2 teaspoon spices

If you want to make multiple pies - the amount of pie that you can bake depends on which of the ingredients you have the “least” of – what will you run out of first?

Limiting ingredient

Ingredient Recipe In pantry # of recipes it can make

flour 3 cups 14 cups 4.67 x

salt 1 tsp 20 tsp 20 x

shortening 1 cup 7 cup 7 x

water ½ cup ∞∞ ∞

pumpkin 2 cups 19 cups 9.5 x

eggs 2 18 9x

sugar ¾ cup 3 cups 4X

spice 2 tsp 21 tsp 10.5 x

Limiting ingredient

Limitingingredient

N2 +3H2→2NH3

You have 4 moles N2 and 9 moles H2

How many moles of NH3 could be produced?

H2 is the limiting reactant and limits how much NH3 can be made

9 moles H2 * 2 mol NH3 = 6 mol NH3

3 mol H2

Limiting reactant

Actual Required Rx

N2 4 1 4

H2 9 3 3

Compare the actual amount of each reactant to the amount required in the balanced equation to determine how many times the “reaction can be run”

Use the amount of the limiting reactant to calculate how much product can be produced

Limiting reactants

How many g of S can be produced if we attempt to react 9.00g of Bi2S3

with 16.00g of HNO3?

9.00 g Bi2S3 * 1 mol Bi2S3 = 0.0175 mol Bi2S3

514.3 g Bi2S3

0.0175 mol Bi2S3 * 1 Rx = 0.0175 allows Rx to run 0.0175 times

1 mol Bi2S3

16.00 g HNO3 * 1 mol HNO3 = 0.254 mol HNO3

63.0 g HNO3

0.254 mol HNO3 * 1 Rx = 0.3175 allows Rx to run 0.0318 times

8 mol HNO3

Therefore Bi2S3 is the limiting reactant

Bi2S3 +8 HNO3 2Bi(NO3)2 +2NO +3S +4H2O

Use Bi2S3 - the limiting reactant to calculate S◦ How many g of S can be produced if we attempt to react 9.00

g of Bi2S3 with 16.00 g of HNO3?

To see how much S can be produced this is gram to mole to mole to gram problem

9.00 g Bi2S3 * 1 mol Bi2S3 = 0.0175 mol Bi2S3

514.3 g Bi2S3

0.0175 mol Bi2S3 * 3 mol S * 32.1 g S = 1.69 g S

mol Bi2S3 mol S

Bi2S3 +8 HNO3 2Bi(NO3)2 +2NO +3S +4H2O

If 3.00 mol H2 reacts with 2.00 mol O2 how many moles of H2O will form?

Solution 1:

◦ 3.00 mol H2 * 1 Rx = 1.50 Rx

2 mol H2

◦ 2.00 mol O2 * 1 Rx = 2.00 Rx

1 mol O2

◦ H2 is limiting reactant so use it to calculate mol of H2O

◦ 3.00 mol H2 * 2 mol H2O = 3.00 mol H2O

2 mol H2

H2 is the limiting reactant so 3.00 mol H2O will theoretically form

Example: 2H2 + O2 2 H2O


Solution 2:

◦ 3.00 mol H2 * 2 mol H2O = 3.00 mol H2O

2 mol H2

◦ 2.00 mol O2 * 2 mol H2O = 4.00 mol H2O

1 mol O2

H2 is the limiting reactant so 3.00 mol H2O will theoretically form

Actual yield is determined experimentally, it is the mass of the product that is measured

Theoretical yield is the calculated mass of the products based on the initial mass or number of moles of the reactants

Percent yield

%100yield ltheoretica

yield actualyield percentage


We just saw that H2 is the limiting reactant so 3.00 mol H2O will theoretically form

If we run this in the lab and end up producing 1.75 mol H2O what is the percent yield?

1.75 mol H2O * 100% = 58.3 %

3.00 mol H2O

Example: 2H2 + O2 2 H2O

Both processes occur together in a single reaction called an oxidation−reduction or redox reaction.

Thus, a redox reaction always has two components, one that is oxidized and one that is reduced

A redox reaction involves the transfer of electrons from one element to another.

Oxidation and reduction

Oxidation is the loss of electrons from an atom.◦ Reducing agents are oxidized

Reduction is the gain of electrons by an atom◦ Oxidizing agents are reduced.

Oxidation and reduction

LEO says GER

Oxidation and reduction reactions

Zn + Cu2+ Zn2+ + Cu

Zn loses 2 e–

•Zn loses 2 e− to form Zn2+, so Zn is oxidized.

•Cu2+ gains 2 e− to form Cu, so Cu2+ is reduced.

Cu2+ gains 2 e−

Oxidation and reduction reactions

Zn + Cu2+ Zn2+ + Cu

Zn loses 2 e–

Cu2+ gains 2 e−

Oxidation half reaction: Zn Zn2+ + 2 e−

Each of these processes can be written as an individual half reaction:

loss of e−

Reduction half reaction: Cu2+ + 2e− Cu

gain of e−

A compound that is oxidized while causing another compound to be reduced is called a reducing agent

Zn acts as a reducing agent because it causes Cu2+ to gain electrons and become reduced

58

Oxidation and Reduction

Zn + Cu2+ Zn2+ + Cu

oxidized reduced

A compound that is reduced while causing another compound to be oxidized is called an oxidizing agent

Cu2+ acts as an oxidizing agent because it causes Zn to lose electrons and become oxidized

59


Zn + Cu2+ Zn2+ + Cu

oxidized reduced

60


61

Oxidation and ReductionIron Rusting

4 Fe(s) + 3 O2(g) 2 Fe2O3(s)

Fe3+ O2–neutral Fe neutral O

Fe loses e– and is oxidized.

O gains e– and is reduced.

62


Inside an Alkaline Battery

Zn + 2 MnO2 ZnO + Mn2O3

neutral Zn Mn4+ Zn2+ Mn3+

Zn loses e− and is oxidized.

Mn4+ gains e− and is reduced.

63


Zn + 2 MnO2 ZnO + Mn2O3

64


Oxidation results in the: Reduction results in the:

•Gain of oxygen atoms

•Loss of hydrogen atoms

•Loss of oxygen atoms

•Gain of hydrogen atoms

Documents

Chemical Reactions. A physical change alters the physical state of a substance without changing its composition ◦ Examples Boiling melting Freezing