Chemical Equations & Reactions

Chemistry 6.0

Chemical ReactionsA. Definition: a process by which 1 or more

substances, called reactants, are changed into 1 or more substances, called products, with different physical & chemical properties.

B. Evidence of a Chemical Reaction1. Color change2. Formation of a precipitate, ppt3. Release of a gas4. Energy change – heat, light, sound5. Odor change

C. Reactions are started by the addition of energy

Chemical EquationA. Form

1. Reactant + Reactant Product + Product2. Symbols: (s), (l), (g), (aq)

NR methanol

Writing Chemical Equations Two moles of water at room temperature are exposed

to an electric current and produces two moles of hydrogen gas and one mole of oxygen gas.

When two moles of aluminum pellets are added to

three moles of a copper(II) chloride solution, 3 moles of copper precipitate out and two moles of aluminum chloride remain in solution.

2 H2O(l) 2 H2(g) + 1 O2(g)

2 Al(s) + 3 CuCl2(aq) 3 Cu(s) + 2 AlCl3(aq)

Characteristics of A Balanced Chemical Equations

1. The equation must represent known facts. All substances have been identified.

2. The equation must contain the correct symbols and/or formulas for the reactants and products

3. Can be either a word equation or a formula equation

4. The law of conservation of mass must be satisfied. This provides the basis for balancing chemical equations. 1st formulated by Antoine Lavoisier

TOTAL MASS REACTANTS = TOTAL MASS PRODUCTS

Number of atoms of EACH element is the SAME on both sides of the equation.

Balancing Chemical Equations1. Balance using coefficients after correct formulas

are written. Coefficients are usually the smallest whole number – required

when interpreted at the molecular level

2. Balance atoms one at a time3. Balance the atoms that are combined and

appear only once on each side.4. Balance polyatomics that appear on both sides5. Balance H and O atoms last

NEVER CHANGE SUBSCRIPTS!!!**Count atoms to be sure that the

equation is balanced**

BALANCING Examples

1. sodium + chlorine sodium chloride

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

3. K(s) + H2O(l) KOH(aq) + H2(g)

HOH

4. AgNO3(aq) + Cu(s) Cu(NO3)2(aq) + Ag(s)

5. C5H10(g) + O2(g) CO2(g) + H2O(g)

Interpretation of a Balanced Equation

2Mg(s) + O2(g) 2MgO(s)

2 atoms of solid magnesium react with 1 molecule of oxygen gas to form

2 formula units of solid magnesium oxide

OR2 moles of solid magnesium react with

1 moles of oxygen gas to form 2 moles of solid magnesium oxide

Reaction Ratios:

Classifying Chemical Reactions Pattern for prediction based

on the kind of reactantsA. Combustion or Burning – complete

combustion always produces carbon dioxide and water!

1. HydrocarbonsCxHy + O2 CO2 + H2O

2. AlcoholsCxHyOH + O2 CO2 + H2O

3. SugarsC6H12O6 + O2 CO2 + H2O

C12H22O11 + O2 CO2 + H2O

Synthesis or Composition2/more reactants 1 product

1. Element + Element CompoundA + B AB

2 Na + Cl2 2 NaCl

4 Al + 3 O2 2 Al2O3

SynthesisCompound + Compound

CompoundEXAMPLE 1: metal oxide + carbon dioxide metal carbonate

CaO + CO2 CaCO3

EXAMPLE 2: metal oxide + water a base (hydroxide) Na2O + H2O 2 NaOH

H(OH)

EXAMPLE 3: nonmetal oxide + water an acidSO3 + H2O H2SO4

**Determine oxidation numbers for molecular compounds and oxyacids**

Decomposition Binary Compounds

1. Binary Compound 2 elements AB A + B

2 H2O 2 H2 + O2

2 HgO 2 Hg + O2

Decomposition - Ternary Compounds

Ternary Compound Compound + Element/Compound

EXAMPLE 1: metal chlorate metal chloride + oxygen2KClO3 2KCl + 3O2

EXAMPLE 2: metal carbonate metal oxide + carbon dioxide CaCO3 CaO + CO2

EXAMPLE 3: metal hydroxide metal oxide + water

Mg(OH)2 MgO + H2O (Except Group IA metals)

EXAMPLE 4: acids nonmetal oxide + water H2CO3 CO2 + H2O

EXAMPLE 5: Hydrogen Peroxide 2H2O2 2H2O + O2

Single Replacement or Single Displacement

Element + Compound New Compound + New Element

1. Metals A + BC AC + B

Active metals displace less active metals or hydrogen from their compounds in aqueous solution. Refer to the Activity Series.

a. 2Al + 3CuCl2 2AlCl3 + 3Cu

b. metal + H2O metal hydroxide + H2 An active metal (top of series to calcium) will react with water to form

the hydroxide of the metal and hydrogen gas.

2Na + 2HOH 2NaOH + H2

Single Replacement or Single Displacement

2. Nonmetals D + EF ED + F Cl2 + 2NaBr 2NaCl + Br2

Many nonmetals displace less active nonmetals from combination with a metal

or other cation. Order of decreasing activity is

F2 Cl2 Br2 I2

Double Replacement/Displacement or Metathesis:

Compound + Compound New Compound + New CompoundAB + CD AD + CB

2AgNO3 (aq) + CaCl2 (aq) 2AgCl (s) + Ca(NO3)2 (aq)

Pb(NO3)2 (aq) + 2NaCl (aq) PbCl2 (s) + 2NaNO3 (aq)

The driving force for these reactions is if it produces a

1. A precipitate (ppt): See Solubility Table

2. Water

3. Gas: Only HCl and NH3 are soluble in water. All other gases (CO2 and H2S) are sufficiently insoluble to force a reaction to occur if they are found as a product.

Double Replacement Reactions Solutions of sodium chromate and aluminum acetate are mixed.

3Na2CrO4(aq) + 2Al(CH3COO)3(aq) + Al2(CrO4)3(s) + 6NaCH3COO(aq)

Magnesium hydroxide and ammonium bromideMg(OH)2 + 2 NH4Br MgBr2 + 2 NH4OH

If NH4OH is produced, it breaks up into ammonia and water

Mg(OH)2 + 2 NH4Br MgBr2 + 2 NH3 + 2 H2O

Sulfuric acid and magnesium carbonateH2SO4 + MgCO3 MgSO4 + H2CO3

If H2CO3 is a product, it breaks up into carbon dioxide and water

H2SO4 + MgCO3 MgSO4 + H2O + CO2

Thermochemistry The study of the changes in energy that

accompany a chemical reaction and physical changes.

Chemical Reactions involve changes in energy that result from Bond breaking that requires energy (absorbs)

from the surroundings. Bond making that produces energy (releases) to

the surroundings.

Changes in energy result in an energy flow or transfer.

Types of Reactions

1. Exothermic Reactions: a reaction that releases heat into their surroundings.

Heat is a product of the reaction and temperature of the surroundings increase.

This occurs during bond formation.

Exothermic Reaction

(system)

surroundings

surroundings

surroundingssurroundings

Types of Reactions

2. Endothermic Reactions: a reaction that absorbs heat from the surroundings.

Heat acts as a reactant and temperature of the surroundings decreases.

This occurs during bond breaking.

Endothermic Reaction(system)

surroundings

surroundings

surroundingssurroundings

Energy & Chemical Equations

Coefficients are always interpreted as moles. Physical states are written – influences the overall energy exchanged. Very

specific!

Exothermic – release energy; E productCaCl2(s) Ca+2 (aq) + 2Cl-1(aq) + 88.0kJ

Combustion reactions are ALWAYS exothermic:

C3H8 + 5O2 → 3CO2 + 4H2O + 2043 kJ

Endothermic– absorbs energy; E reactant2NH4Cl(s) + Ba(OH)2·8H2O(s) + 63.9 kJ

BaCl2(s) + 2NH3(g) + 10H2O Rewrite for 1 mole of Cl-1:½ CaCl2(s) ½ Ca+2 (aq) + 1 Cl-1(aq) + 44.0kJ

Heat and Enthalpy Changes Enthalpy (H): the heat content of a

system at constant pressure. Unit: J

Enthalpy Change (H): is the heat absorbed or released in a physical or chemical change at constant pressure.

H = Hproducts ─ Hreactants This can be measured. H is also known as the heat of the reaction. Difference between the stored energy of the

reactants and the products.

Enthalpy Diagrams

#1 #2

#1 #2

a. Which has a higher enthalpy? Products or Reactants

R P

b. Was heat absorbed or released? R A

c. Is this an endothermic or exothermic reaction? Exo Endo

d. Is ΔH for this reaction positive or negative? - +

e. Would the ΔH be on the left or right side of the yield sign? R L

f. Is the reverse reaction exothermic or endothermic? Endo Exo

Rewrite each equation with the heat term in the reaction as a reactant or product –

THERMOCHEMICAL equation:

#1) C3H8 + 5O2 → 3CO2 + 4H2O + 2043 kJ

#2) C + H2O + 113kJ → CO + H2

Enthalpy Diagrams

H (kJ)

Course of Reaction

reactants

products

∆H = +

H (kJ)

Course of Reaction

reactants

products

∆H = -

ExothermicEndothermic

2NH4Cl + Ba(OH)2 8H2O

BaCl2 + 2NH3 + 10H2O

+63.9 kJ

CaCl2

Ca+2 + 2Cl-

-88.0 kJ

Reaction ProgressA. Collision Theory

1. In order for a reaction to occur, the particles must collide

2. A successful or effective collision occurs when

a) The collision is energetic enough b) The particles collide with the correct orientation

3. During a collision, kinetic energy is converted to potential energy

4. The minimum energy needed for a

successful collision = activation energy (Ea)

Reaction Pathways or Potential Energy (heat content)

Diagrams

Reaction Pathways or Potential Energy (heat content)

Diagrams

Answer the following questions based on the potential energy diagram shown here:

1. Does the graph represent an endothermic or exothermic reaction?

2. Label the position of the reactants, products, and activated complex.

3. Determine the heat of reaction, ΔH, (enthalpy change) for this reaction.

4. Determine the activation energy, Ea for this reaction. 5. How much energy is released or absorbed during the

reaction? 6. How much energy is required for this reaction to occur?

Solution1. The graph represents an endothermic reaction 2.   

3. ΔH = +50 kJ. 4. Ea = +200 kJ 5. 50 kJ of energy are absorbed during this

endothermic reaction (this is the value of ΔH)6. 200 kJ of energy are required for this reaction to

occur (Ea).

Practice Sketch a potential energy curve that

is represented by the following values of ΔH and Ea. You may make up appropriate values for the y-axis (potential energy).

ΔHforward = -20 kJ Earev = 80 kJ Activated Complex = 120 kJ Is this an endothermic or exothermic

reaction?

Solution

Based on your diagram, determine: ΔHforward = -20 kJ

Eaforward = +60 kJ Enthalpy of reactants = 60 kJ Enthalpy of products = 40 kJ

Enthalpy Diagram - Formative Assessment #1 Sketch a potential energy curve that

is represented by the following values of ΔH and Ea.

ΔHreverse = -10 kJ Eaforward = +40 kJ Activated Complex = 50 kJ Is this an endothermic or exothermic

reaction?

Enthalpy Diagram - Formative Assessment #1

Based on your diagram, determine:

1. Endo or Exo?2. ΔHforward =

3. Eaforward =

4. ΔHreverse =

5. Eareverse =

Enthalpy Diagram – FA #1Answer

Based on your diagram, determine:

1. Exothermic2. ΔHforward = -20 kJ

3. Eaforward = +60 kJ

4. ΔHreverse = +20 kJ

5. Eareverse = +80 kJ

Enthalpy Diagram - Formative Assessment #2 Sketch a potential energy curve that

is represented by the following values of ΔH and Ea.

ΔHforward = -100 kJ Eareverse = +150 kJ Activated Complex = 200 kJ Is this an endothermic or exothermic

reaction?

Enthalpy Diagram - FA#2Answer:

Activated Complex = 200 kJ

Eareverse = +150 kJ ΔHforward = -100 kJ ΔHreverse = +100 kJ Eaforward = +50 kJ exothermic

Enthalpy Diagram - Formative Assessment #2

Based on your diagram, determine:

1. Endo or Exo?2. ΔHforward =

3. Eaforward =

4. ΔHreverse =

5. Eareverse =

Enthalpy Diagram - FA#2Answer

Based on your diagram, determine:

1. Endo 2. ΔHforward = +200 kJ

3. Eaforward = +300 kJ

4. ΔHreverse = -200 kJ

5. Eareverse = +100 kJ

Calculating ∆H using Bond Energy 2 H2 + O2 2 H2O

Bonds Formed = exothermic (-)

Bonds Broken = endothermic (+)

Using Bond Energy Table, determine ∆H.

∆H = -482 kJ (482 kJ released = exothermic)

Hess’s Law1. The enthalpy change for a reaction is the

sumsum of the enthalpy changes for a series of reactions that addsadds upup to the overall reaction.

2. This is also called the Law of Heat of Law of Heat of Summation (Summation (ΣΣ))

3. This allows you to determine the enthalpy change for a reaction by indirectindirect means when a direct method cannot be done.

Steps for using Hess’s Law1. Write a balanced equation.2. Identify the compounds.3. Locate the compounds on the Heats of Reaction

Table (or given).4. Write the reaction from the table so the

compound is a reactant or product as it appears in the balanced equation.

5. Write appropriate ΔH for each sub equation.a) If needed, multiply the sub equation and the associated

ΔH’s (coefficients).b) If you reverse the equation, change the sign of the

enthalpy change.6. Add the sub equations to arrive at the desired

balanced equation.7. Add ΔH’s of each sub equation to calculate the

ΔH for the desired balanced equation.

Calculate ΔH for the following example:

#1) XeF2 + F2 XeF4 ΔH = ?

Xe + F2 XeF2 ΔH = -123 kJ

Xe + 2F2 XeF4 ΔH = -262 kJ

#2) C + H2O → CO + H2 ΔH = ?

2CO 2C + O2 ΔH = +222 kJ

2H2 + O2 2H2O ΔH = -484 kJ

XeF2 + F2 XeF4 ΔH = -139 kJ

C + H2O → CO + H2 ΔH = +131 kJ

Calculate ΔH for the following example:

#3) CO + O2 → 2 CO2 ΔH = ?

2C + O2 2CO ΔH = -222 kJ

CO2 C + O2 ΔH = +394 kJ

#4) H2O2 + H2 → 2 H2O ΔH = ?

H2O + ½ O2 H2O2 ΔH = +94.6 kJ

2H2 + O2 2H2O ΔH = -484kJ

CO + O2 → 2 CO2 ΔH = -394 kJ

H2O2 + H2 → 2H2O ΔH = -336.6kJ

Calculate ΔH for the following example:

#1) C(s) + H2O(g) → CO(g) + H2(g)

H2O(g) → H2(g) + 0.5O2(g) ΔH = +242.0kJ

C(s) + 0.5O2 → CO ΔH = -111.0kJ

C(s) + H2O(g) → CO(g) + H2(g) ΔH= +131.0kJ

Thermochemical Equation:C(s) + H2O(g) + 131.0kJ → CO(g) + H2(g)

Calculate ΔH for the following example:

#2) 2 CO(g) + O2(g) → 2 CO2(g)

CO(g) → 0.5C2(g) + C(s) ΔH = +111.0 kJ

C(s) + O2(g) → CO2(g) ΔH = -394.0 kJ

2CO(g) + O2(g) → 2CO2(g) ΔH = -566.0 kJ

Thermochemical Equation:2CO(g) + O2(g) → 2CO2(g) + 566.0kJ

2 CO(g) → 2 C(s) + 1 O2(g) ΔH = +222.0 kJ

1 C(s) + 1 O2(g) → 1 CO2(g) ΔH = -394.0 kJ