1. Chemical Equations & Reactions
Chemistry 6.0

2. Chemical Reactions
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.
Evidence of a Chemical Reaction
Color change
Formation of a precipitate, ppt
Release of a gas
Energy change – heat, light, sound
Odor change
Reactions are started by the addition of energy

4. Chemical Equation Form Reactant + Reactant  Product + Product
Symbols: (s), (l), (g), (aq) NR

6. 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) O2(g)
2 Al(s) + 3 CuCl2(aq)  3 Cu(s) + 2 AlCl3(aq)

7. Characteristics of A Balanced Chemical Equations
The equation must represent known facts. All substances have been identified.
The equation must contain the correct symbols and/or formulas for the reactants and products
Can be either a word equation or a formula equation
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.

9. Balancing Chemical Equations
Balance using coefficients after correct formulas are written.
Coefficients are usually the smallest whole number – required when interpreted at the molecular level
Balance atoms one at a time
Balance the atoms that are combined and appear only once on each side.
Balance polyatomics that appear on both sides
Balance H and O atoms last
NEVER CHANGE SUBSCRIPTS!!!
**Count atoms to be sure that the
equation is balanced**

10. BALANCING Examples sodium + chlorine  sodium chloride
CH4(g) + O2(g)  CO2(g) + H2O(l)
K(s) + H2O(l)  KOH(aq) + H2(g)
HOH
AgNO3(aq) + Cu(s)  Cu(NO3)2(aq) + Ag(s)
C5H10(g) + O2(g)  CO2(g) + H2O(g)

11. 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 OR
2 moles of solid magnesium react with
1 moles of oxygen gas to form
2 moles of solid magnesium oxide
Reaction Ratios:

12. Classifying Chemical Reactions
Pattern for prediction based
on the kind of reactants
Combustion or Burning – complete combustion always produces carbon dioxide and water!
Hydrocarbons
CxHy + O2  CO2 + H2O
Alcohols
CxHyOH + O2  CO2 + H2O
Sugars
C6H12O6 + O2  CO2 + H2O
C12H22O O2  CO2 + H2O

13. Synthesis or Composition
2/more reactants  1 product
Element + Element  Compound
A + B  AB
2 Na + Cl2  2 NaCl
4 Al O2  2 Al2O3

14. Synthesis Compound + Compound  Compound
EXAMPLE 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 acid
SO H2O  H2SO4
**Determine oxidation numbers for molecular compounds and oxyacids**

15. Decomposition Binary Compounds
1. Binary Compound  2 elements
AB  A B
2 H2O  2 H2 + O2
2 HgO  2 Hg + O2

16. Decomposition - Ternary Compounds Ternary Compound  Compound + Element/Compound
EXAMPLE 1: metal chlorate  metal chloride + oxygen
2KClO3  2KCl O2
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 H2O2  2H2O + O2

17. Single Replacement or Single Displacement
Element + Compound  New Compound + New Element
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  2AlCl Cu
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

18. Single Replacement or Single Displacement
2. Nonmetals
D EF  ED F
Cl NaBr  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

19. Double Replacement/Displacement or Metathesis:
Compound + Compound  New Compound + New Compound
AB 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
A precipitate (ppt): See Solubility Table
Water
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.

20. 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 bromide
Mg(OH) NH4Br  MgBr NH4OH
If NH4OH is produced, it breaks up into ammonia and water
Mg(OH) NH4Br  MgBr NH3 + 2 H2O
Sulfuric acid and magnesium carbonate
H2SO4 + MgCO3  MgSO4 + H2CO3
If H2CO3 is a product, it breaks up into carbon dioxide and water
H2SO4 + MgCO3  MgSO4 + H2O + CO2

21. 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.

22. Types of Reactions
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

23. Types of Reactions
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

24. ½ CaCl2(s)  ½ Ca+2 (aq) + 1 Cl-1(aq) + 44.0kJ
Energy & Chemical Equations Coefficients are always interpreted as moles. Physical states are written – influences the overall energy exchanged. Very specific!
Exothermic – release energy; E product
CaCl2(s)  Ca+2 (aq) + 2Cl-1(aq) kJ
Combustion reactions are ALWAYS exothermic:
C3H O2 → 3CO H2O kJ
Endothermic– absorbs energy; E reactant
2NH4Cl(s) + Ba(OH)2·8H2O(s) kJ 
BaCl2(s) + 2NH3(g) + 10H2O
Rewrite for 1 mole of Cl-1:
½ CaCl2(s)  ½ Ca+2 (aq) Cl-1(aq) kJ

25. 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.

26. Enthalpy Diagrams #1 #2

27. #1 #2 R P A Exo Endo - + L Which has a higher enthalpy?
Products or Reactants R P
b. Was heat absorbed or released? 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? L
f. Is the reverse reaction exothermic or endothermic?

28. Rewrite each equation with the heat term in the reaction as a reactant or product – THERMOCHEMICAL equation:
#1) C3H O2 → 3CO H2O kJ
#2) C + H2O kJ → CO + H2

29. Enthalpy Diagrams Endothermic Exothermic -88.0 kJ +63.9 kJ ∆H = -
H (kJ)
Course of Reaction
reactants
products
∆H = +
H (kJ)
Course of Reaction
reactants
products
∆H = -
BaCl2 + 2NH3 + 10H2O
CaCl2
-88.0 kJ
+63.9 kJ
Ca Cl-
2NH4Cl + Ba(OH)2 8H2O
Endothermic
Exothermic

31. Reaction Progress Collision Theory
In order for a reaction to occur, the particles must collide
A successful or effective collision occurs when
The collision is energetic enough
The particles collide with the correct orientation
During a collision, kinetic energy is converted to potential energy
The minimum energy needed for a successful collision = activation energy (Ea)

32. Reaction Pathways or Potential Energy (heat content) Diagrams

33. Reaction Pathways or Potential Energy (heat content) Diagrams

34. Answer the following questions based on the potential energy diagram shown here:
Does the graph represent an endothermic or exothermic reaction?
Label the position of the reactants, products, and activated complex.
Determine the heat of reaction, ΔH, (enthalpy change) for this reaction.
Determine the activation energy, Ea for this reaction.
How much energy is released or absorbed during the reaction?
How much energy is required for this reaction to occur?

35. Solution
The graph represents an endothermic reaction   
ΔH = +50 kJ.
Ea = +200 kJ
50 kJ of energy are absorbed during this endothermic reaction (this is the value of ΔH)
200 kJ of energy are required for this reaction to occur (Ea).

36. 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?

37. Solution Based on your diagram, determine: ΔHforward = -20 kJ
Eaforward = +60 kJ
Enthalpy of reactants = 60 kJ
Enthalpy of products = 40 kJ

38. 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?

39. Enthalpy Diagram - Formative Assessment #1
Based on your diagram, determine:
Endo or Exo?
ΔHforward =
Eaforward =
ΔHreverse =
Eareverse =

40. Enthalpy Diagram – FA #1 Answer
Based on your diagram, determine:
Exothermic
ΔHforward = -20 kJ
Eaforward = +60 kJ
ΔHreverse = +20 kJ
Eareverse = +80 kJ

41. 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?

42. Enthalpy Diagram - FA#2 Answer:
Activated Complex = 200 kJ
Eareverse = +150 kJ
ΔHforward = -100 kJ
ΔHreverse = +100 kJ
Eaforward = +50 kJ
exothermic

43. Enthalpy Diagram - Formative Assessment #2
Based on your diagram, determine:
Endo or Exo?
ΔHforward =
Eaforward =
ΔHreverse =
Eareverse =

44. Enthalpy Diagram - FA#2 Answer
Based on your diagram, determine:
Endo
ΔHforward = +200 kJ
Eaforward = +300 kJ
ΔHreverse = -200 kJ
Eareverse = +100 kJ

45. 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)

46. Hess’s Law
The enthalpy change for a reaction is the sum of the enthalpy changes for a series of reactions that adds up to the overall reaction.
This is also called the Law of Heat of Summation (Σ)
3. This allows you to determine the enthalpy change for a reaction by indirect means when a direct method cannot be done.

47. Steps for using Hess’s Law
Write a balanced equation.
Identify the compounds.
Locate the compounds on the Heats of Reaction Table (or given).
Write the reaction from the table so the compound is a reactant or product as it appears in the balanced equation.
Write appropriate ΔH for each sub equation.
If needed, multiply the sub equation and the associated ΔH’s (coefficients).
If you reverse the equation, change the sign of the enthalpy change.
Add the sub equations to arrive at the desired balanced equation.
Add ΔH’s of each sub equation to calculate the ΔH for the desired balanced equation.

48. 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

49. 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 = kJ
2H2 + O2  2H2O ΔH = -484kJ
CO + O2 → 2 CO2 ΔH = -394 kJ
H2O2 + H2 → 2H2O ΔH = kJ

50. Calculate ΔH for the following example:
#1) C(s) H2O(g) → CO(g) H2(g)
H2O(g) → H2(g) O2(g) ΔH = kJ
C(s) + 0.5O2 → CO ΔH = kJ
C(s) + H2O(g) → CO(g) + H2(g) ΔH= kJ
Thermochemical Equation:
C(s) + H2O(g) kJ → CO(g) + H2(g)

51. Calculate ΔH for the following example:
#2) 2 CO(g) + O2(g) → 2 CO2(g)
CO(g) → 0.5C2(g) + C(s) ΔH = kJ
C(s) + O2(g) → CO2(g) ΔH = kJ
2CO(g) + O2(g) → 2CO2(g) ΔH = kJ
Thermochemical Equation:
2CO(g) + O2(g) → 2CO2(g) kJ
2 CO(g) → 2 C(s) O2(g) ΔH = kJ
1 C(s) + 1 O2(g) → 1 CO2(g) ΔH = kJ