FTCE Chemistry SAEPreparation Course

Session 4

Lisa BaigInstructor

Course OutlineSession 1

Review Pre TestCompetencies 1 & 2

Session 2Competency 5

Session 3Competency 3

Session 4Competency 4

Session 5Competencies 6, 7 and 8

Post Test

Session Norms

• Respect– No side bars– Work on assigned materials only– Keep phones on vibrate– If a call must be taken, please leave the

room to do so

Homework Review

Any questions from last night?

Chemistry Competencies

1. Knowledge of the nature of matter (11%)2. Knowledge of energy and its interaction with

matter (14%)3. Knowledge of bonding and molecular structure

(20%)

4. Knowledge of chemical reactions and stoichiometry (24%)

5. Knowledge of atomic theory and structure (9%)6. Knowledge of the nature of science (13%)7. Knowledge of measurement (5%)8. Knowledge of appropriate laboratory use and

procedure (4%)

Determining Empirical Formulas

• Say you have 65.0g of compound containing Na and Cl.

• Determine the Empirical Formula if the compound is 39.3% Na and 60.7%Cl

Higher Level Practice

• 1st Step: Convert your percentages to mass of each element present

• Na: (.393)(65.0g)= 25.545g Na

• Cl: (.607)(65.0g) = 39.455g Cl

Higher Level Practice

• 2nd Step: Determine number of moles of each element in the sample

25.545g Na 1 mole = 1.11 mol Na 22.989 g/mol

39.455g Cl 1 mole = 1.11 mol Cl 35.453 g/mol

Higher Level Practice

• 3rd Step: Use these moles to determine the smallest whole number ratio of elements to each other. That is your empirical formula!

1.11 mol Na : 1.11 mol Cl1 mol Na : 1 mol Cl

Empirical Formula = NaCl

Balancing Equations

• __ C3H8 + __ O2 __ CO2 + __ H2O

• __ Ca2Si + __ Cl2 __ CaCl2 + __ SiCl4

• __ C7H5N3O6 __ N2 + __ CO + __ H2O + __ C

• __ C2H2 + __ O2 __ CO2 + __ H2O

• __ Fe(OH)2 + __ H2O2 __ Fe(OH)3

• __ FeS2 + __ Cl2 __ FeCl3 + __ S2Cl2

• __ Al + __ Hg(CH3COO)2 __ Al(CH3COO)3 + __ Hg

• __ Fe2O3 + __ H2 __ Fe + __ H2O

• __ NH3 + __ O2 __ NO + __ H2O

Types of Chemical Reactions• Synthesis

– A+B AB• Decomposition

– AB A + B• Combustion

– Burn in the presence of O2, to form dioxide gas, and other products **(CO2 + H2O)

• Single Displacement– ACTIVITY SERIES– AB + C AC + B

• Double Displacement– AB + CD AD + CB

Predict the Product

CaO + H2O

H2SO3 + O2

CaCO3

KClO3

C6H10 + O2

C6H12O6 + O2

Al + CuCl2

Ca + KCl Na2SO4 + CaCl2

KCl + NaOH

Ca(OH)2

H2SO4

CaO + CO2

KCl + O2

CO2 + H2O

CO2 + H2O

AlCl3 + Cu

No ReactionNaCl + CaSO4

KOH + NaCl

Identifying Redox Reactions2 KNO3(s) 2 KNO2(s) + O2(g)

+1 -1 +1 -1 0H2(g) + CuO(s) Cu(s) + H2O(l)

0 -2 +2 0 2(+1) -2NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)

+1 -1 +1 -1 +1 -1 2(+1) -2H2(g) + Cl2(g) 2HCl(g)

0 0 +1 -1SO3(g) + H2O(l) H2SO4(aq)+6 3(-2) 2(+1) -2 2(+1) -2

Redox

Redox

Not Redox

Redox

Not Redox

Balancing Redox Reactions

• The following unbalanced equation represents a redox reaction that takes place in a basic solution containing KOH. Balance the redox reaction.

Br2(l) + KOH(aq) KBr(aq) + KBrO3(aq)

Br2(l) + KOH(aq) KBr(aq) + KBrO3(aq)Ionic Reaction: Br2 Br- + BrO3

-

0 -1 +5 3(-2)-

Reduction ½ Rxn:Br2 Br-

Br2 + 2e- 2Br-

5(Br2 + 2e- 2Br-)Oxidation ½ Rxn:

Br2 BrO3-

12OH- + Br2 2BrO3- + 6H2O + 10e-

Combined Rxn:5Br2 + 12OH- + Br2 + 10e- 10Br- + 2BrO3

- + 6H2O + 10e- 6Br2 + 12KOH 10KBr + 2KBrO3 + 6H2O 3Br2 + 6KOH 5KBr + KBrO3 + 3H2O

Standard Reduction Potentials in Voltaic Cells

Write the overall cell reaction and calculate the cell potential for a voltaic cell consisting of the following half-cells: an Iron electrode in an Iron (III) Nitrate solution, and a Silver electrode in a Silver(I) Nitrate solution.

• Fe3+(aq)+3e-Fe(s) E0=-0.04V• Ag+(aq)+e-Ag(s) E0=+0.80V• E0

cell= E0cathode- E0

anode

• E0cell= (+0.80 V)- (-0.04 V)= +0.84 V

• E0cell= positive = spontaneous

Acid/Base Properties

• Strong Acids and Bases– Will ionize completely in a solvent

• Weak Acids and Bases– Will ionize partially in a solvent

• Buffer Systems– Solution containing a weak acid, and a

salt of the weak acid• Acetic Acid and Sodium Acetate• Carbonic Acid and Bicarbonate

Break Time

Take a 10 minute

break

Mass-Mass Stoichiometry

3 Cu + 8 HNO3 3 Cu(NO3)2 + 4 H2O + 2NO

• Copper Nitrate is used in creation of some light sensitive papers

• Specialty photographic film

• Your company needs 150 grams of Copper nitrate to fill an order. How many grams of Nitric Acid are needed to undergo reaction?

• Step 3: Compute

150g Cu(NO3)2 1 mole 8 mol HNO3 63.012 g =

187.554g 3 mol Cu(NO3)2 1 mole

134 g HNO3

Gas Stoichiometry

Xenon gas reacts with fluorine gas according to the shown reaction. If a researcher needs 3.14L of XeF6 for an experiment, what volumes of Xenon and Fluorine should be reacted? Assume all volumes are measured under the same temperatures and pressures.

Xe (g) + 3 F2 (g) XeF6 (g)

Gas Stoichiometry

• Xenon3.14L XeF6 1mole 1Xe 22.4L =

22.4L 1XeF6 1 mole

3.14L Xe • Fluorine3.14L XeF6 1 mole 3 F2 22.4L =

22.4L 1 XeF6 1 mole

9.42L F2

Solution Stoichiometry

• How many milliliters of 18.0M Sulfuric Acid are required to react with 250mL of 2.50M Aluminum Hydroxide?

• H2SO4 + Al(OH)3 H2O + Al2(SO4)3

• 3 H2SO4 + 2 Al(OH)3 6 H2O + Al2(SO4)3

250mL Al(OH)3 1L 2.5 mol 3 H2SO4 1L 1000mL 1000mL 1 L 2 Al(OH)3 18.0 mol 1L

52.1 mL H2SO4

Titrations• In a titration, 27.4mL of 0.0154M Ba(OH)2 is added

to a 20.0mL sample of HCl solution with unknown concentration until the equivalence point is reached. What is the molarity of the acid solution?

0.0154M Ba(OH)2 x 27.4mL Ba(OH)2 x 2 mol HCl x 1 = 1 1 mol Ba(OH)2 20.0mL

4.22 x 10-2 M HCl

Limiting Reactant

• The reaction of Ozone with Nitrogen Monoxide to form Oxygen and Nitrogen Dioxide in the atmosphere is responsible for the Ozone hole over Antarctica.

• If 0.960g of Ozone reacts with 0.900g of Nitrogen Monoxide, how many grams of Nitrogen Dioxide are produced?

Limiting Reactant

0.960g O3 1 mole 1 NO2 44.0g NO2

48g O3 1 O3 1 mole

0.880g NO2

0.900gNO 1 mole 1 NO2 44.0g NO2

30g O3 1 O3 1 mole

1.32g NO2

Break Time

Take a 10 minute

break

Chemical Equilibrium

• Chemical Equilibrium– Point in a reversible chemical reaction

when the rate of the forward reaction equals the rate of the reverse reaction.

– The concentrations of its products and reactants remain unchanged

• Le Chatelier’s Principle– If a system at equilibrium is stressed, the

equilibrium is shifted in the direction that relieves the stress

How to Affect Equilibrium• Change in Pressure

– Only affects reactions with gases– Increased pressure increases concentration– Decreased pressure decreases concentration

• Change in Concentration– Of reactants or products.

• Increase one- it moves to the other• Decrease one- it moves towards the one you lowered

• Change in Temperature– Exothermic

• Increase temperature will direct in reverse• Decrease temperature will direct forward

– Endothermic• Increase temperature will direct forward• Decrease temperature will direct in reverse

Equilibrium Constant

nA + mB ↔ xC + yD

K= [C]x[D]y

[A]n[B]m

Factors affecting Reaction Rates

Rate LawsA chemical reaction is expressed by the

balanced chemical equation A + 2B C

Three reaction rate experiments yield the following data.

What is the Rate Law for the Reaction?What is the Order of the reaction with

respect to B?

Experiment Number

Initial[A]

Initial[B]

Initial Rate ofFormation of C

1 0.20 M 0.20 M 2.0 x 10-4 M/min

2 0.20 M 0.40 M 8.0 x 10-4 M/min

3 0.40 M 0.40 M 1.6 x 10-3 M/min

Rate Law for the ReactionA + 2B CR = k[A][B]2

Order of the Reaction with respect to BB is of a 2nd order reactionA is of a 1st order reaction

Calculating pH and pOHpH + pOH = 14 pH = -log[H+] pOH = -log[OH-]

• What is the pH of a 2.5x10-6M HNO3 solution?

• pH = -log [2.5x10-6]• pH = 5.6

Homework

• Diagnostic Exam in your AP Chem Prep book- Page 17-26

• Only answer the questions for these Chapters & Questions– Chapter 6 #6-7, 11– Chapter 7 #14, 16– Chapter 8 #20– Chapter 13 #59– Chapter 14 #63– Chapter 15 #66