Honors Chemistry Unit 13: SAT II Prep Welcome To The GowerHour

C H C H I. Organic Chemistry A. Hydrocarbons: Compounds that only contain ______ and _________. 1. Hybridization When carbon forms four single bonds, it is ___ hybridized. It forms a __________ , with a bond angle of _____. It has __ σ bonds and __ π bonds. Examples: CH4, C2H6 carbon hydrogen sp3 tetrahedron 109.5o 4 C H C H

b. When carbon forms two single bonds and one double bond, it is __ hybridized. It forms a _______________ shape, with a bond angle of _____. It has ___σ bonds and ___ π bonds. Example: C2H4 sp2 trigonal planar 120o 3 1 C H

c. When carbon forms a single bond and a triple bond, or two ______ bonds, it is ___ hybridized. It forms a ______ shape, with a bond angle of ____. It has ___σ bonds and ___ π bonds. Example: C2H2 double sp linear 180o 2 2 C H

2. Classes of Hydrocarbons a. Alkanes: Hydrocarbons that contain only ______ bonds. They all have a molecular formula in the general form of: They are named by adding the suffix ____ to a prefix that indicates the number of carbon atoms. Alkanes are said to be __________. (They have the maximum number of ___ bonded to them). # of C atoms Prefix Name of alkane Formula 1 _____ _____________ _______ 2 _____ _____________ _______ 3 _____ _____________ _______ 4 _____ _____________ _______ 5 _____ _____________ _______ 6 _____ _____________ _______ single CnH2n+2 -ane saturated H meth- methane CH4 eth- ethane C2H6 prop- propane C3H8 but- butane C4H10 pent- pentane C5H12 hex- hexane C6H14

b. Alkenes: Hydrocarbons that have _______ bonds b. Alkenes: Hydrocarbons that have _______ bonds. They all have a molecular formula in the general form of: They are named by using the same prefixes as alkanes, except the suffix is _____. Examples: C2H4 ________ C4H8 ________ c. Alkynes: Hydrocarbons that contain ______ bonds. They are named using the same prefixes and alkanes, except the suffix is ____. Alkenes and alkynes are considered to be ___________ hydrocarbons. Examples: C2H2 _________ C3H4 _________ C3H4 structure: double CnH2n -ene ethene butene triple -yne unsaturated ethyne propyne C H

3. Cyclic Hydrocarbons a. Alkanes, alkenes, and alkynes can all form rings. The minimum number of carbons needed to form a ring is ___, though they are not very stable with a ring with fewer than ___ carbons. b. Examples of cyclic hydrocarbons: C6H12 C6H10 C6H8 ____________ ______________ ____________ 4. Aromatics: Certain unsaturated cyclic hydrocarbons are known as aromatics. They have in common that they are cyclic, planar, and have a high degree of stability. The stability comes from the ___________. Example: Benzene 3 5 cyclohexane cyclohexene cyclohexyne resonance C6H6

C H O C H O B. Oxygen-Containing Organic Compounds Alcohols: Contain the functional group –OH. Example: CH3CH2OH (ethanol) CH3OH (methanol)   a. Alcohols have stronger intermolecular forces than hydrocarbons, due to the ________________. Therefore, their boiling points are ______ than the similar hydrocarbon. Example: BP of C3H8 (________) is –42.1°C BP of C3H7OH (________) is 97.4°C C H O • • C H O • • hydrogen bonding higher propane propanol

C H O 2. Ethers: Contain the functional group: -C – O – C- Example: CH3OCH2CH3 (methoxyethane or ethyl methyl ether)   a. Ethers are not capable of hydrogen bonding (O is not bonded to H), so they have _____ boiling points than alcohols. C H O • • lower

3. Aldehydes: Contain the functional group (R group is a Carbon group, ie CH3) ║ R – C – H   4. Ketones: Contain the functional group: O ║ R – C – R

5. Esters: Contain the functional group: ║ R – C – O – R   6. Carboxylic Acids: Contain the functional group: R – C – O – H

C. Organic Review Problems   1. Identify the functional groups in each organic molecule: 1 2 2 3 1 1 1. ketone 1. & 2. ether 1. carboxylic acid 2. alcohol 3. aldehyde

2. What is the hybridization at each carbon in the following molecules:   3. In the above molecules, how many sigma and pi bonds does each carbon have? 1 2 1. sp 2. sp 1 2 1. sp3 2. sp2 1. 4  0  1. 2  2  2. 3  1  2. 2  2 

4. Write the chemical formulas and structural formulas of the following hydrocarbons: (a) butane (b) ethene (c) propyne C H C4H10 C H C2H4 C H C3H4

4. Write the chemical formulas and structural formulas of the following hydrocarbons: Cyclopentane benzene (f) methane H C5H10 C6H6 H C CH4

II. Redox and Electrochemistry A. Oxidation – Reduction: Must occur as a pair. As one substance is oxidized, the other is reduced. 1. Oxidation occurs when a substance _____ electrons, therefore the oxidation state__________. The substance that is oxidized is known as the ________ agent, because it causes something else to be reduced. 2. Reduction occurs when a substance _____ electrons, therefore the oxidation state _________ (reduces). The substance that is reduced is known as the ________ agent, because it causes something else to be oxidized. loses increases reducing gains decreases oxidizing

Rules to assign oxidation numbers: In free elements, each atom has an oxidation state of __. (i.e. H2, Br2, Li, etc.) For monatomic ions, the oxidation state is equal to the ______ on the ion. (i.e. Li+ has an oxidation state of +1) 3. The oxidation state of oxygen in most compounds is ___, but in peroxides its oxidation state is ____. The oxidation state of hydrogen when it is bonded to a non-metal is ___. When it is bonded to a metal its oxidation state is ___. (i.e. HCl = +1; LiH = -1) 5. Fluorine has an oxidation state of ___ in ALL its compounds. Other halogens can have _______ oxidation states when bonded to oxygen. 6. In a neutral compound, the sum of the oxidation states is __. In a polyatomic ion, the sum of the oxidation states is the ______ of the ion. charge – 2 – 1 + 1 – 1 – 1 positive charge

(a) SnCl2 + PbCl4  SnCl4 + PbCl2 3. Example: Assign oxidation numbers to the atoms in the following reactions in order to determine the oxidized and reduced species and the oxidizing and reducing agents: (a) SnCl2 + PbCl4  SnCl4 + PbCl2 (b) Mg + 2 HCl  MgCl2 + H2 + 2 - 1 + 4 - 1 + 4 - 1 + 2 - 1 Sn2+ is oxidized to Sn4+; reducing agent Pb4+ is reduced to Pb2+; oxidizing agent + 1 - 1 + 2 - 1 Mg is oxidized to Mg2+; reducing agent H+ is reduced to H2; oxidizing agent

Balancing Redox Reactions: In order to balance redox reactions, both _____and _______ must be conserved! Each half reaction is balanced separately, then they are added to give the overall reaction. Steps to Balance Redox Reactions in Acidic Solutions a. Assign oxidation numbers to each atom. b. Split the skeleton reaction into half reactions. c. Complete and balance each half reaction. i. Balance all atoms except O and H. ii. Balance O atoms by adding H2O molecules to one side of the rxn. iii. Balance H atoms by adding H+ ions to one side of the rxn. iv. Balance electrons. d. Multiply each half reaction by a factor to balance the electrons. e. Combine half reactions to write the full reaction. Simplify reaction by canceling anything that occurs on both side of the rxn. Steps to Balance Redox Reactions in Basic Solutions a. Balance the rxn as if it were in acidic solution. b. Add as many OH- ions to both sides as there are H+ ions. c. Combine H+ and OH- to form water; simplify the reaction. mass charge

Example: Balance the following redox reaction in an acidic solution: MnO4- + I-  I2 + Mn2+ Oxidation half rxn: Reduction half rxn: + 7 - 2 - 1 + 2 5 2 I-  I2 + 2e- To Balance e- 2 5e- + 8 H+ + MnO4-  Mn2+ + 4 H2O 16 H+ + 2 MnO4- + 10 I-  2 Mn2+ + 5 I2 + 8 H2O

Example: Balance the following redox reaction in an acidic solution: Zn + NO3-  Zn2+ + NH4+   Oxidation half rxn: Reduction half rxn: + 5 - 2 +2 - 3 +1 4 Zn  Zn2+ + 2e- To Balance e- 8e- + 10 H+ + NO3-  NH4+ + 3 H2O 10 H+ + 4 Zn + NO3-  4 Zn2+ + NH4+ + 3 H2O

Example: Balance the following redox reaction in Basic Solution MnO4 + SO32   MnO2 + SO42  Oxidation half rxn: Reduction half rxn: + 7 - 2 + 4 - 2 + 4 - 2 + 6 - 2 3 H2O + SO32–  SO42 + 2 H+ + 2e- 2 3e + 4 H+ + MnO4  MnO2 + 2 H2O In Acidic Soln. 2 H+ + 2 MnO4  + 3 SO32-  2 MnO2 + 3 SO42 + H2O In Basic Soln. 2 H+ + 2 OH + 2 MnO4  + 3 SO32-  2 MnO2 + 3 SO42 + H2O + 2 OH 2 H2O 1 H2O 2 MnO4  + 3 SO32-  2 MnO2 + 3 SO42 + 2 OH H2O +

B. Electrochemical Cells: A contained system in which a redox reaction occurs in conjunction with the passage of an electric current.   1. Galvanic Cells (Voltaic Cells): A redox reaction that occurs _____________ ΔG = __; EMF = __ a. These redox reactions can supply ______ and are used to do _____. b. The oxidation and reduction reactions are placed in separate containers called half-cells. The half-cells are connected by an apparatus that allows for the flow of __. spontaneously – + energy work e–

c. Example of a galvanic cell: Oxidation = anode Reduction = cathode

d. The salt bridge contains an inert salt (ie KCl) which allows for the movement of charged particles (ions). Without the salt bridge, there would be a buildup of electrons in the ________, and excess positive charge would buildup on the ______. e. During the reaction, electrons flow from the zinc anode and through the wire and voltmeter. This electric current can be used for energy. f. Example of a galvanic cell: __________ car battery. cathode anode lead – acid

2. Electrolytic Cells: A redox reaction that is _______________ ΔG = ____; EMF = ____ a. Electrical energy is required to induce reaction (ie battery). b. Oxidation and reduction reactions are generally placed in one container. non-spontaneous + 

c. Example: Electrolysis of Water

Oxidation rxn: 2 H2O  O2 + 4 H+ + 4e Reduction rxn: 4 H+ + 4 e  2 H2 Overall rxn: d. Michael Faraday was the first scientist to define certain quantitative principles governing the behavior of electrolytic cells. Example 1: How many Farradays (F) are required for the reduction of 1 mol of Ni2+ to Ni(s)? Example 2: The gold-plating process involves the following reaction: Au3+ + 3e  Au(s). If 0.600 g of Au is plated onto a metal, how many Coulombs are used? 2 H2O  O2 + 2 H2 1 e- = 1.6 x 10-19 Coulombs (C), amount of charge 1 mol e- = (1.6 x 10-19 C)(6.022 x 1023) = 96,487 C = 1 Faraday (F) 2 mol e = 2 F Ni2+  Ni + 2e 1 mol Au 3 mol e- 96487 C 0.600 g Au x x x = 882 C 196.97 g 1 mol Au 1 mol e-

C. Reduction Potentials and the Electromotive Force (EMF) 1. Electromotive Force is the difference in electric potential between two points, measured in volts.

2. The higher the reduction potential (voltage), the _____ likely the substance will be reduced and act as an ________ agent. Example: Order the following oxidizing agents by increasing strength: NO3, H2O2, Cl2 3. The oxidation potential is the _________ of the reduction potential. Example: Write the half reaction for copper metal oxidizing to the cupric ion. Determine the oxidation potential for this reaction. 4. Using the standard voltages (as in Table 19.1), the voltage of the electrochemical cell can be determined: more oxidizing NO3 < Cl2 < H2O2 + 0.96V + 1.36V + 1.77V negative Cu  Cu2+ + 2 e Eox = 0.34V Eºcell = Eºred + Eºox

5. If the E°cell is positive, the electrochemical cell is spontaneous, therefore it will be a ________ cell. If the E°cell is negative, the electrochemical cell is not spontaneous, therefore it requires an input on energy to react. This is known as an __________ cell. Example 1: Calculate the E°cell for the following reaction and determine if it is a galvanic or electrolytic cell . Al + Cu2+  Al3+ + Cu Example 2: F2 + 2e-  2 F- Ered = +2.87 V Ca2+ + 2e-  Ca Ered = -2.87 V When the above half-reaction are combined in a galvanic cell, which species will be reduced and which will be oxidized? galvanic electrolytic ox. 2 Al  Al3+ + 3e- Eox = +1.66 V red. 3 2e- + Cu2+  Cu Ered = +0.34 V 2 Al + 3 Cu2+  2 Al3+ + 3 Cu Ecell = +2.00 V (galvanic) F2 will be reduced Ecell = +2.87 V + 2.87 V Ca will be oxidized = 5.74 V

Example 3: A galvanic cell is constructed with the following half reactions: Ag+ + e-  Ag Ered = +0.80 V Cu2+ + 2e-  Cu Ered = +0.34 V Calculate the standard cell potential. Example 4: Calculate the Ecell of the electrolysis of water. (Hint: Use the half reactions on page 8). red. 2 e- + Ag+  Ag Ered = +0.80 V ox. Cu  Cu2+ + 2e- Eox = -0.34 V 2 Ag+ + Cu  Cu2+ + 2 Ag Ecell = +0.46 V ox. 2 H2O  O2 + 4 H+ + 4 e- Eox = – 1.23 V red. 4 H+ + 4 e-  2 H2 Ered = 0.00 V 2 H2O  O2 + 2 H2 Ecell = – 1.23 V energy must be added

D. Thermodynamics of Redox Reactions 1. The ΔG of the reaction can be determined using the following equation: ΔG = Gibb’s Free Energy (Must be in J) ΔG = -nFEcell n = # mol e- F = Faraday’s constant = 96,487 C/mol e- Ecell = cell potential (V) (NOTE: 1 C = J/V) 2. Example: What is Ecell for a reaction where the ΔG = -553.91 kJ and 2 electrons are transferred? – 553, 910 J = – (2 mol e-)(96487 J/V mol e-)Ecell Ecell = + 2.87 V

3. Example: Calculate the standard free-energy change for the following reaction at 25°C. Will this reaction occur spontaneously? What type of electrochemical cell is needed? 2 Au + 3 Ca2+  2 Au3+ + 3 Ca ΔG = -nFEcell ox. 2 Au  Au3+ + 3e- Eox = –1.50 V red. 3 2e- + Ca2+  Ca Ered = –2.87 V Ecell = –4.37 V ΔG = –(6 mol e-)(96,487 J/V mol e-)(-4.37 V) ΔG = + 2529889 J = + 2.53 x 103 kJ non-spontaneous  electrolytic cell

4. Example: Calculate the ΔG for the following reaction at 25°C 4. Example: Calculate the ΔG for the following reaction at 25°C. Will this reaction occur spontaneously? What type of electrochemical cell is needed? 2 Al3+ + 3 Mg ↔ 2 Al + 3 Mg2+ ΔG = -nFEcell ox. 2 3e- + Al3+  Al Eox = –1. 66 V red. 3 Mg  Mg2+ + 2e- Ered = +2.37 V Ecell = +0.71 V ΔG = –(6)(96487)(0.71) ΔG = – 411034.6 J = – 4.1 x 102 kJ spontaneous  galvanic cell

III. Colligative Properties of Solutions: Properties that depend only on the ________ of solute particles in solution and NOT on the nature of the solute particles. A. Types of Colligative Properties: 1. Vapor-Pressure Lowering (Raoult’s Law) The greater the number of solute particles, the _____ the vapor pressure of the solvent. 2. Freezing-Point Depression: Dissolving a solute in a solvent will _______ the freezing point of the solvent. Example: salt on icy roads. 3. Boiling-Point Elevation: Dissolving a solute in a solvent will ________ the boiling point of the solvent. Example: antifreeze 4. Osmotic Pressure: When a container is separated into two compartments by a _______________ membrane (one side contains pure water and the other side contains water with a dissolved solute), ______ will flow from the side with __________ to the side with the _______. number lower decrease increase semi-permeable water pure water solution

Only H2O can cross the semi-permeable membrane Pure water solution

a. It is more ________ to equalize their concentrations. b. The water level will ____ in the side with the solution, and _____ in the side with pure water. c. The water level will not rise indefinitely. At some point the pressure due to raised water level will oppose the ______ of water. This pressure is known as the _______ pressure. d. Osmotic pressure is _______ proportional to the molarity of the solution. As long as there is a difference in concentrations between the two sides, _______ will occur. e. If the two solutions are of equal concentration, they are called _______. f. If the two solutions are of unequal concentrations, the more concentrated solution is called _________ and the less concentrated solution is called _________. g. Water will always flow from the LESS concentrated (_________) solution to the MORE concentrated (__________) solution. favorable rise lower influx osmotic directly osmosis isotonic hypertonic hypotonic hypotonic hypertonic

Hypo Isotonic Hyper Hyper Hypo Cell swells Cell shrinks

h. In the above figure, (a) is a cell in an isotonic solution, (b) is a cell in a hypotonic solution, and (c) is a cell in a hypertonic solution. i. Example: When making jam or jelly, a large quantity of sugar is used in the preservation process. This is done to help kill bacteria that may cause botulism. When the bacterial cell is placed in a hypertonic solution, it will ______, and kill the cell. This is known as _________. shrink crenation

B. Calculations with Colligative Properties 1. Vapor Pressure Lowering (Raoult’s Law): P1 = vapor pressure of the solution X1 = mol fraction of the solvent P°1 = vapor pressure of the pure solvent ΔP = the decrease in vapor pressure X2 = mol fraction of the solute 2. Freezing Point Depression: ΔTf = the decrease in freezing point ΔTf = Kfm Kf = the freezing pt. depression constant (°C/m) m = molality of the soln (mol solute/kg solvent)

3. Boiling-Point Elevation: ΔTb = the increase in boiling point ΔTb = Kbm Kb = the boiling pt. elevation constant (°C/m) m = molality of the soln (mol solute/kg solvent) 4. Osmotic Pressure: π = Osmotic Pressure π = MRT M = molarity of the solution R = 0.0821 L atm/ mol K T = Temperature (K)

C. Example Calculations 1. Calculate the vapor pressure lowering when 100.0 g sucrose (C12H22O11) is dissolved in 1000.0 g of water if the vapor pressure at 25.0°C is 23.8 mmHg. 55.49 mol H2O 0.2921 mol sucrose P = (0.005237)(23.8 mmHg) P = 0.125 mmHg

2. (a) At what temperature does a 1. 0 m sugar water solution freeze 2. (a) At what temperature does a 1.0 m sugar water solution freeze? boil? (b) At what temperature does a 3.0 m sugar water solution freeze? boil? (a) ΔTf = Kfm = (1.86 oC/m)(1.0 m) = 1.86 oC – 1.86 oC ΔTb = Kbm = (0.52 oC/m)(1.0 m) = 0.52 oC 100.52 oC (b) 101.56 oC ; – 5.58 oC

3. What is the osmotic pressure of a 0.100 M glucose solution at 25°C?      4. Calculate the FP and BP when 25.0 g glucose (C6H12O6) is dissolved in 90.0 g H2O. π = MRT π = (0.100 M)(0.0821 L•atm/mol•K)(298 K) π = 2.45 atm ΔTf = Kfm = (1.86)(1.542) = 2.87 oC – 2.87 oC ΔTb = Kbm = (0.52)(1.542) = 0.80 oC 100.80 oC

5. What is the molality and boiling point of a solution that freezes at -2.79°C?   6. Calculate the vapor pressure of a solution made by dissolving 218 g of glucose (MW = 180.2 g/mol) in 460.0 mL of water at 30°C. The vapor pressure of pure water at this temperature is 31.82 mmHg. Assume the density of water is 1.00 g/mL. ΔTf = Kfm ΔTb = Kbm = (0.52)(1.50) = 0.78 oC 2.79 = (1.86)m m = 1.50 m BP = 100.78 oC 1.210 mol glucose 25.53 mol H2O P = (0.04525)(31.82 mmHg) P = 1.44 mmHg = 0.04525 P1 = 31.82 – 1.44 = 30.38 mmHg