name: ap chemistry 5: thermodynamics and applications of ... · 3/5/2020 · 7 3/16 3/17 unit test...

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Name: Class: AP Chemistry

5: Thermodynamics and Applications of Thermodynamics

Topics/ Daily Outline:

Day A B Content TEXT CW HW 1 2/26 2/27 Measuring heat flow 6.1, 6.2 1 1

2 2/28 3/2 Hand warmer design 6.2 2 --

3 3/3 3/4 Finding ∆H without experiment 6.3, 6.4 3 2 4 3/5 3/6 Spontaneity 17.1 – 17.3 4 3

5 3/9 3/11 Free energy 17.4 – 17.6 5 4

6 3/12 3/13 Free energy and chemical reactions 17.7 – 17.10 6 -- 7 3/16 3/17 Unit Test (CW 1 to CW 6) -- -- --

8 3/18 3/19 Redox reactions 4.9 – 4.10 7 5

9 3/20 3/23 Galvanic cells, Standard reduction potentials 18-1 – 18.2 8 6 10 3/24 3/25 Cell potential, Electrical work, Free energy 18.3 9 7

11 3/26 3/27 Electrolysis 18.7 10 8

12 3/30 3/31 Unit Test (CW 7 to CW 10) -- -- --

For tutorials and additional resources: www.leffellabs.com

If you are absent, use this sheet to determine what you missed and collect the appropriate materials from your instructor. Get help from a friend, the link above, or the instructor.

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HW Assignments

HW 1: Section 6.1 and 6.2 Cornell Notes Due 2/28 (A) & 3/2 (B) Read Sections 6.1 and 6.2 of the text and complete Cornell Notes.

HW 2: Heat and Enthalpy Due: 3/5 (A) & 3/6 (B) Complete through OWL.

HW 3: Spontaneity Due: 3/9 (A) & 3/11 (B) Complete through OWL.

HW 4: Practice Test and Review Due: 3/16 (A) & 3/17 (B) Complete through AP Classroom.

HW 5: Redox Review Due: 3/20 (A) & 3/23 (B) Complete through OWL.

HW 6: Describing Galvanic Cells Due: 3/24 (A) & 3/25 (B) Complete worksheet, hand in during class

HW 7: Spontaneity of Electrochemical Processes Due: 3/26 (A) & 3/27 (B) Complete through OWL.

HW 8: Practice Test and Review Due: 3/30 (A) & 3/31 (B) Complete through AP Classroom.

Important Due Dates

o Hand Warmer Design, 3/16 (A Day) and 3/17 (B Day) o AP Classroom Progress Check 6, 3/9 (A Day) and 3/11 (B Day) o Quarterly Assessment

o Take home test, 4/1 (A Day) and 4/2 (B Day) o Study plan and corrections, 4/3 (A Day) and 4/6 (B Day)

o AP Classroom Progress Check 9, 3/30 (A Day) and 3/31 (B Day)

OWL assignments will close at 11:45 PM on 4/3

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Drills

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CW 1: Energy and Heat Flow

Important Terms and Concepts Energy The capacity to do work or to produce heat.

o Heat, q: Energy transfer between two objects due to a difference in temperature. 𝑞 = 𝑚𝑐∆𝑇

o Work, w: Force acting over a distance. In the diagram below, the energy due to the position of A is used to push the ball up the smaller hill (work).

For a gas, work can be found from the product of pressure and volume change.

𝑤 = −𝑃∆𝑉 Potential Energy, PE Energy due to position, arrangement, or composition

o Water behind a dam: PE is converted to work as water runs turbines to make electricity. o Attractive and repulsive forces: The energy change of a chemical reaction is due to

differences in attractive forces between the nuclei and electrons in the reactants and products.

o During a reaction, bonds are broken and formed, resulting in a change in PE. Kinetic Energy, KE Energy due to motion, depends on mass and velocity.

𝐾𝐸 =1

2𝑚𝑣2

State Function/ Property A property of the system that depends only on its present state – it doesn’t matter how you arrived at the present state.

Altitude is a state function; distance is not

∆PE is a state function

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The Universe We divide the universe into two parts:

o System: What we are studying, often the reactants and products of a chemical reaction. o Surroundings: Everything else.

Endothermic A system, reaction, or process that absorbs heat from the surroundings.

Exothermic A system, reaction, or process that releases heat to the surroundings.

Enthalpy, H Energy that is usable by the system. It is the sum of internal energy and work. At constant pressure, the enthalpy change (∆H) is equal to the heat change (q). Entropy, S Energy that is not usable by the system because it is converted into random thermal motion. Any time the “randomness” of the system increases, what is being indicated is that less energy is usable by the system.

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First Law of Thermodynamics (Law of Conservation of Energy) The total energy of the universe is constant. Energy is not created or destroyed, only transferred between different types of energy, such as heat or work. Second Law The entropy of the universe is always increasing. This is because concentrated energy tends to become dispersed energy in the form of random thermal motion. Third Law Entropy is a measure of the energy that is unable to do work in a system. It has units of change in energy per degree, typically j/K (as a system heats up, random thermal motion increases). Calorimeter Insulated device designed to measure heat flow during a chemical reaction or process.

o Calorimeter constant: The amount of heat absorbed per degree by the device itself; used as a correction factor, units of J/°C.

o Constant pressure calorimetry: Measurement of heat flow under constant pressure (usually of the atmosphere).

o Constant volume calorimetry: Measurement of heat flow under constant volume (achieved using a bomb calorimeter). Because volume is constant, no PV work is done.

A constant pressure calorimeter; open to the atmosphere

A constant volume (bomb) calorimeter. Often used to study combustion reactions. A sample is combusted via electrical ignition wires in the inner “bomb;” this energy is transferred to the water.

Specific heat, c The amount of energy required to raise one gram of a substance by one degree (J/g°C) or the amount of energy required to raise one mole of a substance by one degree (J/mol°C).

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Heat Flow Problems Complete the following on a separate sheet of paper.

1. Calculate the change in internal energy (∆E) for a system undergoing an endothermic process in which 15.6 kJ of heat flows and 1.4 kJ of work is done on the system.

2. Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm.

3. If the internal energy of a thermodynamic system is increased by 300. J while 75 J of expansion work is done, how much heat was transferred and in which direction?

4. When 1 mol of methane (CH4) is burned with oxygen at constant pressure, 890 kJ of energy is released as heat. Calculate the change in enthalpy (∆H) if 5.8 g of methane is burned at constant temperature.

5. When Ba(NO3)2 solution is reacted with Na2SO4 solution, the white solid BaSO4 forms. Use the data below to determine the enthalpy change per mole of BaSO4 formed.

Ba(NO3)2 solution 50.0 mL of 1.00 M Ba(NO3)2

Na2SO4 solution 50.0 mL of 1.00 M Na2SO4 Tinitial (both solutions) 25.0°C

Tfinal (final solution, mixed) 28.1°C

Specific heat of final solution 4.184 J/g°C (dilute solution = same as H2O)

Density of final solution 1.0 g/mL (dilute solution = same as H2O)

6. Are the following processes endothermic or exothermic? a. When solid KBr is dissolved in water, the solution gets colder. b. Water condensing on a cold pipe c. CO2(g) → CO2(s) d. F2(g) → 2F–(g)

7. When 40.0 mL of water at 60.0 °C is added to 40.0 mL at 25.0 °C water in a calorimeter, the final temperature is 40.0 °C. What is the calorimeter constant?

8. A 1.013 g sample of vanillin (C8H8O3) is burned in the same calorimeter, causing the temperature of the 1000.0 g water inside to increase from by 6.02°C. What is the energy of combustion per mole of vanillin?

9. How many kJ are absorbed by 25 g of H2O as the temperature rises from 90°C to 110°C? Useful Constants

𝑐𝑠𝑡𝑒𝑎𝑚 = 1.996𝐽

𝑔℃

cwater = 4.184J

g℃

cice = 2.087J

g℃

∆𝐻𝑓𝑢𝑠 = 6.02𝑘𝐽

𝑚𝑜𝑙

∆Hvap = 40.7kJ

mol

1 mol H2O = 18.0 g

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CW 2: Designing a Hand Warmer

Challenge You will use chemistry to design an effective, safe, environmentally benign, and inexpensive hand warmer. The ideal hand warmer increases in temperature by 20°C (but no more) as quickly as possible, uses about 50 grams of water, costs as little as possible to make, and uses chemicals that are as safe and environmentally friendly as possible. You will carry out an experiment to determine which substances, in what amounts, to use to make a hand warmer that meets these criteria.

Background People put hand warmers inside their gloves on cold days to keep their fingers warm. They are popular with people who work outside in winter or engage in winter sports. One type of hand warmer consists of a packet that contains water in one section and a soluble substance in another section. When the packet is squeezed the water and the soluble substance are mixed, the solid dissolves and the packet becomes warm. Breaking chemical bonds or particulate attractions absorbs energy from the surroundings, while forming new bonds and particulate attractions releases energy to the surroundings. When an ionic solid dissolves in water, ionic bonds between cations and anions and hydrogen bonds between water molecules are broken. New attractions between water molecules and the cations and anions are formed. The amount of energy involved to break these bonds and form new ones depends on the chemical properties of the anions and cations. Therefore, when some ionic solids dissolve, more energy is required to break the ion-ion attractions than is released in forming the new water-ion attractions, and the overall process absorbs energy in the form of heat. If this occurs at constant pressure, the change in enthalpy is the same as the heat change (q), and the process is endothermic. When other ionic compounds dissolve, bond making releases more energy than bond breaking absorbs, and the process releases heat and is exothermic. In this experiment, you will collect data to calculate the enthalpy of solution (∆Hsoln, kJ/mol solute) occurring in aqueous solution. The data necessary to calculate the heat of solution can be obtained using a calorimeter, a device designed to isolate the system and surroundings from the rest of the universe so that heat flow may be studied. This is accomplished by insulating the calorimeter to prevent heat loss.

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In the calorimeter shown, a chemical reaction (system) occurs in water (surroundings). According to the law of conservation of energy, energy cannot be created or destroyed, only changed between forms or transferred from one system to another. The thermometer measures the temperature of the water, which can be used to determine the heat absorbed or released by the reaction.

∆𝐻𝑤𝑎𝑡𝑒𝑟 = −∆𝐻𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 Recall that at constant pressure, the enthalpy change (∆H) is equal to the heat change (q).

𝑞𝑤𝑎𝑡𝑒𝑟 = −∆𝐻𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛

𝑚𝑤𝑎𝑡𝑒𝑟 (4.184𝐽

𝑔℃) ∆𝑇𝑤𝑎𝑡𝑒𝑟 = −∆𝐻𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛

During the experiment, the calorimeter itself may absorb (or release) a small amount of heat. The amount of heat gain (or loss) by the calorimeter can be used as a correction factor in heat calculations, known as the calorimeter constant. The calorimeter constant is determined by mixing hot and cold water in the calorimeter. When hot and cold water are mixed, the hot water transfers some of its heat to the cool water. The law of conservation of energy dictates that the amount of heat lost by the hot water, qhot, is equal to the heat gain of the cool water, qcold, but opposite in sign. Using the mass of water, the temperature change, and the specific heat of water (4.184 J/g°C), qhot and qcold may be determined.

𝑞ℎ𝑜𝑡 = −𝑞𝑐𝑜𝑙𝑑

𝑚ℎ𝑜𝑡 (4.184𝐽

𝑔℃) ∆𝑇ℎ𝑜𝑡 = −𝑚𝑐𝑜𝑙𝑑 (4.184

𝐽

𝑔℃) ∆𝑇𝑐𝑜𝑙𝑑

The difference in these two numbers is due to heat loss to the calorimeter. The calorimeter constant (C) is found by dividing the heat loss to the calorimeter by the temperature change of the calorimeter.

𝐶 =𝑒𝑛𝑡ℎ𝑎𝑙𝑝𝑦 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑎𝑙𝑜𝑟𝑖𝑚𝑒𝑡𝑒𝑟

𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑎𝑙𝑜𝑟𝑖𝑚𝑒𝑡𝑒𝑟

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Guided Inquiry Design and Procedure 1. Obtain a calorimeter from your teacher and collect the following data using the

temperature probe and a scale. Use it to find the calorimeter constant. a. Obtain and wear goggles. b. Record the mass of an empty cup. Half fill the cup with cold water and record its

mass. Record the temperature of the cold water. Repeat for a cup of hot water. c. Put the temperature probe in the hot water, wait until the temperature remains

constant. Click to begin data collection. d. Pour the cold water into the hot water, stirring with the temperature probe until

the temperature stops changing. Click to finish collecting data.

e. Click the statistics button, . The final and initial temperatures are listed in the floating box on the graph.

Mass of hot cup (g)

Mass of hot cup and hot water (g)

Mass of hot water (g)

Temperature of hot water just before mixing (°C)

Mass of cold cup (g)

Mass of cold cup and cold water (g)

Mass of cold water (g)

Temperature of cold water just before mixing (°C)

Final temperature, mixed hot and cold (°C)

ΔT for hot water

ΔT for cold water

Heat lost by hot water (J)

Heat gained by cold water (J)

Heat gained by calorimeter (J)

Calorimeter constant (J/°C) (use ΔT for hot water)

2. Should this be added to or subtracted from the calculated heat of solution? Why?

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3. Determine the heat of solution of baking soda (NaHCO3) a. Obtain and wear goggles. b. Measure and record the mass of an empty cup. Half fill the cup with distilled

water. Measure and record the mass of the cup and distilled water. c. Connect the probe to the computer. Open LoggerPro. Prepare the program for

data collection by opening the file “04 Heat of Fusion” from the Chemistry with Vernier folder.

d. Place the temperature probe into the distilled water in the Styrofoam cup. Wait

until the temperature remains constant. Click to begin data collection. e. Add about 10 g of NaHCO3(s) to the water, stirring with the temperature probe

until the temperature stops changing. Click to finish collecting data.

f. Click the statistics button, . The final and initial temperatures are listed in the floating box on the graph.

g. Rinse, dry, and neatly return all materials to the bin.

Mass empty cup (g)

Mass cup with water (g)

Mass of water (g)

Mass of solid NaHCO3 (g)

Moles of solid NaHCO3 (mol)

T1 of water (°C)

T2 of water (°C)

ΔT of water (°C)

Heat transferred to/from water (kJ) q = mcΔT c(water) = 4.18 J/g°C

Corrected heat transferred (using the calorimeter constant)

Heat of Solution for NaHCO3 (kJ/mol)

4. How many moles of this solid are required to get a ΔT of 10.0°C when added to 50.0 g of

H2O?

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Handwritten Lab Notebook PRE-LAB PRE-LAB PRE-LAB PRE-LAB PRE-LAB PRE-LAB PRE-LAB

1. Introduction: Explain theory and applications behind the lab, as well as any chemical equations, graphs, diagrams, or mathematical equations that will be required.

o Explain the criteria for the hand warmer design o Describe the hand warmer packet, why mixing the sections releases heat o Explain calorimetry, why and how to find the calorimeter constant

2. Safety Data Sheets: Look up the SDS for all chemicals used during the lab use the table below to list major hazards (typically in the Hazards Identification section).

Chemical Eye Irritant Skin Irritant Respiratory Irritant Other Info NH4NO3

Na2CO3

CaCl2 NaCl

LiCl NaC2H3O2

Summary of Safety Precautions: 3. Materials and Methods: Complete the corresponding classwork in the unit packet.

Write a procedure listing all chemicals and equipment. Do not worry about sizes of glassware, concentrations/ amounts of chemicals, or detailed LoggerPro instructions.

o Write a procedure for finding the calorimeter constant o Write a separate procedure for determining ∆Hsoln for each solid

PRE-LAB PRE-LAB PRE-LAB PRE-LAB PRE-LAB PRE-LAB PRE-LAB

4. Procedural Notes: Record details of the procedure, exact sizes of glassware, concentrations or amounts of chemicals, instructions for using LoggerPro, and any other notes provided by your instructor or group members. Finalize your procedure.

5. Data Collection: Neatly record any measurements, written with the correct precision based on the equipment, detailed observations and possible sources of error.

o Calorimeter Constant Data: mass of hot water, mass of cold water, initial temperature of hot and cold water, final temperature of the mixture

o Hand Warmer Data: solid, cost per gram of solid, molar mass of solid, exact mass of water, initial water temperature, final water temperature

6. Calculations: Show one handwritten worked example for each major required calculation. Include the unrounded answer and the answers with correct sig figs. You do not need to show sample calculations for finding averages or standard deviations.

o Calculate your calorimeter constant. This should have units of J/°C. o Calculate the heat of solution for each of your solids, taking the calorimeter

constant into consideration. This should in in units of kJ/mol of solid. o Using the actual values supplied by the instructor, find the percent error in the

heat of solution for each salt.

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Typed Lab Report 7. Typed Conclusion: Include the following in narrative form:

o Discussion of what you did, including the purpose/ goal of the experiment. Describe any changes to the procedure from what you wrote in your lab notebook.

o Based on the heat of solution data and hand warmer criteria, describe your hand warmer design. Be specific with amounts and include calculations for the mass of solid required.

o Comment on error, both inherit error (due to precision of equipment) and experimental/ human error. Remember that “the lab would have been better if we did it correctly” or “human errors were made” are not discussion of error. Errors should be clearly described along with how they impacted the calculations in the data analysis.

o Describe refinements or future experiments. What can we study next and why? 8. Cover Sheet: Create a cover sheet using the exact table below. Place cover sheet on top

of items 4 to 8 from above. Report Title: Hand Warmer Design

My Name:

Partners: Class Period:

Due Date:

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CW 3: Finding ∆H without Experiment

For a reaction studied under constant pressure, we can use calorimetry to find the enthalpy change. Sometimes it is difficult to use this process. The reaction may be too slow, or it may be an intermediate step in a series of reactions. These pen and paper methods provide another way to find ΔH.

Hess’ Law Hess’s law of heat summation allows you to determine the heat of reaction indirectly using known heats of reaction of two or more thermochemical equations. This is possible because enthalpy is a state function. Consider the conversion of diamond to graphite. This reaction is too slow to measure directly.

C(s, diamond) → C(s, graphite) ΔH = ? (a) The enthalpy change is known for the following reactions:

C(s, diamond) + O2(g) → CO2(g) ΔH = -393.5 kJ (b) C(s, graphite) + O2(g) → CO2(g) ΔH = -395.4 kJ (c)

Write reaction (c) in reverse, starting with products and creating reactants. Because you reversed the direction of the reaction, you must also reverse the sign of the ΔH value.

ΔH = (d)

Write reaction (b) and the reversed reaction (d) in the boxes below. Combine and cancel reactions to get reaction (a) and find the value of ΔH.

(b)

ΔH =

(d)

ΔH =

(a)

ΔH =

Normal algebraic relationships may be applied to chemical equations. For example, 1 mole of A left of the arrow cancels 1 mole of A right of the arrow, leaving behind 2 moles of A.

3A + B → C + D C + D → 1A + E 2A + B → E

You may multiply a reaction by any whole positive number, so long as you multiply the ΔH by the same whole number.

A + B → C ΔH = 10 kJ 2A + 2B → 2C ΔH = 20 kJ

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1. Given the following data

Calculate ∆H for the reaction

2. Given the following data:

2ClF(g) + O2(g) ⇌ Cl2O(g) + F2O(g) ∆H = 167.4 kJ 2ClF3(g) + 2O2(g) ⇌ Cl2O(g) + 3F2O(g) ∆H = 341.4 kJ

2F2(g) + O2(g) ⇌ 2F2O(g) ∆H = –43.4 kJ

Calculate the ∆H for the reaction: ClF(g) + F2(g) ⇌ ClF3(g)

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Standard Enthalpy of Formation The standard enthalpy of formation (∆Hf

°) of a compound is defined as the change in enthalpy that accompanies the formation of 1 mole of a compound from its elements with all substances in their standard states (denoted by the degree sign). The standard state for a substance is a precisely defined reference state. Because enthalpy is a state function, the difference in these standard reference states can be used to determine the enthalpy of a reaction. The change in enthalpy for a reaction can be calculated by subtracting the enthalpies of formation of the reactants from the enthalpies of formation of products. Remember to multiply the enthalpies of formation by their molar coefficients from the balanced chemical equation. An important note is that ∆Hf

° for an element in its standard state (whatever state it would be at 1 atm and 25°C) is zero.

∆𝐻𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛° = ∑ 𝑛𝑝∆𝐻𝑓

° (𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠) − ∑ 𝑛𝑟∆𝐻𝑓° (𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠)

Here are some standard enthalpies for formation for several compounds. A more comprehensive table is found in Appendix 4 of your text.

3. Calculate the standard enthalpy change for the reaction that occurs when ammonia is burned in air to form nitrogen dioxide and water. This is the first step in the manufacture of nitric acid.

4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(l)

4. Calculate the standard enthalpy change for the thermite reaction. 2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(s)

Compound ∆Hf° (kJ/mol)

NH3(g) –46 NO2(g) 34

H2O(l) –286 Al2O3(s) –1676

Fe2O3(s) –826

CO2(g) –394 CH3OH(l) –239

C8H18(l) –269

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Bond Enthalpy and Energy Diagrams The bond enthalpy is the energy required to break a chemical bond, given in units of kJ/mol. The table below contains some common bond enthalpies.

Single Bonds (kJ/mol)

C–H 413 N–H 391 O–H 463 F–F 155 C–C 348 N–N 163 O–O 146 C–N 293 N–O 201 O–F 190 Cl–F 253 C–O 358 N–F 272 O–Cl 203 Cl–Cl 242 C–F 485 N–Cl 200 O–I 234 C–Cl 328 N–Br 243 Br–F 237 C–Br 276 Br–Cl 218 C–I 240 H–H 436 S–H 339 Br–Br 193 C–S 259 H–F 567 S–F 327 H–Cl 431 S–Cl 253 I–Cl 208 Si–H 323 H–Br 366 S–Br 218 I–Br 175 Si–Si 226 H–I 299 S–S 266 I–I 151 Si–C 301 S–N 464 Si–O 368

Multiple Bonds (kJ/mol)

C=C 614 N=N 418 O=O 495 C≡C 839 N≡N 941 C=N 615 N=O 607 S=O 523 C≡N 891 S=S 418 C=O 799 C≡O 1072

To find the ΔH of a reaction by bond enthalpies, you need only to apply the correct sign and add the ΔH of the bonds being broken and the bonds being formed.

• Bond breaking is an endothermic process (+ΔH)

• Bond forming is an exothermic process (-ΔH)

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5. Consider the combustion of propane, shown below. a. Use bond enthalpies to determine the ΔH for the combustion of propane.

Endothermic Change for Bond Breaking Exothermic Change for Bond Forming

Bond # Bond Enthalpy Product Bond # Bond Enthalpy Product

C-H

-----

C=O -----

C-C

O-H

O=O

TOTAL:

TOTAL:

b. Complete the energy diagram for the combustion of propane. Include the activation energy, ΔH, and the energy released during the formation of bonds.

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6. Find the ΔH and complete the enthalpy diagram for the reaction below.

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CW 4: Spontaneity

Many physical and chemical processes proceed naturally in one direction, but not the other. Consider the reaction between zinc and hydrochloric acid:

Zn(s) + HCl(aq) ⇌ ZnCl2(aq) + H2(g) When zinc is added to the acid, it begins reacting immediately, forming H2 gas bubbles and ions in solution. We do not observe the reverse reaction – it would seem very odd for the bubbles to form at the surface of the solution and sink down as the metal zinc strip reformed. Reactions are spontaneous in the direction in which they naturally occur. What factor(s) determine the direction in which reactions are spontaneous?

A Ball Rolls Downhill

1. Considering the diagram above: a. Which is the lower energy state: the ball at the top of the hill or at the bottom?

b. Is the change in potential energy for this process positive, negative or zero?

c. Why doesn’t the ball roll uphill?

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Solid NaCl is Formed from Gaseous Ions

2. Based on the diagram: a. Which is the lower energy state: 1 mol of solid NaCl or 1 mol of Na+ and Cl– ions?

b. Is the H for this process Na+(g) + Cl– (g) ⇌ NaCl(s) positive, negative, or zero?

c. Why doesn’t a salt crystal suddenly become gaseous sodium and chloride ions?

3. Write a chemical equation that describes the freezing of water, and indicate whether ΔH for the process is positive, negative, or zero.

a. Under what temperature conditions does this process naturally occur?

b. Is it possible to determine whether a process will occur naturally solely by examining the sign of ΔH for the process? Explain.

c. What other factor(s) besides the sign of ΔH must be considered to determine if a process will occur spontaneous under a given set of conditions?

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Disorder Drives the Universe Every developers of thermodynamics thought that a process would be spontaneous if it was exothermic. This explanation is not complete, as it cannot account for the melting of ice at room temperature, an endothermic process. If this is the answer, why does the melting of ice occur spontaneously at room temperature, an endothermic process? The characteristic common to all spontaneous processes is an increase in entropy (S). Entropy can be viewed as a measure of molecular randomness or disorder. The natural order is from low entropy to higher entropy (more disorder). Entropy is a thermodynamic function that describes the number of arrangements that are available to a system and is closely related to probability. We can use this to explain why the gas below spontaneously expands to fill both bulbs.

4. Considering the arrangements above:

a. Which arrangement would you expect to find in the real world?

b. Which arrangement has higher entropy?

c. How do Questions 4a and 4b relate to each other and to probability? Explain.

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Arrangement Microstates # Available

States Probability

I (4 in Left)

1 1/16 = 0.0625

II (3 in Left

1 in Right)

4

III (2 in Left

2 in Right)

6

IV (1 in Left

3 in Right)

4

V (4 in Right)

1

5. The diagram above shows all possible arrangements (states) of four gas particles in a

two-bulb flask. a. Which arrangement has the highest entropy? I II III IV V

b. Calculate the probability of encountering each possible arrangement. Which of

these states is most probable?

c. If two more particles are added, how would the probability for encountering arrangement I (all on the left) change? Arrangement III (evenly distributed)?

d. Return to Question 4. Why is B more likely? Explain.

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We have seen that processes are spontaneous when they result in an increase in disorder. Nature always moves towards the most probable state available to it. This leads to the second law of thermodynamics: the entropy of the universe is increasing. We can divide the universe into system and surroundings:

∆𝑆𝑢𝑛𝑖𝑣 = ∆𝑆𝑠𝑦𝑠 + ∆𝑆𝑠𝑢𝑟𝑟

o If ΔSunvi is positive, the entropy of the universe increases, and the process is spontaneous in the direction written.

o If ΔSunvi is negative, the process will be spontaneous in the opposite direction. o If ΔSunvi is zero, the process has no tendency to occur, and the system is at equilibrium.

6. In a living cell, large molecules are assembled from simple ones. Is this process

consistent with the second law of thermodynamics?

Molecules in a Box

7. Which system has the greatest entropy?

8. Suppose that at some temperature the process above starts with A and ends with C.

a. Is SA > SC or is SC > SA?

b. Is ΔS for this naturally occurring process positive or negative?

c. Suppose that at some different temperature, the process starts with C and ends with A. Is ΔS for this naturally occurring process positive or negative?

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The following generalizations are useful to find the change in entropy (ΔS) for a process: o As the number of particles increases, disorder increases. (+ΔS) o As the volume in which particles can move increases, disorder increases. (+ΔS) o As the temperature (KE) of particles increases, disorder increases. (+ΔS)

9. For each of the following, is ΔS positive or negative? Explain.

a. C4H8(g) ⇌ 2C2H4(g)

b. C4H8(g) ⇌ C4H8(s)

c. C4H8(g, 298K, 1 atm) ⇌ C4H8(g, 298K, 0.5 atm)

d. C4H8(g, 298K, 1 atm) ⇌ C4H8(g, 398K, 1 atm)

Sodium Chloride Dissolves in Water

10. Considering the dissolving of solid NaCl in water: a. Is this process endothermic or exothermic?

b. Does entropy increase or decrease? Explain in terms of the generalizations above.

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The Relationship Between ΔH, ΔS, and T ΔH ΔS Occurs @ Higher T? Occurs @ Lower T?

+ + Yes No + – No No

– + Yes Yes

– – No Yes

11. In the table above: a. Which row above corresponds to the melting of ice? b. Which row above corresponds to the freezing of ice? c. Is the data consistent with the statement that exothermic reactions tend to be

spontaneous? Explain.

d. For what values of ΔS do chemical reactions occur spontaneously? How does this relate to the second law of thermodynamics?

12. Under what conditions will an endothermic reaction be spontaneous?

13. For the following, indicate the driving forces that make the process spontaneous.

Process Entropy Enthalpy Both When NH4NO3(s) dissolves in water the temperature of the solution decreases.

When concentrated sulfuric acid is added to water, the acid solution is diluted and the temperature of the solution increases.

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Recall that the change entropy of the universe is found from the change in entropy of the system and surroundings:

∆𝑆𝑢𝑛𝑖𝑣 = ∆𝑆𝑠𝑦𝑠 + ∆𝑆𝑠𝑢𝑟𝑟

Case ΔSsys ΔSsurr ΔSunvi A + + +

B – – –

C + – ? D – + ?

14. Considering each case:

a. Why is case A spontaneous? Why is case B nonspontaneous? Explain.

b. Can you determine if cases C and D are spontaneous or not? Explain.

15. The boiling of water can be expressed as: H2O(l) → H2O(g) a. Is the boiling of water endothermic or exothermic?

b. Is the change in entropy for this system (ΔSsys) positive or negative?

c. What values of ΔSsurr allow this process to occur spontaneously?

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CW 5: Free Energy

Another Way to Predict Spontaneity So far, we have used the change in entropy to predict if a process occurs naturally (spontaneously). Another thermodynamic function, free energy (G), is useful in dealing with the temperature dependence of spontaneity. For a process that occurs at constant temperature, the change in free energy is found by:

∆𝐺 = ∆𝐻 − 𝑇∆𝑆

Where ΔG: Change in free energy ΔH: Change in enthalpy T: Kelvin temperature ΔS: Change in entropy

A process (at constant T and P) is spontaneous for all values of T that result in a negative free energy. These conditions of constant T and P are encountered often by chemists, so this equation is very useful.

Melting Ice The data below is for the melting of ice at three different temperatures:

T (°C) T (K) ΔH° (J/mol) ΔS° (J/mol∙K) TΔS° (J/mol) ΔG° = ΔH° – TΔS°

–10

6030 22.1

0

6030 22.1

+10

6030 22.1

Note: the degree sign (°) indicates that all substances are in their standard states (CW 3).

1. Complete the chart to find ΔG. a. Based on your calculations, at which temperature is this process spontaneous?

b. Does this match observations you have about melting ice? Explain.

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2. Use the free energy equation to explain how a process can be … a. endothermic and spontaneous.

b. exothermic and nonspontaneous.

3. Which of the following processes are spontaneous under the given conditions? a. ΔH = +25 kJ, ΔS = +5.0 J/K, T = 300. K b. ΔH = +25 kJ, ΔS = +100. J/K, T = 300. K c. ΔH = –10. kJ, ΔS = +5.0 J/K, T = 298 K d. ΔH = –10. kJ, ΔS = –40. J/K, T = 200. K

4. At what temperatures will the following processes be spontaneous? a. ΔH = –18 kJ and ΔS = –60 J/K b. ΔH = +18 kJ and ΔS = +60 J/K c. ΔH = +18 kJ and ΔS = –60 J/K d. ΔH = –18 kJ and ΔS = +60 J/K

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Entropy Changes in Aqueous Solutions Recall the concept of like dissolves like. This implies that an ionic solute (such as NaCl) should dissolve in a polar solvent (such as H2O). This process can be described as a series of steps:

o Breaking solute-solute attractions (endothermic), such as bond energy

o Breaking solvent-solvent attractions (endothermic), such as H–bonding

o Forming solvent-solute attractions (exothermic), in solvation

5. Thermodynamic data for several solution processes are given below.

Solute Solvent ΔHsoln ΔS T ΔG Polar Polar + or – + 298 –

Nonpolar Nonpolar + or – + 298 – Nonpolar Polar + or – – 298 +

Polar Nonpolar + or – – 298 +

a. Which processes are spontaneous?

b. Based on the data, which is a better predictor of the sign of ΔG: ΔH or ΔS?

c. Why does NaCl dissolve in water despite having ΔHsoln = 3 kJ/mol?

d. In terms of ΔS, use the diagram below to explain why a nonpolar solute won’t dissolve in a polar solvent.

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6. Thermodynamic data is given below for the dissolving process of several salts. Process ΔS ΔH Soluble in H2O? ΔG

KCl → K+(aq) + Cl–(aq) + +

LiF → Li+(aq) + F–(aq) – –

CaS → Ca2+(aq) + S2–(aq) – +

a. Use solubility rules to identify each salt as soluble or insoluble in water.

b. Based on your previous answer, will ΔG will be + or – for each process?

c. Explain why some salts are soluble despite negative entropy.

d. How does increasing the temperature effect the value of ΔG? Does this increase or decrease the solubility of a salt?

Entropy Changes in Chemical Reactions Because entropy is a state function of a system, the change in entropy for a given chemical reaction can be calculated by taking the difference between the standard entropy values of products and reactants:

∆𝑆𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛° = ∑ 𝑛𝑝 𝑆𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠

° − ∑ 𝑛𝑟 𝑆𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠°

7. Calculate ΔS° at 25°C for the reaction using the given standard entropy values.

2NiS(s) + 3O2(g) → 2SO2(g) + 2NiO(s)

Substance S° (J/mol∙K)

SO2(g) 248 NiO(s) 38

O2(g) 205 NiS(s) 53

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Finding ΔG for Chemical Reactions

Method 1: Free Energy Equation

8. Consider the reaction 2SO2(g) + O2(g) → 2SO3(g)

carried out at 25°C and 1 atm. Calculate ΔH°, ΔS°, ΔG° using the following data: Substance Hf° (kJ/mol) S° (J/mol∙K)

SO2(g) –297 248

SO3(g) –396 257 O2(g) 0 205

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Method 2: Like Hess’ Law

Free energy is a state function (like enthalpy), allowing us to find ΔG° much like Hess’ law.

9. Determine the free energy change for the reaction: 2CO(g) + O2(g) → 2CO2(g)

from the following data: 2CH4(g) + 3O2(g) → 2CO(g) + 4H2O(g) ΔG° = –1088 kJ

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) ΔG° = –801 kJ

Method 3: Standard Free Energy of Formation

∆𝐺𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛° = ∑ 𝑛𝑝∆𝐺𝑓

° (𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠) − ∑ 𝑛𝑟∆𝐺𝑓° (𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠)

10. Methanol is a high-octane fuel used in high-performance racing engines. Calculate ΔG°

for the reaction: 2CH3OH(g) + 3O2(g) → 2CO2(g) + 4H2O(g)

Given the following free energies of formation:

Substance ΔGf° (kJ/mol) CH3OH(g) –163

O2(g) 0

CO2(g) –394 H2O(g) –229

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CW 6: Thermodynamics and Equilibrium

Free Energy and Equilibrium We have seen that when the products are at a lower enthalpy than the reactants (–ΔH°, exothermic), a chemical reaction is energetically favored. We have also seen that when the products are more disordered than the reactants (+ΔS°), a chemical reaction is entropically favored.

Reaction ΔH° (kJ/mol) Enthalpy

favorable? ΔS° (J/mol∙K)

Entropy favorable?

NaCl(s) ⇌ Na+(aq) + Cl–(aq) 3.86 43.3

NH4NO3(s) ⇌ NH4+(aq) + NO3

–(aq) 28.07 108.6

Zn(s) + Cu2+(aq) ⇌ Cu(s) + Zn2+(aq) –218.67 –21.0 2Cl–(aq) + Br2(l) ⇌ 2Br–(aq) + Cl2(g) 91.23 106.6

CH3COOH ⇌ CH3COO–(aq) + H+(aq) –0.25 –92.2

1. For each of the reactions in the table… a. indicate if the reaction is favorable with respect to the change in enthalpy and the

change in entropy. Enter Y or N into the table.

b. Can you predict, from the given data, which processes are spontaneous?

c. Are there any reactions where both the enthalpy and entropy factors are favorable?

d. Are there any reactions where both the enthalpy and entropy factors are favorable?

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Reaction ΔH° (kJ/mol) T × ΔS°

(kJ/mol) ΔH° – T × ΔS°

(kJ/mol) K

NaCl(s) ⇌ Na+(aq) + Cl–(aq) 3.86 12.9 –9.00 38 NH4NO3(s) ⇌ NH4

+(aq) + NO3–(aq) 28.07 32.38 –4.31 5.7

Zn(s) + Cu2+(aq) ⇌ Cu(s) + Zn2+(aq) –218.67 –6.3 –212.4 1.6 × 1037

2Cl–(aq) + Br2(l) ⇌ 2Br–(aq) + Cl2(g) 91.23 31.78 59.45 3.9 × 10–11 CH3COOH ⇌ CH3COO–(aq) + H+(aq) –0.25 –27.43 27.18 1.7 × 10–5

2. Considering the thermodynamic and equilibrium data above…

a. Which reactions appear to be favorable (K>1)?

b. Which reactions appear to be unfavorable (K<1)?

c. Which factor is used to predict if the reaction is favorable or unfavorable? i. ΔH°

ii. T × ΔS° iii. ΔH° – T × ΔS°

d. When K>1, is the factor you identified positive or negative?

e. When K<1, is the factor you identified positive or negative?

f. What is the quantitative relationship between the factor you identified and the

magnitude of K?

3. Without referring to tables to calculate ΔH° and ΔS°, predict whether the equilibrium constant at room temperature for the following reaction will be greater than, less than, or equal to 1. Explain your reasoning.

C3H8(g) + 5O2(g) ⇌ 3CO2(g) + 4H2O(g)

4. For what combinations of ΔH° and ΔS° will a chemical reaction always have K>1? Always have K<1?

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The Meaning of ΔG for a Chemical Reaction We have learned that a system at constant temperature and pressure will proceed spontaneously in the direction that lowers its free energy. This is why reactions proceed until they reach equilibrium. The equilibrium position represents the lowest free energy value available to a reaction system. This leads to the following equations:

For a reaction NOT at equilibrium: ∆𝐺 = ∆𝐺° + 𝑅𝑇𝑙𝑛(𝑄)

(This can also be used to determine the change in free energy of a reaction at

nonstandard temperatures and pressures)

For a reaction at equilibrium: ∆𝐺° = −𝑅𝑇𝑙𝑛(𝐾)

(When a reaction is at equilibrium, ΔG = 0, and Q = K. Substitute this into the previous

equation and solve to get this one.) Where:

o ΔG = Change in free energy at specified (nonstandard state) conditions o ΔG° = Change in free energy at standard state conditions o R = Gas law constant = 8.3145 J/mol∙K o T = Kelvin temperature o Q = Reaction quotient o K = Equilibrium constant

Recall that we can write these for both pressure and concentration.

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5. One method for synthesizing methanol involves reacting carbon monoxide and hydrogen gases:

CO(g) + 2H2(g) → CH3OH(l) a. Calculate ΔG at 25°C for this reaction where carbon monoxide at 5.0 atm is mixed

with hydrogen gas at 3.0 atm to produce liquid methanol. (HINT: Find ΔG° using Appendix 4)

b. Based on the value for ΔG, would you expect this reaction to be more spontaneous at high or low pressures?

c. How does this relate to Le Châtelier’s principle?

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6. Consider the ammonia synthesis reaction: N2(g) + 3H2(g) ⇌ 2NH3(g)

Where ΔG° = –33.3 kJ/mole of N2 consumed at 25°C. For each of the following mixtures of reactants and products at 25°C, predict the direction in which the system will shift to reach equilibrium.

a. 𝑃𝑁𝐻3= 1.00 𝑎𝑡𝑚, 𝑃𝑁2

= 1.47 𝑎𝑡𝑚, 𝑃𝐻2= 1.00 × 10−2 𝑎𝑡𝑚

b. 𝑃𝑁𝐻3= 1.00 𝑎𝑡𝑚, 𝑃𝑁2

= 1.00 𝑎𝑡𝑚, 𝑃𝐻2= 1.00 𝑎𝑡𝑚

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7. The overall reaction for the corrosion (rusting) of iron by oxygen is: 4Fe(s) + 3O2(g) ⇌ 2Fe2O3(s)

Using the following data, calculate the equilibrium constant for this reaction at 25°C. Substance ΔHf° (kJ/mol) S° (J/mol∙K)

Fe2O3(s) –826 90 Fe(s) 0 27

O2(g) 0 205

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CW 7: Oxidation-Reduction Review

Oxidation States Oxidation states are a method to keep track of electrons in redox reactions, especially when covalent substances are involved. Recall that electrons are shared in covalent bonds. The oxidation states of such atoms are determined by assigning electrons to atoms. For a covalent bond between two identical atoms, the electrons are spilt equally. If the two atoms are not identical, the shared electrons are assigned to the atom that has a stronger attraction for electrons. The following rules should be used to determine oxidation states.

The oxidation state of… Summary Examples

An atom in elemental form is zero Element = 0 Na(s), O2(g), O3(g), Hg(l)

A monatomic ion is the same as its charge Monatomic ion = charge of ion

Na+, Cl–

Fluorine is –1 in its compounds Fluorine = –1 HF, PF3

Oxygen is usually –2 in its compounds (exception: peroxides, containing O2

2–, in which oxygen is –1) Oxygen = –2 H2O, CO2

Hydrogen is +1 in its covalent compounds Hydrogen = +1 H2O, HCl, NH3

Remember that… The sum of oxidation states is zero for an electrically neutral compound

For an ion, the sum of the oxidation states must equal the charge of the ion.

1. Identify the following reactions are either redox or non-redox using oxidation numbers as evidence.

a. Pb(NO3)2(aq) + 2NaI(aq) → PbI2(s) + 2NaNO3(aq)

b. 2H2O → 2H2 + O2

c. CH4 + 2O2 → 2H2O + CO2

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Half Reactions The process of oxidation and reduction can be thought of as a transfer of electrons from one atom to another. Thus, one atom gives up electrons and the other atom gains them. As a result of this process, the oxidation numbers of both atoms change. All redox reactions can be divided up into two reactions: an oxidation half-reaction and a reduction half-reaction. This allows for better understanding of the electron transfer process.

A. Zn(s) + Cu2+(aq) → Zn2+(aq)+ Cu(s) o Ox: Zn → Zn2+ + 2e– o Red: Cu2+ + 2e– → Cu

B. 2I–(aq) + S2O8

2–(aq) →I2(s) + 2SO42–(aq)

o Ox: 2I– → I2 + 2e– o Red: S2O8

2– + 2e– → 2SO42–

C. 4Fe(s) + 3O2(g) → 2Fe2O(s)

o Ox: Fe → Fe2+ + 2e– o Red: O2 + 4e– → 2O2–

D. 4H+(aq) + MnO4

–(aq) + 3Fe2+(aq) → 3Fe3+(aq) + MnO2(aq) + H2O(l) o Ox: Fe2+ → Fe3+ + e– o Red: 2H+ + MnO4

– + 3e– →MnO2 + 2H2O

2. Considering the oxidation half reactions:

a. Which of the following types of particles undergo oxidation? Neutral atoms/ molecules Cations Anions

b. Are electrons gained or lost during the process of oxidation?

c. When an atom is oxidized, does its oxidation number increase or decrease?

3. Considering the reduction reactions: a. Which of the following types of particles undergo reduction?

Neutral atoms/ molecules Cations Anions

b. Are electrons gained or lost during the process of reduction?

c. When an atom is reduced, does its oxidation number increase or decrease?

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4. Identify the oxidation and reduction half reactions for the following. Then balance each redox reaction.

a. I–(aq) + NO2–(aq) → I2(s) + NO(g) (acidic solution)

b. MnO4–(aq) + Cl–(aq) → Mn2+(aq) + Cl2(g) (basic solution)

Steps to Balance a Redox Reaction 1. Spilt the reaction into two half reactions. 2. Balance elements in the equation other than O and H. 3. Balance the oxygen atoms by adding the appropriate number of water (H2O) molecules

to the opposite side of the equation. 4. Balance the hydrogen atoms (including those added in step 2 to balance the oxygen

atom) by adding H+ ions to the opposite side of the equation. 5. Add up the charges on each side. Make them equal by adding enough electrons (e–) to

the more positive side. (Rule of thumb: e– and H+ are almost always on the same side.) 6. The e– on each side must be made equal; if they are not equal, they must be multiplied

by appropriate integers (the lowest common multiple) to be made the same. 7. If the reaction occurs in basic conditions, add OH– ions to each side to balance out H+

ions and form water. If conditions are acidic or neutral, skip this step. 8. The half-equations are added together, canceling out the electrons to form one

balanced equation. Common terms should also be canceled out.

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The Activity Series One of the chemical properties of metals is that they form cations in chemical reactions. This requires them to lose their electrons (be oxidized). However, electrons need to go somewhere, so some metals will tend to stay in their neutral state. Which metals are better at making cations and which are better staying neutral?

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5. Considering the experiments shown: a. All solutions were made from nitrate salts. Why are there no NO3

– ions pictured?

b. List two pieces of evidence from Experiment A that indicate a reaction has occurred.

6. For each experiment that resulted in a reaction: a. Write a single replacement reaction under the beakers in each experiment to

illustrate the reaction. Ignore spectator ions. b. Circle the species in each of the reactant beakers that underwent oxidation. c. Draw an arrow to show the transfer of electrons during the reaction.

7. Find the two experiments in that involve aluminum and magnesium.

a. Will aluminum give its electrons to magnesium? Y N b. Will magnesium give its electrons to aluminum? Y N c. Which metal is better at giving up electrons? Al Mg

8. Find the two experiments in that involve aluminum and copper.

a. Will aluminum give its electrons to copper? Y N b. Will copper give its electrons to aluminum? Y N c. Which metal is better at giving up electrons? Al Cu

9. Predict what might happen if you put a piece of metal in a solution of its own ion (for

example, copper metal in a Cu2+ solution or magnesium metal in a Mg2+ solution).

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10. An active metal readily gives up its electron(s) and is oxidized. Rank the metals Mg, Al, Cu, and Zn from least to most active. Support with evidence from the experiments.

11. Complete the reaction diagrams below to illustrate whether a reaction occurs or not.

a. Zinc metal in Mg2+ solution

b. Zinc metal in Cu2+ solution

12. Although hydrogen is not a metal, it is included in the activity series. Predict whether the reaction shown will be spontaneous. If so, complete and balance the equation.

a. Zn(s) + HCl(aq) →

b. Fe(s) + HCl(aq) →

c. Cu(s) + HCl(aq) →

Li Most Reactive

Least Reactive

Rb Zn

Cr Fe

Sn Pb

H

Cu Ag

Hg Pt

Au

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CW 8: The Electrochemical Cell

Diagram of a Galvanic Cell A galvanic cell is a device that allows the half reactions of a redox reaction to occur in separate locations (cells). Reduction occurs at the cathode while oxidation occurs at the anode. A salt bridge (or sometimes a porous disk) allows ions to slowly migrate from one beaker to the other to maintain electrical neutrality in each half-cell, leading to a voltage (or potential) between the two electrodes. If the temperature is 298 K, and the solutions are 1 M, each system is a standard half-cell.

When the switch is closed, the following are observed:

o The mass of the Cu electrode increases and the concentration of Cu2+(aq) decreases. o The mass of the Zn electrode decreases and the concentration of Zn2+ increases o Electrons flow through the wire o The voltage measures 1.10 V

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1. Based on the observations from the galvanic cell, which of these chemical reactions describes the actual process that is occurring? Why? Cu(s) + Zn2+(aq) ⇌ Zn(s) + Cu2+(aq) Cu2+(aq) + Zn(s) ⇌ Zn2+(aq) + Cu(s)

2. Considering the diagram: a. Label the cathode and anode.

b. Indicate the flow of electrons using an arrow.

c. Electrons flow from the negative electrode to the positive electrode. Which

electrode, Zn or Cu, is the negative electrode?

d. What species is oxidized in the overall process?

e. What species is reduced in the overall process?

f. What happens when the switch is opened?

g. What happens if the salt bridge is removed?

3. Comparing each half cell: a. What is the half-reaction occurring in the copper half-cell?

b. What is the half-reaction occurring in the zinc half-cell?

c. Based on your answers, which has a greater attraction for electrons, Cu2+ or Zn2+?

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Electron Pulling Strength The chemical processes taking place in a galvanic cell is a "tug-of-war" for electrons between the two half-cells. The "winner" is the stronger oxidizing agent. It gains electrons and gets reduced. The voltage is a measure of the difference in electron-pulling strength.

When a galvanic cell is constructed from the half-cells above, the electrons flow to the Cu(s) and the voltage is 0.62 V.

4. Considering a galvanic cell made from the electrodes pictured: a. Write the half-reaction occurring in the cobalt half-cell.

b. Write the half-reaction occurring in the copper half-cell.

c. Using the observations for this galvanic cell, which has a greater attraction for electrons, Cu2+ or Co2+? Explain.

d. Refer to the first galvanic cell. Note that the voltage is 1.10 V. Which has a stronger pull on electrons, Zn2+ or Co2+? Explain.

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Standard Reduction Potentials The “pull” on the electrons in a galvanic cell is called the cell potential (Ecell), or the electromotive force (emf) of the cell. The unit of electrical potential is the volt (V), which is defined as 1 joule of work per coulomb of charge transferred.

The cathode used above is known as a standard hydrogen electrode (SHE). It consists of a chemically inert platinum electrode in contact with 1 M H+ ions and bathed by H2 gas at 1 atm. Although we can measure the total potential of this cell (0.76 V), there is no way to measure the potentials of the individual half-cells. To get around this, we assign the standard hydrogen electrode a potential of zero volts. For example, if we assign the reaction

2H+ + 2e– → H2 Where [H+] = 1.0 M and PH2 = 1 atm a potential of exactly zero volts, then the reaction

Zn → Zn2+ + 2e– will have a potential of 0.76 V because

𝐸𝑐𝑒𝑙𝑙∘ = 𝐸2𝐻+→𝐻2

∘ + 𝐸𝑍𝑛→𝑍𝑛2+∘

0.76 V 0 V 0.76 V where the symbol ° indicates that standard states are used. By setting the standard potential for the half-reaction 2H+ + 2e– → H2 equal to zero, we can assign values to all other half-reactions.

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5. Use the data to determine the standard reduction potential for each of the following half reactions.

Cathode Anode Ecell (V) Cu│Cu2+ Zn│Zn2+ 1.10

Cu│Cu2+ SHE 0.34 Br2│Br– Zn│Zn2+ 1.85

Zn│Zn2+ K│K+ 2.16

Cl2│Cl– Ag│Ag+ 0.56 Ag│Ag+ K│K+ 3.72

a. Cl2 + 2e– ⇌ 2Cl–

b. Br2 + 2e– ⇌ 2Br–

c. Ag+ + e– ⇌ Ag

d. Cu2+ + 2e– ⇌ Cu

e. K+ + e– ⇌ K

The accepted convention is to give the potentials of half-reactions as reduction processes. The values of Ecell with all solutes at 1 M and all gases at 1 atm are called standard reduction potentials. A table of these values is found in the references section of the unit packet. To combine half reactions into a balanced redox reaction:

o One of the reduction half reactions must be flipped (so that one substance is oxidized). The half reaction with the largest positive potential will run as written (as a reduction) and the other half reaction will run in reverse, flipping the sign of the potential.

o Electrons must be balanced by multiplying the half reactions by integers. The potential for the cell is not multiplied by this number because a standard reduction potential is an intensive property (it does not depend on how many times the reaction occurs).

6. Consider a galvanic cell based on the reaction

Al3+(aq) + Mg(s) → Al(s) + Mg2+(aq) The relevant half-reactions are:

Al3+ + 3e– → Al E° = –1.66 V Mg2+ + 2e– → Mg E° = –2.37 V

Give the balanced cell reaction and calculate E° for the cell.

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7. Consider a galvanic cell based on the reaction Fe3+(aq) + Cu(s) → Cu2+(aq) + Fe(s)

The relevant half-reactions are: Fe3+ + e– → Fe2+ E° = 0.77 V Cu2+ + 2e– → Cu E° = 0.34 V


8. Consider a galvanic cell based on the reaction MnO4

–(aq) + H+(aq) + ClO3–(aq) → ClO4

–(aq) + Mn2+(aq) + H2O(l) The relevant half-reactions are:

MnO4– + 5e– + 8H+

→ Mn2+ + 4H2O E° = 1.51 V ClO4

– + 2H+ + 2e– → ClO3– + H2O E° = 1.19 V


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Line Notation Line notation is often used to describe a galvanic cell. In this notation, anode components are listed on the left, and cathode components are listed on the right, separated by double vertical lines (which indicate the salt bridge or porous disk). A single vertical line is used to denote the phase difference in each electrode. Here is line notation for the cell from Question 6:

Mg(s)│Mg2+(aq)││Al3+(aq)│Al(s) If all the components of a cell are ions, a nonreacting conductor must be used because the ions cannot act as an electrode. Platinum is the usual choice for this electrode. Here is the line notation for a cell with one iron electrode and one platinum electrode.

Fe(s)│Fe2+(aq)││MnO4–(aq), Mn2+(aq)│Pt(s)

9. Return to Question 7 and 8 and use line notation to describe each galvanic cell.

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CW 9: Cell Potential, Free Energy, Equilibrium

Question Notes Important terms, constants, and units

What is the maximum voltage possible for this cell?

A galvanic cell was made based on the given half reactions. Cu2+ + 2e– → Cu Ecell = 0.34 Au3+ + 3e– → Au Ecell = 1.50

Calculate the theoretical amount of electrical work and the actual amount of electrical work done by this cell.

When complete, 1.50 mol of electrons were passed through this cell at an average potential of 1.08 V.

Calculate the efficiency of this cell.

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Question Notes How does electrical work relate to free energy?

What is meant by a “dead” battery?

What is a concentration cell?

When and how do you use the Nernst equation?

SUMMARY

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Practice Problems 1. Using standard reduction potentials, calculate ΔG° for the reaction:

Cu2+(aq) + Fe(s) → Cu(s) + Fe2+(aq) Is this reaction spontaneous?

2. Use the standard reduction potentials to predict whether 1 M HNO3 will dissolve gold metal to form a 1 M Au3+ solution.

Au(s) + NO3–(aq) + 4H+(aq) → Au3+(aq) + NO(g) + 2H2O(l)

3. For the reaction 2Al(s) + 3Mn2+(aq) → 2Al3+(aq) + 3Mn(s) E°cell = 0.48 V

Predict whether Ecell is larger or smaller than E°cell for the following cases. a. [Al3+] = 2.0 M, [Mn2+] = 1.0 M

b. [Al3+] = 2.0 M, [Mn2+] = 1.0 M

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4. A galvanic cell was constructed based on the following half reactions and conditions.

VO2+ + 2H+ + e– → VO2+ + H2O E°cell = 1.00 V T = 25°C

[VO2+] = 2.0 M

[H+] = 0.50 M

[VO2+] = 1.0 × 10–2 M [Zn2+] = 1.0 × 10–1 M Zn2+ + 2e– → Zn E°cell = –0.76 V

a. Write and balance the cell reaction.

b. Calculate Ecell under these (nonstandard) conditions.

c. Complete the diagram by drawing in the cathode and anode, components of each solution, the flow of electrons, and the measured voltage.

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CW 10: Electrolysis

A galvanic cell produces current when an oxidation-reduction reaction proceeds spontaneously. An electrolytic cell does the opposite: it uses electrical work to cause an otherwise nonspontaneous chemical reaction to occur. The process of electrolysis makes use of this in many practical ways: charging a battery, producing aluminum, and electroplating.

1. Circle the differences between the galvanic cell (a) and the electrolytic cell (b).

2. Based on your findings, how can you convert a galvanic cell into an electrolytic cell?

3. Which electrode would gain mass in each cell? (Hint: is this electrode oxidized or reduced?)

Zn2+(aq) + 2e– → Zn E°cell = –0.76 Cu2+(aq) + 2e– → Cu E°cell = 0.34

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A common question is how much chemical change occurs with the flow of a given current for a specified time. The unit for current is ampere (amp, A), which is one coulomb of charge per second. To solve this stoichiometry problem, we need the following steps:

4. How many grams of copper can be plated out when a current of 10.0 amps is passed for 30.0 minutes through a solution containing Cu2+?

a. Multiply the current (C/s) by seconds to get

coulombs (C) of charge passed into the Cu2+ solution.

b. One mole of electrons carries a charge of 1 faraday, or 96486 coulombs. Use this to convert charge in coulombs into moles of electrons.

c. Each Cu2+ requires two electrons to become solid copper: Cu2+ + 2e– → Cu. Use this as a mole ratio to find moles of copper.

d. Convert moles of copper into grams (molar mass = 63.546 g/mol).

current and time

aquanitity of

charge in coulombs

bmoles of electrons

cmoles of copper

dgrams of copper

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5. Consider the electrolysis of a solution of Ag+ using 5.00 A. a. How long must the 5.00 A current be applied to the solution to produce 10.5 g of

silver metal?

b. If the electrolysis runs for 10.0 minutes, how much solid silver can be plated out?

6. A solution in an electrolytic cell contains the ions Cu2+, Ag+, and Zn2+. If the voltage is initially low and increases gradually, in what order will the metals plate out in? (Hint: which is the most spontaneous as written?)

Ag+ + e– → Ag E°cell = 0.80 V Cu2+ + 2e– → Cu E°cell = 0.34 V

Zn2+ + 2e– → Zn E°cell = –0.76 V

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Reference Materials

Polyatomic Ions Ion Name Ion Name

H3O+ hydronium CrO42– chromate

Hg22+ dimercury (I) Cr2O7

2– dichromate NH4

+ ammonium MnO4– permanganate

C2H3O2–

CH3COO– acetate

NO2– nitrite

NO3– nitrate

C2O42– oxalate O2

2– peroxide CO3

2– carbonate OH– hydroxide

HCO3– hydrogen (bi)carbonate CN– cyanide

PO43– Phosphate SCN– thiocyanate

ClO– hypochlorite SO32– sulfite

ClO2– chlorite SO4

2– sulfate ClO3

– chlorate HSO4– hydrogen sulfate

ClO4– perchlorate S2O3

2– thiosulfate

Solubility Guidelines Ions that form Soluble Compounds

Exceptions Ions that form Insoluble Compounds

Exceptions

Group 1 ions (Li+, Na+, etc.)

-- Carbonate (CO3

2–) When combined with Group 1 ions or NH4

+

Ammonium (NH4

+) --

Chromate (CrO4

2–) When combined with Group 1 ions, Ca2+, Mg2+ or NH4

+

Nitrate (NO3–) --

Phosphate (PO4

3–) When combined with Group 1 ions or NH4

+

Acetate (C2H3O2–

or CH3COOH) --

Sulfide (S2–)

When combined with Group 1 ions or NH4

+

Hydrogen carbonate (HCO3

–) --

Hydroxide (OH–)

When combined with Group 1 ions, Ca2+, Ba2+, Sr2+ or NH4

+

Chlorate (ClO3–) --

Perchlorate (ClO4

–) --

Halides (Cl–, Br–, I–)

When combined with Ag+, Pb2+, Hg2

2+

Sulfates (SO42–)

When combined with Ag+, Ca2+, Sr2+, Ba2+, Pb2+

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Standard Reduction Potentials

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Average Bond Enthalpies Single Bonds (kJ/mol) C–H 413 N–H 391 O–H 463 F–F 155 C–C 348 N–N 163 O–O 146 C–N 293 N–O 201 O–F 190 Cl–F 253 C–O 358 N–F 272 O–Cl 203 Cl–Cl 242 C–F 485 N–Cl 200 O–I 234 C–Cl 328 N–Br 243 Br–F 237 C–Br 276 Br–Cl 218 C–I 240 H–H 436 S–H 339 Br–Br 193 C–S 259 H–F 567 S–F 327 H–Cl 431 S–Cl 253 I–Cl 208 Si–H 323 H–Br 366 S–Br 218 I–Br 175 Si–Si 226 H–I 299 S–S 266 I–I 151 Si–C 301 S–N 464 Si–O 368 Multiple Bonds (kJ/mol)

C=C 614 N=N 418 O=O 495 C≡C 839 N≡N 941 C=N 615 N=O 607 S=O 523 C≡N 891 S=S 418 C=O 799 C≡O 1072

Specific Heats Substance J/(g∙°C) cal/(g∙°C)

Water (liquid) 4.18 1.00 Water (gas) 1.9 0.45

Water (solid) 2.1 0.50 Ethanol 2.4 0.58

Chloroform 0.96 0.23

Aluminum 0.90 0.21 Iron 0.46 0.11

Silver 0.24 0.057 Mercury 0.14 0.033

Heats of Physical Change

Substance ∆Hfus

(kJ/mol) ∆Hvap

(kJ/mol) Ammonia

(NH3) 5.66 23.3

Ethanol (C2H6O)

4.93 38.6

Hydrogen (H2) 0.12 0.90

Methanol (CH4O)

3.22 35.2

Oxygen (O2) 0.44 6.82 Water (H2O) 6.01 40.7

name: ap chemistry 5: thermodynamics and applications of ... · 3/5/2020 · 7 3/16 3/17 unit test...

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