Chemistry 203 Term Test 1 1 Lecture Slides Booklet Solutions - - PowerPoint PPT Presentation

Chemistry 203 – Term Test 1

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Questions? Suggestions? E­mail me and I will try to find time to reply. psedach@learnfaster.ca All the best!

• Lecture Slides
• Booklet Solutions
• Test Solutions
• Extra notes and advice
• Advice on which exams and questions to write
• Corrections

Available February 9th at: http://learnfaster.ca/blog/chem­203­term­test­1/

Friday 3:30 to 5:00 Intro Stoichiometry and Gases 5:00 to 5:30 Break 5:30 to 6:45 Energy and Thermochemistry 6:45 to 7:00 Break 7:00 to 8:20 Equilibrium 8:20 to 8:30 Break 8:30 to 9:30 Review

What is on the Exam? (Part 1)

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Chem 203 is cumulative, however topics get expanded into Equilibrium, Kinetics and Electrochemistry. Some topics may not be tested directly in the future but are foundational to future concepts.

Stoichiometry Percent yield Grams to atoms (Avogadro’s number) Solutions (moles/L and Liters) Stoichiometry through pictures Stoichiometry through gas laws Redox Balancing (pre­requisite) Gases Working with ideal gases PV = nRT Graphs of P vs. V, P vs. T, etc. y =

1 x or y = x

Partial Pressure Mole fraction P

a = χaPtotal

Mass density

𝑜 𝑊 × 𝑁𝑋 = 𝑛 𝑊

Moles/liter from gas equation Stoichiometry: Initial and Final P given constant V 2 sets of conditions:

P1V1 P2V2 = n1R1T1 n2R2T2

Partial pressure and mass density through pictures Qualitative – if system expands, system DOES PV work If system contracts, work is done ON system Non­ideal gases At high pressures, molecular size counts, Vreal > Videal At low temperatures, IMFs are felt and Preal < Pideal

What is on the Exam? (Part 2)

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Thermodynamics Understand exothermic/endothermic diagrams Understand how rearranging an equation changes ΔH Hess’s Law: adding reactions adds enthalpy Define standard enthalpy of formation (compound made from elements) Calculate enthalpy of reaction using heats of formation (P­R) Math with Q = nΔH Qualitative treatment of heat capacity (based on Q = mcΔT) Bond Dissociation Energy (R­P) Kinetic Molecular Theory Average vs. Total Kinetic Energy Molecules vs. Speed Graphs Molecules vs. Kinetic Energy Graphs How can a smaller container increase pressure? (more F/A)

Stoichiometry is a prerequisite.

Breakdown of old exams

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In recent years, for TT1, there has been a shift away from kinetics and more towards a broad overview of thermochemistry, work, heat and equilibrium. Gases and stoichiometry are foundational to this exam. The majority of the grades are from basic stoichiometry, gases and thermochemistry. Do your best on this exam and don’t fall behind – term test 2 and the final are progressively more difficult and are cumulative with the knowledge on this exam.

IF you did not get a topic on TT1, you have a month to review before TT2. MAKE SURE you get it! (Cumulative Exams!)

Winter 2016 Fall 2015 Fall 2013 Spring 2013 Stoichiometry 7% 9% 43% 45% Kinetic Molecular Theory 11% 9% 3% 3% Gases & Partial Pressure 33% 23% 7% 33% Equilibrium 0% 23% 47% 6% Calorimetry and Enthalpy 33% 26% 9% Work and Energy 15% 11% 3%

What we offer

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Lecture Slides and Extra Questions 2 Copies Exam Package Use the 2nd copy to write closed book exams and grade yourself. learnfaster.ca Full solutions Format: Cover every concept at least once through the lecture. Every question is based off exams If we finish a section early, we practice some exam questions.

x2

Significant Figures (F2015 Q2)

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Leading zeroes (red) don’t add information so they don’t count. 0.000011500 = 1.1500 × 10−5 Zeroes on the right tell us precision. Ex. 50 means 50 ± 0.5 whereas 50.0 means 50.0 ± 0.05 Addition and Subtraction ­ Least number of decimals: 100 − 0.1547869 = 99.8452131 = 100 If two numbers use different scientific notation, convert to same notation: 17.2 × 103 − 5.5 × 102 = 172 × 102 − 5.5 × 102 = 166.5 = 167 Multiplication ­ Least number of sig. figs. 1230 × 16 = 19680 = 2.0 × 104 Evaluate sig figs. at every step but round only at the end. Practice: 24.71 − 24.0 = 24.71 × 24.0 =

Ion Concentrations (F2013 Q3)

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If 50.0 mL of 0.100 M K2Cr2O7 and 100 mL of 0.20 M K3PO4 are combined with 50.0 mL of NaF, what is the concentration of potassium ions in solution?

Basic Stoichiometry

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I have 6 wheels and 4 frames, given the following equation: 2 wheels + 1 frame  1 bicycle How many bicycles can I make? Given that 1 bicycle was broken during assembly, what was my percent yield?

Tips to Balancing Reactions

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• 1. Balance, 2. Convert to moles, 3. Use a molar ratio, 4. Convert moles to unit required

If I react 15.0 g of Paraffin (mostly C31H64) with 40.0 g of O2, what will my yield of CO2 be? C31H64 + O2 → CO2 + H2O

H2 N2 O2 F2 Cl2 Br2 I2

*When solving, we get 1.250 mol of O2 and 0.0343 mol paraffin. After doing a mol:mol ratio, we get 0.825 mol CO2 (36.3 g) from the oxygen and 1.065 mol of CO2 from the paraffin. The oxygen is therefore the limiting reagent. If my actual yield of CO2 g was 15.0 g, what was my percent yield? 41.3% How much of my excess reagent (in grams) was left? 3.38 g or 0.00774 mol of paraffin will be left over.

Stoichiometry with Pictures (F2013 Q4)

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Whenever you encounter picture questions, ALWAYS convert the number of dots to moles and write them below the image: Given: 2 A + 3

2 B2 → A2B3

Which of the containers below will have the highest final yield of A2B3?

= A = B2

Review (W2016 Q2)

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• 1. Balance, 2. Convert to moles, 3. Use a molar ratio, 4. Convert moles to unit required

If we react 5.25 moles of sodium hydroxide with 6.75 moles of sulfuric acid, what is the final amount of all reagents?

all values in moles

2 NaOH s + H2SO4 l → 2 H2O l +Na2SO4 s Initial 5.25 6.75 Change Final

*Sometimes professors refer to “Final” as “Result” or “End” – the idea is the same – this is NOT an equilibrium ICE table, it’s an ICF table used for stoichiometry and one of the concentrations WILL go to zero! *Remember this example – the above is how you evaluate the species present at ANY step in a titration OR an all gas system!

PV = nRT ) Pressure P × Volume V = moles n × ideal gas constant R × Temperature in Kelvin (T How many atoms of chlorine are present in are present in 3.45 g of PCl5 g ?

Stoichiometry (W2016 Q1)

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Calculating the final pressure in a container (W2016 Q3)

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For a gas system, PV = nRT and, if volume and temperature are constant we get: If I have a container with 1.50 atm of CH4 g and 12.00 atm of O2 g , what is the initial total pressure in the container? What is the total pressure AFTER reaction?

The Ideal Gas Law PV=nRT

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If you have a mixture of CO2 g (44.01 g/mol), N2 g (28.02 g/mol) and C12H22O11 g (MW=342 g/mol) they all behave the same! For instance, I have a container with 1 mol of CO2 g , 1 mol of N2 g and 2 mol of C12H22O11 g (all at the same temperature) and my total pressure is 24 atmospheres, what portion of the pressure is each gas responsible for? moles pressure CO2 g 1 mol N2 g 1 mol C12H22O11 g 2 mol total 24 atm This is because we assume all gases are ideal because they follow the “Kinetic Molecular Theory” (KMT). KMT says that regardless of the size of the gas molecule, they’re so small that the volume of a gas is zero. KMT also says that intermolecular forces (you know how water sticks to itself ?) are ZERO for any gas.

Kinetic Molecular Theory

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Gas molecules ­can be treated like spheres of the same size. ­negligible volume compared to volume of space between gas molecules. ­not true at high pressures! (VOLUME is GREATER THAN IDEAL) ­still have mass (a more dense gas than another can settle to the bottom of a container) Gases expand to fill their container and collide with the container walls. This exerts pressure on the walls of the container: Pressure = Force Area = N m2 Gases travel in constant, random, linear (straight­line) motion. gas molecule collisions are elastic (momentum and energy (EK) are conserved). gases never lose energy (compressed gas can be used years later to run a barbecue despite constant collisions!) Molecules of gas experience no intermolecular forces, therefore each molecule moves at constant speed between collisions. ­ not true at low temperatures! (PRESSURE is LOWER THAN IDEAL)

Review (W2016 Q4/5/6)

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At a given temperature, ALL gases (e.g. N2 g vs. H2 g ) have the same average EK. HOWEVER, EK = 1

2 𝑛𝑤2 where m is

Molecular Weight. We have two containers, 1 L of N2 g at 25 °C and 1L of H2 g at 25 °C, which gas would move faster?

What does a Look like?

Other Gas Laws (W2012 Q6) *assume other variables constant

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3 Scenarios: Direct Proportion y = x Inverse Proportion y =

1 x OR xy = 1

Geometric y = x2 𝑦 𝑧 𝑦 𝑧 𝑦 𝑧 V vs. T PV = nRT P vs. V PV = nRT P vs. T PV = nRT

Why? Why?

­300 ­200 ­100 100 200

KMT : Temperature IS Kinetic Energy (W2012 Q5)

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EKaverage = 3 2 RT = 1 2 𝑛𝑤2,

Can Energy ever be negative? Should temperature ever be negative? What happens to the energy of the gas at absolute zero? According to PV = nRT, What happens to the volume of a gas at absolute zero? Volume (Liters) Degrees Celsius ℃ Kelvin Volume (Liters) 100 200 300 400 500 Does negative energy exist? From this we get ‘absolute zero is when a V vs. T graph crosses the x­Axis’

Number of Molecules vs. Speed/Energy Curves (W2016 Q4/5/6)

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The area under both curves is equal. The following curves describe a system of nitrogen and hydrogen gas at the same

• temperature. Which gas is which?

Speed System 1 System 2 # Number of Molecules What does a graph of molecules vs. kinetic energy look like? Which formula explains this? System 1 System 2 # Number of Molecules Kinetic Energy

𝐹𝑙 =

3 2 𝑆𝑈 = 1 2 𝑛𝑤2 (W2016 Q4/5/6)

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Speed System 1 System 2 Kinetic Energy System 1 System 2 # Number of Molecules # Number of Molecules Lighter/Heavier/The Same System 1 has ________ particles than System 2 Slower/Faster/Same Speed System 1 has ________ particles than System 2 Lower/Higher/The Same System 1 has ________ temperature as System 2

Overlap

The area under both curves is equal

Slower/Faster/Same Speed System 1 has ________ particles than System 2 Lower/Higher/The Same System 1 has ________ temperature as System 2 Lighter/Heavier/The Same System 1 has ________ particles than System 2

𝐹𝑙 =

3 2 𝑆𝑈 = 1 2 𝑛𝑤2 (W2016 Q4/5/6)

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Speed System 1 System 2 Kinetic Energy # Number of Molecules System 1 System 2 # Number of Molecules The area under both curves is equal

Real Gases (W2016 Q15C)

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Conclusion 1: Gases do not behave ideally at high pressures; molecular volume means Vreal > Videal Conclusion 2: Gases do not behave ideally at low temperatures; Intermolecular Forces at lower T cause Preal < Pideal Conclusion 3: Gases are only ideal at low pressure and high temperature. Positive deviation due to molecular volume at high P, Vreal > Videal Negative deviation due to IMFs at moderate P OR low Temperature, Preal < Pideal

Pressure (atm) PV nRT N2 @ 200K N2 @ 500K N2 @ 1000K Ideal Gas

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Partial Pressures (F2015 Q4)

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Consider a system with 14 moles of N2, 8 moles of Xe and 6 moles of CO2. The total pressure is 14 atm. If all the gases are ideal and contribute to the pressure equally, what is the partial pressure of each gas? What is the mole fraction of Xenon?

Ptotal = P

a + Pb + ⋯

and ntotal = na + nb + ⋯

Pa Ptotal = na ntotal = χa Pb Ptotal = nb ntotal = χb

Therefore And P

a = χaPtotal

Partial Pressures with Pictures (F2015 Q4)

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The following containers are at the same temperature. *When you encounter pictures, convert the symbols into moles and write them down!! = 1 mol N2 = 1 mol H2 Which container has the highest pressure? Which container has the highest partial pressure of N2? Which container has the highest mole fraction of N2?

2 Liters 1 Liter

Thermochemistry

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Like the session so far? Share it with your friends! https://www.facebook.com/prep101uofc It’s great seeing you all out today!

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Closed Systems and Constant Pressure (Isobaric)

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∆U = q + w ∆U=change in energy, q=heat, w=work For us, ∆U always refers to the system. But according to the first law of thermodynamics energy is neither created nor destroyed SO – when our system gains energy, where does it come from?

A Closed System System Matter Energy Surroundings

When the system studied gains energy, it is endothermic. When a system studied loses energy, it is exothermic. You use an electric stove to boil water. The water is ____________ (∆U is ____). The stove is _____________ (∆U is ____). Water vapor condenses on a window. The water is ____________ (∆U is ____). The window is _____________ (∆U is ____).

Review ­ ∆U = q + w (W2013 Q13)

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In a chemical laboratory, a technician applies 340 J of heat to a gas system while the surroundings do an additional 140 J of work

• n the gas. What is the change in internal energy for this gas system?

a) +140 J b) ­200 J c) +200 J d) +480 J From the question above, the process can be defined as: a) Endothermic b) Exothermic c) State dependent d) Taking place in an isolated system

Pressure and Volume Work (W2016 Q7)

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Consider the following reactions: 2 H2 g + O2 g → 2 H2O g 2 NaN3 s → 3 N2 g + 2 Na s 4 NH3 g + 5 O2 g → 4 NO g + 6 H2O g C2H5OH l + C2H5COOH l ⇌ C2H5COOC2H5 l + H2O l CaCO3 s ⇌ CO2 g + CaO s How many of these reactions involve the system doing pressure­volume work on the environment?

• a. 0
• b. 1
• c. 2
• d. 3

Calorimetry (W2016,Q11) based on Q = mcΔT

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When a hot object contacts a cold object, they equilibrate to the same temperature Which glass experiences the greatest temperature change? After equilibrium, which glass is the warmest? 5 g Fe (c = 0.449

J g°C)

100 °C 5 g Al (c = 0.897

J g°C)

100 °C 5 g Cu (c = 0.385

J g°C)

−5 °C 5 g tin (c = 0.227

J g°C)

−5 °C

A B

100 mL 30 °C 100 mL 30 °C

A B

100 mL 30 °C 100 mL 30 °C

Calorimetry (W2016,Q11) based on Q = mcΔT

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Two glasses of water, both at 30 °C, have a block of metal dropped in them. After equilibrium, which glass is the warmest? 5 g Fe (c = 0.449

J g°C)

100 °C 7.5 g Fe (c = 0.449

J g°C )

100 °C A 100 mL A 100 mL

Review

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A gas is enclosed in a container fitted with a piston. The pressure of the gas is maintained at 8000 Pa while heat is slowly added. As a result, the piston is pushed up and the volume of the gas increased by 4 L. If 42 J of heat was added to the system during this expansion, what was the change in internal energy for the piston? a) ­10 J b) 0 J c) +10 J d) +42 J

*In class, your professors have put a lot of emphasis on saying that kPa × L or Pa × m3 equal Joules! WATCH OUT: When using 𝐱 = −𝐐∆𝐖 ALWAYS use kilopascals for Pressure! Volume is in Liters. kPa × L = Joule

Review

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Calculate the change in enthalpy for the following gas phase reaction at 27 °C if the change internal energy was ­1000 J. Note: R = 8.314

J mol K

a) ­3.49 kJ b) +3.49 kJ c) ­22.6 kJ d) +22.6 kJ

Review

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A gas confined by a piston expands at 100 kPa. When 20,000 J of heat is absorbed by the system, its volume increases from 100 L to 250 L. Which of the following statements is correct? a) The internal energy of the system increased. b) All heat added to the system was converted into work. c) Work was done on the system by the surroundings. d) The temperature of the system decreased.

CH4 g + 2 O2 g CO2 g + 2 H2O g CO2 g + 2 H2O g

Reaction Coordinate Diagrams (F2015 Q10)

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35 Energy Reaction Coordinate ∆H = +800 kJ Endothermic (energy is taken in) Exothermic (energy is released) Energy Reaction Coordinate CH4 g + 2 O2 g ∆H = −800 kJ A → B ∆H = +100 kJ B → C ∆H = −40kJ A → C ∆H = +60kJ Energy Reaction Coordinate ∆H = +100 kJ

𝐁 B 𝐃

∆H = −40 kJ ∆H = +60 kJ

Hess’s Law Given: A + B ⇌ C, ∆H = +50 kJ

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C ⇌ D, ∆H = −120 kJ A + B ⇌ D, ∆H = Reaction 1 Reaction 2 Net If stoichiometric coefficients of a reaction step are divided by a factor, then the ΔH for that step must be divided by that same factor. A + B ⇌ C ∆H = If a reaction step is reversed, then the sign of the ΔH of that step must also be reversed. C ⇌ A + B ∆H = A + B ⇌ C, ∆H = +50 kJ Energy Reaction Coordinate If stoichiometric coefficients of a reaction step are multiplied by a factor, then the ΔH for that step must be multiplied by that same factor. A + B ⇌ C ∆H =

Using the equations below:

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C s + O2 g → CO2 g ∆H° = −390 kJ Mn s + O2 g → MnO2 s ∆H° = −520 kJ What is ∆H° (in kJ) for the following reaction? MnO2 s + C s → Mn s + CO2 g a) 910 kJ b) 130 kJ c) ­130 kJ d) ­910 kJ *Review – what is ∆𝑜𝑕𝑏𝑡 for the above system?

Calculating ∆H using Heats of Formation

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Calculate the enthalpy of CaCO3(s) → CO2 g + CaO s Given ∆Hf CO2 g

°

= −393.5 kJ mol ∆Hf CaCO3(s)

°

= −1207.6 kJ mol ∆Hf CaO s

°

= −634.9 kJ mol ∆Hrxn

°

= ෍ 𝑜∆Hf(Products)

°

− ෍ 𝑜∆Hf Reactants

°

*Review – what is ∆𝑜𝑕𝑏𝑡 for the above system?

Given

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C8 NO2 8(s) → 8 CO2 g + 4 N2 g What is the molar enthalpy for the above reaction if the enthalpy decreased by 800 kJ through the combustion of 2.5 moles of octanitrocubane at STP? What is q sys? What is the enthalpy per mole of octanitrocubane? What is the enthalpy change if 5 moles of octanitrocubane reacted? If done in a closed container, what would happen to the pressure? At STP, how much PV work does the decomposition of 10 moles of ONC due on the surroundings?

Enthalpy and Stoichiometry (W2016 Q10)

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2 Al s + Fe2O3 s → 2 Al2O3 s + Fe s , ΔH° = −849 kJ Given the above reaction, how much heat is released when 22 moles of aluminum are reacted? How many grams of iron would have been produced in a reaction that released 16.0 MJ of energy? If the heat of formation of Fe2O3 s is ­824.2 kJ/mol, what is the heat of formation of Al2O3 s ?

Enthalpy of Formation

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What is the ∆𝐼

𝑔 0 of Br2 l ? (negative or positive?)

Is this process endothermic or exothermic? What reaction are we describing? What is the ∆𝐼

𝑔 0 of Br l ? (negative or positive?)

Is this process endothermic or exothermic? What reaction are we describing? What is the ∆𝐼

𝑔 0 of Br2 s ? (negative or positive?)

Is this process endothermic or exothermic? What reaction are we describing?

∆𝐼

𝑔 is the enthalpy change when one mole of a substance forms from its pure elements at standard state. The standard state of an element is the phase (solid, liquid,

• r gas) of that pure element at 25℃ and 1 atm. Most of the elements on the periodic table are solids at standard state (25℃ and 1 atm)!

Mercury and Bromine are the ONLY liquids at standard state (Hg l and Br2 l ) The homonuclear diatomics are H2 g , N2 g , O2 g , F2 g , Cl2 g , Br2 l , I2 s . The only gases on the periodic table are H2 g , N2 g , O2 g , F2 g , Cl2 g , and the noble gases. For elements in standard state e.g., ∆Hf

° = 0 kJ/mol

How does this look on an energy diagram?

Energy Rxn Coordinate Energy Rxn Coordinate Energy Rxn Coordinate

Review

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Which one of the following thermochemical equations represents the standard formation reaction for solid mercury (I) iodate? a) Hg aq

+

+ IO3(aq)

−

→ HgIO3 s b) HgI aq +

2 3 O2 g → HgIO3 s

c) Hg l +

1 2 I2 s + 3 2 O2 g → HgIO3 s

d)

1 2 Hg2 l + 1 2 I2 s + 2 O2 g → HgIO3 s

Review

43

What is the standard reaction enthalpy for the reaction below? 2 CO2 g → 2 C s + O2 g a) −∆Hf

° CO2 g

b) −2∆Hf

° CO2 g

c) +2∆Hf

° CO2 g

d) +2∆Hf

° C s

+ ∆Hf

° O2 g

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Bond Dissociation Energy (W2016 Q12, W2013 Q23)

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44 Given the following bond dissociation energies ∆Hrxn

°

= ෍ 𝐶𝑝𝑜𝑒𝑡 𝑆𝑓𝑏𝑑𝑢𝑏𝑜𝑢𝑡 − ෍ 𝐶𝑝𝑜𝑒𝑡 𝑄𝑠𝑝𝑒𝑣𝑑𝑢𝑡 What is the enthalpy of the following reaction? CH4 g + 2 O2 g → CO2 g + 2 H2O g Bond BDE (kJ/mol) C—H 413 C—O 358 O=O 498 O—H 467 H—H 432 C=O 799 Statement on energy in bonds:

Equilibrium

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45

Equilibrium is a thermodynamic property – it lets you know WHETHER a reaction will happen: Diamonds ⇌ Graphite K =

Products Reactants = Graphite Diamond = 1025

The above process is favored due to equilibrium. But does it happen?

Chemical Equilibrium – The ‘Forever Cake’

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Reversible Reaction H2O l ⇌ H2O g Condensation  Evaporation, position of equilibrium is temperature dependent Irreversible Reaction egg  omelet Irreversible Imagine you put a cake in the oven… and regardless of how long you bake it, there’s always batter – it NEVER gets done! THIS is the concept of equilibrium – a system that maintains a mixture of reactants and products

Equilibrium

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all values in

𝐧𝐩𝐦𝐟𝐭 𝐌𝐣𝐮𝐟𝐬

H2 g + I2 g ⇌ 2 HI g Initial 1.0 2.0 Change Equilibrium 1.5

time time Concentration (mol/L) Rate of Reaction H2 g [I2 g ] HI g Forward Rate Reverse Rate 2.5 1.5 1.0 0.5 2.0 Forward reaction: H2 g + I2 g ⇌ 2 HI g Reverse reaction: 2 HI g ⇌ H2 g + I2 g 2 HI g H2 g + I2 g Energy Reaction Coordinate ∆H

Kconcentration and Kpressure for an all Gas System

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An ‘all gas’ system is simply one where there are no aqueous and only gas reagents: 4C g + 3 E s → 2D g + 3F l ∆ngas = Kc= Kp= Let’s say the above system has a Kc of 1500. What is its Kp? We use the formula Kp = Kc RT ∆ngas but how is it derived? PV = nRT P RT = n V Kp RT ∆ngas = Kc OR Kp = Kc RT ∆ngas

Le Chatelier’s Principle – Qualitative Analysis

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A reaction maintains the equilibrium constant (K) at a specified temperature by shifting right (to products) or shifting left (to reactants) PCl3 g + Cl2 g ⇄ PCl5 g , K=1.00 If PCl3 g = Cl2 g = PCl5 g = 1.0 M and [Cl2 g ] is decreased to 0.5 M, which way does the system ‘shift’ to compensate for the change in concentration?

• 1. Think of equilibrium as being on a well­balanced see­saw:

The system is initially at equilibrium if we remove Cl2(g) then the see­saw is unbalanced we shift left to correct for the lack of Cl2(g) PCl3 g + Cl2 g ⇄ PCl5 g

Equilibrium Mathematics

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Q is the “Reaction Quotient”

Q = Productsnot at equilibrium Reactantsnot at equilibrium K = Products Reactants

• 1. If K > Q, the system shifts to products

(shifts right)

• 2. If K = Q, the system is at equilibrium

(no shit)

• 3. If K < Q, the system shifts to reactants

(shifts left) For the following reaction: P4 g + 5 O2 g ⇌ P

4O10 g , K = 3

In which direction must the reaction proceed to reach equilibrium if: (a) [P4 g ], [O2 g ], and [P

4O10 g ] = 1.0 M

(b) [P4 g ] = 8.0 M, [O2 g ] = 0.70 M, and [P

4O10 g ] = 5.0 M

Is the value of K greater than or less than 1?

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The __________________ are favored and K is ______________ than 1. (products/reactants/both) (greater/less/equal)

⇌

Time 1 Time 2 Time 3 Time 4

[] vs. t Graphs and Bar Graphs

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all values in moles/L

CH3Br g + Cl g

−

→ CH3Cl g +Br g

−

Initial 1.00 1.00 0.20 0.20 Change Equilibrium time Concentration (mol/L) Initial [] Equilibrium []

1.00 0.60 0.40 0.20 0.80

If K=0.25, which way does the system shift and what is the concentration of reagents at equilibrium?

CH3Br g Cl g

−

CH3Cl g Br g

−

CH3Br g Cl g

−

CH3Cl g Br g

−

CH3Cl g Br g

−

CH3Br g [Cl g

− ]

Manipulating Equilibrium Constants

53

(C) Pavel Sedach Learnfaster.ca

Reaction 1 A + B ⇌ C K1 Reaction 2 C ⇌ D K2 Net A + B ⇌ D K3 = K1 × K2 Reaction 1 A + B ⇌ C K1= Reaction 2 C ⇌ D K2= Net A + B + C ⇌ D + C K3= × 1 = A + B ⇌ C K1 × 2 = 2A + 2B ⇌ 2C K2 × 3 = 3A + 3B ⇌ 3C K3 × −1 = C ⇌ A + B K−1 = 1

K

× −

1 2 = 1 2 C ⇌ 1 2 A + 1 2 B

K− 1

2 =

1 K1/2 = 1 K

Manipulation of Equation Effect on Equilibrium Constant K =

Review ­ Manipulating Equilibrium Constants

(C) Pavel Sedach Learnfaster.ca

54 1 2 N2 g + O2 g → NO2 g , K1 = 1.2 × 10−4 2 NO2 g → N2O4 g , K2 = 52 What is the equilibrium constant for: N2O4 g → N2 g + 2O2 g

3 3

4 4

5 5

6 6

7 7

8 8

9 9

12 12

• ❏
• ❏
• ❏
• 13

13

• ⇌

⇌ ½⇌

• Marks

2 Marks 2

• ❏
• ❏
• ❏
• 23

23

• ·−·− ·−·−
• Δ ·
• Δ ·
• Δ ·
• Δ ·

Δ

• Δ −·
• Δ −
• Δ
• Δ
• 27

27