General Chemistry M R NaimiJamal Faculty of Chemistry

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General Chemistry M. R. Naimi-Jamal Faculty of Chemistry Iran University of Science & Technology

General Chemistry M. R. Naimi-Jamal Faculty of Chemistry Iran University of Science & Technology

Contents 5 -1 5 -2 5 -3 5 -4 5 -5 5 -6 5

Contents 5 -1 5 -2 5 -3 5 -4 5 -5 5 -6 5 -7 5 -8 5 -9 Getting Started: Some Terminology Heats of Reaction and Calorimetry Work The First Law of Thermodynamics Heats of Reaction: U and H The Indirect Determination of H: Hess’s Law Standard Enthalpies of Formation Fuels as Sources of Energy

Thermochemistry: Basic Terms • Thermochemistry is the study of energy changes that occur during

Thermochemistry: Basic Terms • Thermochemistry is the study of energy changes that occur during chemical reactions. • System: the part of the universe being studied. • Surroundings: the rest of the universe.

Types of Systems • Open: energy and matter can be exchanged with the surroundings.

Types of Systems • Open: energy and matter can be exchanged with the surroundings. • Closed: energy can be exchanged with the surroundings, matter cannot. • Isolated: neither energy nor matter can be exchanged with the surroundings. A closed system; energy (not matter) can be exchanged.

Terminology • Energy, U – The capacity to do work. • Work – Force

Terminology • Energy, U – The capacity to do work. • Work – Force acting through a distance. • Kinetic Energy – The energy of motion.

Energy • Kinetic Energy ek = 1 2 mv 2 kg m 2 =

Energy • Kinetic Energy ek = 1 2 mv 2 kg m 2 = J [ek ] = 2 s • Work w = Fd kg m m = J [w ] = 2 s

Energy • Potential Energy – Energy due to condition, position, or composition. – Associated

Energy • Potential Energy – Energy due to condition, position, or composition. – Associated with forces of attraction or repulsion between objects. • Energy can change from potential to kinetic.

Energy and Temperature • Thermal Energy – Kinetic energy associated with random molecular motion.

Energy and Temperature • Thermal Energy – Kinetic energy associated with random molecular motion. – In general proportional to temperature. – An intensive property. • Heat and Work – q and w. – Energy changes.

Thermal Equilibrium • Heat transfer ALWAYS occurs from a hotter object to a cooler

Thermal Equilibrium • Heat transfer ALWAYS occurs from a hotter object to a cooler object (directionality) • Transfer of heat continues until both objects are at the same temperature (thermal equilibrium) • The quantity of heat lost by a hotter object and the quantity of heat gained by a cooler object when they are in contact are numerically equal (required by the law of conservation of energy)

Units of Heat • Calorie (cal) heat required to raise the temperature of 1.

Units of Heat • Calorie (cal) heat required to raise the temperature of 1. 00 g of pure liquid water from 14. 5 to 15. 5 o. C. • Joule (J): SI unit for heat 1 cal = 4. 184 J

Heat Capacity • The quantity of heat required to change the temperature of a

Heat Capacity • The quantity of heat required to change the temperature of a system by one degree. – Specific heat capacity, c. q = mc T • System is one gram of substance – Heat capacity, C. • C = Mass x specific heat (mc). q = C T

Many metals have low specific heats. The specific heat of water is higher than

Many metals have low specific heats. The specific heat of water is higher than that of almost any other substance.

Example: How much heat, in joules and in kilojoules, does it take to raise

Example: How much heat, in joules and in kilojoules, does it take to raise the temperature of 225 g of water from 25. 0 to 100. 0 °C? Example: What will be the final temperature if a 5. 00 -g silver ring at 37. 0 °C gives off 25. 0 J of heat to its surroundings? The specific heat of silver is 0. 235 Jg-1°C-1.

Example: An Estimation Example Without doing detailed calculations, determine which of the following is

Example: An Estimation Example Without doing detailed calculations, determine which of the following is a likely approximate final temperature when 100 g of iron at 100 °C is added to 100 g of 20 °C water in a calorimeter, the specific heat of iron is 0. 449 Jg -1°C-1: (a) 20 °C (b) 30 °C (c) 60 °C (d) 70 °C.

Conservation of Energy • In interactions between a system and its surroundings the total

Conservation of Energy • In interactions between a system and its surroundings the total energy remains constant. energy is neither created nor destroyed. qsystem + qsurroundings = 0 qsystem = -qsurroundings

First Law: Sign Convention • Energy entering a system carries a positive sign: –

First Law: Sign Convention • Energy entering a system carries a positive sign: – heat absorbed by the system, or – work done on the system • Energy leaving a system carries a negative sign – heat given off by the system – work done by the system

Determination of Specific Heat

Determination of Specific Heat

Example: Determining Specific Heat from Experimental Data. Use the data presented on the last

Example: Determining Specific Heat from Experimental Data. Use the data presented on the last slide to calculate the specific heat of lead. qlead = -qwater = mc T = (50. 0 g)(4. 184 J/g °C)(28. 8 - 22. 0)°C qwater = 1. 4 x 103 J qlead = -1. 4 x 103 J = mc T = (150. 0 g)(c)(28. 8 - 100. 0)°C clead = 0. 13 Jg-1°C-1

Heats of Reaction and Calorimetry • Chemical energy. – Contributes to the internal energy

Heats of Reaction and Calorimetry • Chemical energy. – Contributes to the internal energy of a system. • Heat of reaction, qrxn. – The quantity of heat exchanged between a system and its surroundings when a chemical reaction occurs within the system, at constant temperature.

Heats of Reaction (qrxn) • An exothermic reaction gives off heat – In an

Heats of Reaction (qrxn) • An exothermic reaction gives off heat – In an isolated system, the temperature increases. – The system goes from higher to lower energy; – qrxn is negative. • An endothermic reaction absorbs heat – In an isolated system, the temperature decreases. – The system goes from lower to higher energy; – qrxn is positive.

Bomb Calorimeter qrxn = -qcal = qbomb + qwater + qwires +… Define the

Bomb Calorimeter qrxn = -qcal = qbomb + qwater + qwires +… Define the heat capacity of the calorimeter: qcal = ∑mici T = C T i heat

Example: Using Bomb Calorimetry Data to Determine a Heat of Reaction. The combustion of

Example: Using Bomb Calorimetry Data to Determine a Heat of Reaction. The combustion of 1. 010 g sucrose, in a bomb calorimeter, causes the temperature to rise from 24. 92 to 28. 33°C. The heat capacity of the calorimeter assembly is 4. 90 k. J/°C. (a) What is the heat of combustion of sucrose, expressed in k. J/mol C 12 H 22 O 11 (b) Verify the claim of sugar producers that one teaspoon of sugar (about 4. 8 g) contains only 19 calories.

Example: Calculate qcalorimeter: qcal = C T = (4. 90 k. J/°C)(28. 33 -24.

Example: Calculate qcalorimeter: qcal = C T = (4. 90 k. J/°C)(28. 33 -24. 92)°C = (4. 90)(3. 41) k. J = 16. 7 k. J Calculate qrxn: qrxn = -qcal = -16. 7 k. J per 1. 010 g

Example: Calculate qrxn in the required units: -16. 7 k. J qrxn = -qcal

Example: Calculate qrxn in the required units: -16. 7 k. J qrxn = -qcal = = -16. 5 k. J/g 1. 010 g qrxn 343. 3 g = -16. 5 k. J/g 1. 00 mol = -5. 65 x 103 k. J/mol (a) Calculate qrxn for one teaspoon: qrxn 4. 8 g 1. 00 cal )= -19 kcal/tsp = (-16. 5 k. J/g)( )( 4. 184 J 1 tsp Note: in food industry we say calorie instead of kcal! (b)

Internal Energy • Internal Energy, U. – Total energy (potential and kinetic) in a

Internal Energy • Internal Energy, U. – Total energy (potential and kinetic) in a system. • Translational kinetic energy. • Molecular rotation. • Bond vibration. • Intermolecular attractions. • Chemical bonds. • Electrons.

First Law of Thermodynamics • A system contains only internal energy. – A system

First Law of Thermodynamics • A system contains only internal energy. – A system does not contain heat or work. – These only occur during a change in the system. U = q + w q = heat gained by system w = work gained (+) by or done on system (-) • Law of Conservation of Energy – The energy of an isolated system is constant

Example: A gas does 135 J of work while expanding, and at the same

Example: A gas does 135 J of work while expanding, and at the same time it absorbs 156 J of heat. What is the change in internal energy? U = q + w = 156 – 135 = 21 J

State Functions • The state of a system: its exact condition at a fixed

State Functions • The state of a system: its exact condition at a fixed instant. • State is determined by the kinds and amounts of matter present, the structure of this matter at the molecular level, and the prevailing pressure and temperature. • A state function is a property that has a unique value that depends only on the present state of a system, and does not depend on how the state was reached (does not depend on the history of the system).

State Functions • Water at 293. 15 K and 1. 00 atm is in

State Functions • Water at 293. 15 K and 1. 00 atm is in a specified state. • d = 0. 99820 g/m. L • This density is a unique function of the state. • It does not matter how the state was established.

Functions of State • U is a function of state. – Not easily measured.

Functions of State • U is a function of state. – Not easily measured. • U has a unique value between two states. – Is easily measured.

Path Dependent Functions For example: Fuel-Consume to go from one point to the another

Path Dependent Functions For example: Fuel-Consume to go from one point to the another one is NOT a function of state, but a path dependent function

Internal Energy Change at Constant Volume • For a system where the reaction is

Internal Energy Change at Constant Volume • For a system where the reaction is carried out at constant volume, V = 0 and U = q. V. • All thermal energy produced by conversion from chemical energy is released as heat; no P-V work is done.

Work • In addition to heat effects chemical reactions may also do work. •

Work • In addition to heat effects chemical reactions may also do work. • Gas formed pushes against the atmosphere. • Volume changes. • Pressure-volume work.

Heats of Reaction: U and H Reactants → Products Ui Uf U = Uf

Heats of Reaction: U and H Reactants → Products Ui Uf U = Uf - Ui U = qrxn + w In a system at constant volume: P V=0 U = qrxn + 0 = qrxn = qv But we live in a constant pressure world! How does qp relate to qv?

lnternal Energy Change at Constant Pressure • For a system where the reaction is

lnternal Energy Change at Constant Pressure • For a system where the reaction is carried out at constant pressure, U = q. P – P V or U + P V = q. P • Most of thermal energy is released as heat. • Some work is done to expand the system against the surroundings (push back the atmosphere).

Heats of Reaction

Heats of Reaction

Heats of Reaction U = q. P + w We know that w =

Heats of Reaction U = q. P + w We know that w = - P V (minus because done by system) therefore: U = q. P - P V q. P = U + P V These are all state functions, so define a new function. Let H = U + PV Then H = Hf – Hi = U + PV= U + P V + V P If we work at constant pressure and temperature: V P = 0 H = U + P V = q. P

Standard States and Standard Enthalpy Changes • Define a particular state as a standard

Standard States and Standard Enthalpy Changes • Define a particular state as a standard state. • Standard enthalpy of reaction, H° – The enthalpy change of a reaction in which all reactants and products are in their standard states. • Standard State – The pure element or compound at a pressure of 1 atm and at the temperature of interest (usually 25 °C).

Enthalpy Diagrams

Enthalpy Diagrams

Indirect Determination of H: Hess’s Law • H is an extensive property. – Enthalpy

Indirect Determination of H: Hess’s Law • H is an extensive property. – Enthalpy change is directly proportional to the amount of substance in a system. N 2(g) + O 2(g) → 2 NO(g) ½N 2(g) + ½O 2(g) → NO(g) H = +180. 50 k. J H = +90. 25 k. J • H changes sign when a process is reversed NO(g) → ½N 2(g) + ½O 2(g) H = -90. 25 k. J

Hess’s Law • Hess’s law of constant heat summation – If a process occurs

Hess’s Law • Hess’s law of constant heat summation – If a process occurs in stages or steps (even hypothetically), the enthalpy change for the overall process is the sum of the enthalpy changes for the individual steps. ½N 2(g) + ½O 2(g) → NO(g) H = +90. 25 k. J NO(g) + ½O 2(g) → NO 2(g) H = -57. 07 k. J ½N 2(g) + O 2(g) → NO 2(g) H = +33. 18 k. J

Example: Calculate the enthalpy change for reaction (a) given the data in equations (b),

Example: Calculate the enthalpy change for reaction (a) given the data in equations (b), (c), and (d). (a) 2 C(graphite) + 2 H 2(g) C 2 H 4(g) DH = ? (b) C(graphite) + O 2(g) CO 2(g) Hb = – 393. 5 k. J (c) C 2 H 4(g) + 3 O 2 2 CO 2(g) + 2 H 2 O(l) Hc = – 1410. 9 k. J (d) H 2(g) + ½ O 2 H 2 O(l) Hd = – 285. 8 k. J (a) = 2 (b) + 2 (d) – (c) & Ha = 2 Hb + 2 Hd - Hc

Standard Enthalpies of Formation Hf° • The enthalpy change that occurs in the formation

Standard Enthalpies of Formation Hf° • The enthalpy change that occurs in the formation of one mole of a substance in the standard state from the reference forms of the elements in their standard states. • The standard enthalpy of formation of a pure element in its reference state is 0.

Standard Enthalpy of Formation When we say: “The standard enthalpy of formation of CH

Standard Enthalpy of Formation When we say: “The standard enthalpy of formation of CH 3 OH(l) is 238. 7 k. J”, we are saying that the reaction: C(graphite) + 2 H 2(g) + ½ O 2(g) CH 3 OH(l) has a value of ΔH of – 238. 7 k. J. We can treat ΔHf° values as though they were absolute enthalpies, to determine enthalpy changes for reactions. Question: What is ΔHf° for an element in its standard state [such as O 2(g)]? Hint: since the reactants are the same as the products …

Standard Enthalpies of Formation

Standard Enthalpies of Formation

Enthalpy of Reaction Hrxn = ∑ Hf°products - ∑ Hf°reactants

Enthalpy of Reaction Hrxn = ∑ Hf°products - ∑ Hf°reactants

Table 7. 3 Enthalpies of Formation of Ions in Aqueous Solutions

Table 7. 3 Enthalpies of Formation of Ions in Aqueous Solutions

Example: Synthesis gas is a mixture of carbon monoxide and hydrogen that is used

Example: Synthesis gas is a mixture of carbon monoxide and hydrogen that is used to synthesize a variety of organic compounds. One reaction for producing synthesis gas is 3 CH 4(g) + 2 H 2 O(l) + CO 2(g) 4 CO(g) + 8 H 2(g) ΔH° = ? Use standard enthalpies of formation from Table 6. 2 to calculate the standard enthalpy change for this reaction. Example: The combustion of isopropyl alcohol, common rubbing alcohol, is represented by the equation 2 (CH 3)2 CHOH(l) + 9 O 2(g) 6 CO 2(g) + 8 H 2 O(l) ΔH° = – 4011 k. J Use this equation and data from Table 6. 2 to establish the standard enthalpy of formation for isopropyl alcohol.

Fuels as Sources of Energy • Fossil fuels. – Combustion is exothermic. – Non-renewable

Fuels as Sources of Energy • Fossil fuels. – Combustion is exothermic. – Non-renewable resource. – Environmental impact.

Chapter 5 Questions 6, 8, 17, 23, 25, 27 30, 32, 34, 36, 39

Chapter 5 Questions 6, 8, 17, 23, 25, 27 30, 32, 34, 36, 39 47, 53, 60, 63