General Chemistry M R NaimiJamal Faculty of Chemistry
- Slides: 54
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 -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 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. • 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 acting through a distance. • Kinetic Energy – The energy of motion.
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 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. – 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 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. 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 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 that of almost any other substance.
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 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 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: – 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
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 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 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 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 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. 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 = = -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 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 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 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 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 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. • 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 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 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. • Gas formed pushes against the atmosphere. • Volume changes. • Pressure-volume work.
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 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 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 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
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 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), (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 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 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
Enthalpy of Reaction Hrxn = ∑ Hf°products - ∑ Hf°reactants
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 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 resource. – Environmental impact.
Chapter 5 Questions 6, 8, 17, 23, 25, 27 30, 32, 34, 36, 39 47, 53, 60, 63
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