Créer une présentation
Télécharger la présentation

Télécharger la présentation
## AP Chemistry

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**AP Chemistry**Thermodynamics 9.24.10**Bell work**• Turn in your precipitation lab. • By titration, 15.0 mL of 0.108 M sodium hydroxide is needed to neutralize a 0.2053-g sample of an organic acid. What is the molar mass of the acid if it is monoprotic? (b) An elemental analysis of the acid indicates that is is composed of 5.89% H, 70.6% C, and 23.5% O by mass. What is its molecular formula?**Agenda**• Bell work • Turn in lab • Review chapter 4 homework problems • Review unit 1 test (focus on stoichiometry) • Quiz – solubility rules/predicting products, stoichiometry, naming (50 points) • Homework: Chapter 4 # 91, 92, 96, 114 and stoichiometry revisit – due Friday**Chapter 4 Homework**• P-set 1: 16, 18, 22, 24, 27, 34, 36, 38, 40, 44 • P-set 2: 50, 52, 56, 70, 80, 86**Bell work**• Give acid/base definitions on Friday • 70, 80, and 86 in class • Test 41, 42, 43, 44**9/24 Bell work**• Turn in stoichiometry revisit and chapter 4 # 91, 92, 96, 114 • A solution is made by mixing 12.0 g of NaOH and 75.0 mL of 0.200 M HNO3. After the reaction takes place, how many moles of each ion remain in solution? • *** You have a quiz today. When you finish the bell work, study for your quiz!!*****Agenda**• Bell work • Chapter 4 quiz • When you finish, begin working on your homework—Read chapter 5 and complete chapter 5 overview (due Tuesday) • Homework: Read chapter 5 and complete chapter 5 overview**Quiz**• Predicting products #1 – change iron (II) sulfate to calcium sulfate • As far as solubility goes, treat nitrite like nitrate • Unless a redox reaction is indicated, keep oxidation numbers the same. We will get to more complicated cases later.**Homework – due next Tuesday**• 5.1, 5.2, 5.23, 5.26, 5.33, 5.38, 5.50, 5.59, 5.60, 5.62, 5.64, 5.68, 5.70, 5.74, 5.75**Bell work**• Give acid/base definitions on Friday • 70, 80, and 86 in class • Test 41, 42, 43, 44**Bell work**• Turn in your chapter 5 overview. • What is thermodynamics? What is your experience with thermodynamics in the past? Which of the topics from chapter 5 were familiar? Which were new? • On a scale of 1-10 (1 being easy, 10 being impossible), how was chapter 5?**Agenda**• Bell work • Chapter 5 discussion • Chapter 5 overview with practice problems built in • HW: Try the sample problems (NOT the homework problems listed at the top of the sheet, but the sample problems) Chapter 4 #2, 3, 4, 24, 25, 33, 35, 37, 43, 45 Have your parents sign your grade sheet**Thermodynamics**• Thermodynamics: study of energy and its transformations • Chapter 5 focuses on thermochemistry: chemical reactions and energy changes involving heat**Energy**• The ability to do work or transfer heat. • Work: Energy used to cause an object that has mass to move. • Heat: Energy used to cause the temperature of an object to rise.**Different types of energy**• Kinetic energy • Potential energy • Electrostatic potential energy • Chemical energy • Thermal energy**How can matter possess energy and how can that energy be**transferred? • Kinetic energy – the energy of motion • Depends on mass and speed • KE= ½ mv2 • Potential energy • Based on virtue of an object’s position or chemical composition. • PE = mgh • m=mass; h=height; g=gravitational constant (9.8m/s2)**Where is the bicyclist’s potential energy the highest?**What happens to his potential energy as he goes down the hill?**Units of energy**• SI unit: • Joules • A joule is equal to what? • Use KE= ½ mv2 • mkg, vm/s, v2 = m2/s2 • 1 J= kg-m2/s2 • We will often use kJ when discussing energies associated with chemical reactions**Systems and surroundings**• System: the portion of the universe we single out for study • Surroundings: everything else • Closed system: can exchange energy (in the form of heat and work) but not matter with its surroundings**System and surroundings**• Here we have hydrogen and oxygen gases confined in a cylinder with a movable piston. • What is the system? What are the surroundings? • Has the system lost or gained mass? Why or why not? 2 H2 (g) + O2 2 H2O (g) + energy The system includes the molecules we want to study (here, the hydrogen and oxygen molecules). The surroundings are everything else (here, the cylinder and piston).**Transfer of energy**• Energy is transferred between systems and surroundings as work or heat • Work: • Heat: • Energy used to cause an object that has mass to move. • Energy transferred when an object is moved by a force • Energy used to cause the temperature of an object to rise.**Work**• w = F x d • We perform work when we lift an object against the force of gravity or when we bring two like charges together • In these two situations, what is the system and what are the surroundings? • What is the transfer of energy? Into or out or system?**Heat**• Heat: energy transferred from a hotter object to a colder one • Energy transferred between a system and its surroundings as a result of the difference in temperature • Example: Burning of natural gas combustion reaction • What is the system and what are the surroundings? • What is the transfer of energy?**Example**• The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall. • As the ball falls, its potential energy is converted to kinetic energy. • When it hits the ground, its kinetic energy falls to zero (since it is no longer moving); some of the energy does work on the ball, the rest is dissipated as heat.**Example 2**• A bowler lifts a 5.4-kg (12-lb) bowling ball from ground level to a height of 1.6 m (5.2 feet) and then drops the ball back to the ground. • What happens to the potential energy of the bowling ball as it is raised from the ground? • What quantity of work, in J, is used to raise the ball? • After the ball is dropped, it gains kinetic energy. If we assume that all of the work done in part (b) has been converted to kinetic energy by the time the ball strikes the ground, what is the speed of the ball at the instant just before it hits the ground? (Note: The force due to gravity is F = mg, where m is the mass of the object and g is the gravitational constant; g = 9.8 m/s2.)**What happens to the potential energy of the bowling ball as**it is raised from the ground? • Because the bowling ball is raised to a greater height above the ground, its potential energy increases.**What quantity of work, in J, is used to raise the ball?**The ball has a mass of 5.4 kg, and it is lifted a distance of 1.6 m. To calculate the work performed to raise the ball, we use both Equation 5.3 and F = mg for the force that is due to gravity: w = F x d w = m x g x d w = (5.4 kg) (9.8 m/s2) (1.6 m) w = 85 kg-m/s2 w = 85 J**After the ball is dropped, it gains kinetic energy. If we**assume that all of the work done in part (b) has been converted to kinetic energy by the time the ball strikes the ground, what is the speed of the ball at the instant just before it hits the ground? (Note: The force due to gravity is F = mg, where m is the mass of the object and g is the gravitational constant; g = 9.8 m/s2.) • When the ball is dropped, its potential energy is converted to kinetic energy. At the instant just before the ball hits the ground, we assume that the kinetic energy is equal to the work done in part (b), 85 J: • How can we find the speed?**We can now solve this equation for v:**Check: Work must be done in part (b) to increase the potential energy of the ball, which is in accord with our experiences. The units are appropriate in both parts (b) and (c). The work is in units of J and the speed in units of m/s. In part (c) we have carried an additional digit in the intermediate calculation involving the square root, but we report the final value to only two significant figures, as appropriate.**Practice exercises**• (a) What is the KE, in J, of an Ar atom moving with a speed of 650 m/s? • (b) What is the KE, in J, of a mole of Ar atoms moving with a speed of 650 m/s? (Hint 1 amu= 1.66 x 10-27 kg)**First law of thermodynamics**• Energy cannot be created or destroyed • It is also the law of conservation of energy: simply put, energy is conserved • The total energy of the universe is constant • Any energy lost by the system must be gained by the surroundings and vice versa • This can be expressed quantitatively**Internal Energy**• Internal energy: sum of all the kinetic and potential energies of all its components • Denoted as U or E (we’ll use E) • It is very difficult to know precisely the internal energy of a system (includes the motion of molecules through space, their rotations, and internal vibrations) • Rather, we hope to know the change in E • By definition, the change in internal energy, E, is the final energy of the system minus the initial energy of the system: E = Efinal−Einitial**Values of ΔE**• Negative Δ E: Efinal < Einitial • System has lost energy to surroundings • Positive Δ E: Efinal > Einitial • System has gained energy from surroundings Negative Δ E Positive Δ E**Signs of Δ E**• Tell you how heat flows in relation to the system • Out of the system negative • Into the system positive**Changes in Internal Energy**Energy diagram • If E > 0, Efinal > Einitial • Therefore, the system absorbed energy from the surroundings. • This energy change is called endergonic.**Changes in Internal Energy**• If E < 0, Efinal < Einitial • Therefore, the system released energy to the surroundings. • This energy change is called exergonic.**Relating ΔE to Heat and Work**• When a system undergoes a physical or chemical change, the change in internal energy is given by the heat added to or liberated from the system plus the work done on or by the system • Any energy entering the system has a positive sign • Therefore, q or w would have a positive value Δ E = q + w heat work**Example**• Two gases, A(g) and B(g), are confined in a cylinder-and-piston arrangement. Substances A and B react to form a solid product: As the reactions occurs, the system loses 1150 J of heat to the surrounding. The piston moves downward as the gases react to form a solid. As the volume of the gas decreases under the constant pressure of the atmosphere, the surroundings do 480 J of work on the system. What is the change in the internal energy of the system? A (g) + B (g) C (s)**We first determine the signs of q and w (Table) and then use**Equation 5.5, E = q + w, to calculate E. • Heat is transferred from the system to the surroundings, and work is done on the system by the surroundings, so q is negative and w is positive: q –1150 J and w 480 kJ. Thus, E is • Did the system lose or gain energy?**Practice exercise**• Calculate the change in the internal energy of the system for a process in which the system absorbs 140 J of heat from the surroundings and does 85 J of work on the surroundings.**Endothermic vs Exothermic**• When heat is released by the system to the surroundings, the process is exothermic. • When heat is absorbed by the system from the surroundings, the process is endothermic.**State functions**• The internal energy of a system is independent of the path by which the system achieved that state • Because of this, internal energy is a state function • It depends only on the present state of the system, not on the path by which the system arrived at that state. • And so, E depends only on Einitial and Efinal.**Suppose you are traveling between Chicago and Denver.**Chicago is 596 ft above sea level; Denver is 5280 ft above sea level. No matter what route you take, the altitude will change 4684 feet. The distance you travel, however, will depend on your route. Altitude is analogous to a state function because the change in altitude is independent of the path taken. Distance traveled is not a state function. • In what ways is the balance in your checkbook a state function?**A battery in a flashlight can be discharged by producing**heat and light. The same battery in a toy car is used to produce heat and work. • The change in internal energy of the battery is the same (ΔE – a state function) • But q and w are different in the two cases (q and w are not state functions)**Work**• When a process occurs in an open container, commonly the only work done is a change in volume of a gas pushing on the surroundings (or being pushed on by the surroundings). • We can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston. This is called pressure-volume work. w = −PV**Enthalpy**• If a process takes place at constant pressure (as the majority of processes we study do) and the only work done is this pressure-volume work, we can account for heat flow during the process by measuring the enthalpy of the system. • Enthalpy is the internal energy plus the product of pressure and volume: H = E + PV