C H E M 1 CHEM 101-GENERAL CHEMISTRY CHAPTER 9 ENERGY & CHEMISTRY INSTR : FİLİZ ALSHANABLEH

Transcription

1 C H E M 1 CHEM 101-GENERAL CHEMISTRY CHAPTER 9 ENERGY & CHEMISTRY 0 1 INSTR : FİLİZ ALSHANABLEH

2 CHAPTER 9 ENERGY & CHEMISTRY Defining Energy Energy Transformation & Conservation of Energy Heat Capacity & Calorimetry Enthalpy Hess s Law & Heats of Reaction Heats of Reactions from Heat of Formation 2

3 Forms of Energy Two broad categories of energy: potential energy and kinetic energy. Potential energy - associated with the relative position of an object. Kinetic energy - associated with motion. Kinetic energy = 1 2 mv2 3

4 Forms of Energy Internal energy - the combined kinetic and potential energies of atoms and molecules that make up an object or system. Chemical energy - energy released or absorbed during a chemical reaction. Other forms of energy include radiant, mechanical, thermal, electrical, and nuclear. Thermochemistry - the study of the energetic consequences of chemistry involving heat flow. 4

5 Heat and Work Heat is the flow of energy between two objects because of a difference in temperature. Heat always flows from the warmer object to the cooler object. Work is the transfer of energy accomplished by a force moving a mass some distance against resistance. Pressure-volume work (PV-work) is the most common work type in chemistry. Releasing an inflated balloon before it is tied off illustrates an example of PV-work. 5

6 Energy Units Energy = Work done = Force x distance W = mass acceleration distance = kg m s 2 m The Joule is the SI unit of energy. 1 Joule = 1 kg m 2 /s 2 Other energy units include the Btu and the calorie. 1 Btu is the energy required to raise 1 lb of water 1 o F. 1 Btu = 1055J 1 calorie is the energy required to raise 1 g water from 14.5 to 15.5 o C. 1 calorie = J 6

7 Energy Transformation and Conservation of Energy During energy transformation, the total energy must be conserved. The sum of all energy conversions and energy transfers must equal the total energy present which must remain constant. To account for energy transformations and conversions, the system and surroundings must be specified. System - the part of the universe being considered. Surroundings - the remainder of the universe. System + Surroundings = Universe System and surroundings are separated by a boundary. 7

8 Energy Transformation and Conservation of Energy For a system or surroundings, the only possible forms of energy flow are heat, q, and work, w. The delta,, means change in and is defined as the difference in the final and initial states. E = q + w E = E final E initial 8

9 Energy Transformation and Conservation of Energy The sign resulting from the difference in the final and initial states indicates the direction of the energy flow. Negative values indicate energy is being released. Positive values indicate energy is being absorbed. q (+) q (-) SYSTEM w (+) w (-) SURROUNDING 9

10 Example Problem 9.1 If 515 J of heat is added to a gas that does 218 J of work as a result, what is the change in the energy of the system? q = J (heat added to the system) w = J (work done by the system) E = J + ( J) = J 10

11 Problem 9.23 Which sytem does more work: (a) E = J, q = 400 J; or (b) E = 317 J, q = 347 J Answer (a) E = J, q = 400 J (b) E = 317 J, q = 347 J w = J w = - 30 J System (a) does more work 11

12 Problem 9.24 In which case is heat added to the sytem: (a) E = - 43 J, w = 40 J; or (b) E = 31 J, w = 34 J Answer (a) E = - 43 J, w = 40 J (b) E = 31 J, w = 34 J q = - 83 J q = - 3 J In non of the cases heat is added to the system 12

13 Energy Transformation and Conservation of Energy First law of thermodynamics states that energy can be transformed from one form to another but cannot be created or destroyed. E universe = E surroundings + E system = 0 13

14 Waste Energy A common way to obtain work from a system is to heat the system. Heat flows in and is converted to work. It is impossible to completely convert all heat to work. Heat not converted to work is considered waste energy, which may contribute to thermal pollution. The efficiency of conversion from heat to work can be expressed as a percentage. Increases in energy consumption can be offset by increasing energy efficiencies. 14

15 Waste Energy Typical efficiencies of some common energy conversion devices. 15

16 Heat Capacity and Specific Heat c : The specific heat capacity, or specific heat, is a physical property of a substance that describes the amount of heat required to raise the temperature of one gram of a substance by 1 º C. C p : The molar heat capacity is a physical property of a substance that describes the amount of heat required to raise the temperature of one mole of a substance by 1 º C. The amount of heat energy absorbed can be quantified. q = mc T q = nc p T 16

17 Heat Capacity and Specific Heat Specific heat and molar heat capacities for some common substances. 17

18 Problem 9.31 A metal radiator is made from 26.0 kg of iron. The specific heat of iron is J/g º C. How much heat must be supplied to the radiator to raise its temperature from 25.0 to 55.0 º C. Answer: T = T final - T initial = = 30 º C q = mc T = (26.0x10 3 g)(0.449 J/g º C)(30 º C) = 3.50x10 5 J 18

19 Problem 9.31 A copper nail and an iron nail of the same mass and initially at the same room temperature are both put into a vessel containing boiling water. Which one would you expect to reach 100 º C first? Why? c iron = J/g º C and c copper = J/g º C. Answer: Copper will reach 100 º C first. Because its heat capacity that is specific heat less than specific heat of iron. Copper required less energy to reach expected temp. 19

20 Example Problem 9.4 A glass contains g of warm water at 78.0 º C. A piece of gold at 2.30 º C is placed in the water. The final temperature reached by this system is 76.9 º C. What was the mass of gold? The specific heat of water is J/g º C, and that of gold is J/g º C. Answer: q gold = q water m gold x c gold x T = m water x c water x T m gold (0.129 J/gºC)( ) = (250.0g) (4.184 J/gºC)( ) m gold = 120 g 20

21 Calorimetry Heat flow is measured using a calorimeter. A calorimeter measures the heat evolved or absorbed by the system of interest by measuring the temperature change in the surroundings. q system = q surroundings q gained = q lost q system Reaction System q surrounding Calorimetry 21

22 Calorimetry Two types of Calorimeter: Coffee-Cup Calorimeter Bomb Calorimeter Actual Measurement - temperature change for the calorimeter and the calorimeter constant are used to determine the amount of heat released by a reaction. q calorimeter = C calorimeter T calorimeter q reaction = q calorimeter 22

23 Coffee-Cup Calorimeter q reaction = q alorimeter q reaction = q water q reaction = = m water x c water x T 23

24 Bomb Calorimeter q reaction = q alorimeter q reaction = C calorimeter x T 24

25 Example Problem g of a particular fuel is burned in bomb calorimeter which has calorimeter constant (heat capacity) 118 J/ º C and the temperature increases by 5.28 º C. Calculate the energy density of this fuel, which is the amount of energy liberated per gram of fuel burned. Answer: q reaction = q calorimeter q reaction = C calorimeter x T q calorimeter = C calorimeter x T = (118 J/ o C )(5.28 o C) = J q reaction = q calorimeter = J heat evolved when g fuel burned Energy density = J / g = 1.40x 10 3 J/g = 1.40 kj/g 25

26 Enthalpy Enthalpy is the heat flow under conditions of constant pressure. The enthalpy change can be expressed as H = q p ΔH = H products H reactants State function ΔH = q at constant pressure 26

27 Enthalpy When a system releases heat, the process is said to be exothermic. The value of H is less than zero; the sign on H is negative. CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) ΔH < 0 When a system absorbs heat, the process is said to be endothermic. The value of H is greater than zero; the sign on H is positive. H 2 O(s) H 2 O(l) ΔH > 0 27

28 H of Phase Changes Phase changes occur under constant pressure conditions. The heat flow during a phase change is an enthalpy change. During a phase change, temperature does not change with heat flow due to formation or breaking of intermolecular attractive forces. 28

29 H of Phase Changes The heat required to convert a liquid to a gas is the heat of vaporization, H vap. H vap is endothermic with a positive value. The heat released to convert a gas to a liquid is the heat of condensation, H cond. H cond is exothermic with a negative value. H cond = H vap The value of H for a phase change is compound specific and has units of kj/mol. The heat flow can be calculated using the number of moles of substance, n, and the value of the enthalpy change. H = n H phase change 29

30 H of Phase Changes Standard molar enthalpies and temperatures for phase changes of water. 30

31 Problem 9.47 If 14.8 kj of heat is given off when 1.6 g of HCl condenses from vapor to liquid, what is H cond for this substances? (H:1.008, Cl:35.45) Answer: n HCl = 1.6 g / (36.46 g/mol) = mol HCl gives off 14.8 kj of heat H cond = 14.8 kj / (0.044 mol) H cond = kj / mol 31

32 Heat of Reaction Enthalpy changes can be calculated for chemical reactions, in addition to temperature changes and phase transitions that is commonly referred to as the heat of reaction. The enthalpy change for a reaction can be estimated using bond energies. During a chemical reaction, reactant bonds are broken and product bonds are made. Breaking bonds requires energy. Making bonds releases energy. If the amount of energy released making product bonds is greater than the amount of energy required to break reactant bonds, the reaction is exothermic. If the energy released is less than the energy required, the reaction is endothermic. 32

33 Bonds and Energy The combustion of methane breaks 4 C-H bonds and 2 O=O bonds. 2 C=O bonds and 4 O-H bonds are made. 33

34 Thermochemical Equations A thermochemical equation summarizes the overall energetics for a chemical reaction. The sign on the H indicates whether the reaction is endothermic or exothermic CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) ΔH rxn = 891 kj H 2 O(s) H 2 O(l) ΔH rxn = kj H 2 O(l) H 2 O(s) ΔH rxn = 6.0 kj 34

35 Thermochemical Equations The combustion of methane is an exothermic reaction and releases 891 kj of heat energy when 1 mole of methane reacts with 2 moles of oxygen. CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) ΔH rxn = 891 kj For thermochemical equations, if the stoichiometric coefficients are multiplied by some factor, the heat of reaction must also be multiplied by the same factor. 2CH 4 (g) + 4O 2 (g) 2CO 2 (g) + 4H 2 O(l) ΔH rxn = 1,782 kj 35

36 Hess s Law and Heats of Reaction Hess s law: the enthalpy change for any process is independent of the particular way the process is carried out. In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. Enthalpy is a state function. A state function is a variable whose value depends only on the state of the system and not its history. 36

37 Hess s Law Conceptual diagram representing Hess s law. Enthalpy is a state function, so any convenient path can be used to calculate the enthalpy change. 37

38 An Example on Hess s Law N 2 (g) + 2O 2 (g) 2NO 2 (g) ΔH 1 = 68 kj This reaction also can be carried out in two distinct steps, with enthalpy changes designated by ΔH 2 and ΔH 3. N 2 (g) + O 2 (g) 2NO(g) 2NO(g) + O 2 (g) 2NO 2 (g) N 2 (g) + 2O 2 (g) 2NO 2 (g) ΔH 2 = 180 kj ΔH 3 = 112 kj ΔH 2 + ΔH 3 = 68 kj ΔH 1 = ΔH 2 + ΔH 3 = 68 kj 38

39 Example Problem Consider the following data: NH( g) N( g) + H( g) H = 46 kj 2 H ( g) + O ( g) 2 H O( g) H = 484 kj Calculate ΔH for the reaction: Answer Reverse the two reactions: N ( g) + 6 H O( g) 3 O ( g) + 4 NH ( g) N( g) + H( g) NH( g) H = 46 kj 2 H O( g) 2 H ( g) + O ( g) H = +484 kj Multiply reactions to give the correct numbers of reactants and products: 4x 3x N( g) + H( g) NH( g) H = 46 kj 2 H O( g) 2 H ( g) + O ( g) H = +484 kj

40 Example Problem (continued) Final reactions: 2 N ( g) + 6 H ( g) 4 NH ( g) H = 184 kj H O( g) 6 H ( g) + 3 O ( g) H = kj Desired reaction: 2 N ( g) + 6 H O( g) 3 O ( g) + 4 NH ( g) ΔH = kj 40

41 Heats of Reaction for Some Specific Reactions Some classes of chemical reactions are given their own labels for heats of reactions. Heat of combustion, H comb Heat of neutralization, H neut Heat of formation, H f, is the heat of reaction for formation of substances. C(s) O 2(g) CO(g) Fractional coefficients are allowed for formation reactions because only one mole of product can be formed. 41

42 Heats of Reactions from Heat of Formation A formation reaction is the chemical reaction by which one mole of a compound is formed from its elements in their standard states. The standard state is the most stable form of an element at room temperature, 25 o C, and pressure, 1 atm, indicated with a superscript o. H f o = 0 for an element in its standard state. The enthalpy change for a reaction can be calculated using heats of formation. H o = o o v i H f (products) i v j H f (reactants) j i j 42

43 Example Problem Calculate H for the following reaction: 2Na(s) + 2H 2 O(l) 2NaOH(aq) + H 2 (g) Given the following information: H f (kj/mol) H 2 O(l) 286 NaOH(aq) 470 Answer: H = [2( H f ) NaOH + 1 ( H f ) H2 ] - [2( H f ) Na + 2( H f ) H2O ] H = [2( -470) + 1 (0.0) ] - [2(0.0) + 2(-286) ] H = 368 kj 43

44 Example Problem 9.9 Ethanol, C 2 H 5 OH, has a heat of combustion of kj/mol. What is the heat of formation of ethanol? Answer: Given the following information: H f (kj/mol) H 2 O(l) CO 2 (g) C 2 H 5 OH(l) + 3O 2 (g) 2 CO 2 (g) + 3 H 2 O(l) H = kj H = [2 H f (CO 2 ) + 3 H f (H 2 O)] [ H f (C 2 H 5 OH) + 3 H f (O 2 )] kj = [2x (-393.5) + 3x(-285.5)]- ( H f (C 2 H 5 OH) + 3x(0.0)] H f (C 2 H 5 OH) = kj/mol 44

45 Energy Density and Fuels When deciding the economic merits of a fuel, several factors must be considered. Technology available to extract the fuel. The amount of pollution released by its combustion. The fuel s relative safety. The ease of transporting the fuel. The fuel s energy density. 45

46 Energy Density and Fuels Energy density is the amount of energy that can be released per gram of fuel burned. The higher the energy density of a fuel, the less fuel that must be transported to the customer. 46