Chapter 7: Thermochemistry. Contents

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1 General Chemistry Principles and Modern Applications Petrucci Harwood Herring 9 th Edition Chapter 7: Thermochemistry Dr. Chris Kozak Memorial University of Newfoundland, Canada General Chemistry: Chapter 7 Slide 1 Contents 7-1 Getting Started: Some Terminology 7-2 Heat 7-3 Heats of Reaction and Calorimetry 7-4 Work 7-5 The First Law of Thermodynamics 7-6 Heats of Reaction: U and H 7-7 The Indirect Determination of H: Hess s Law General Chemistry: Chapter 7 Slide 2

2 Contents 7-7 The Indirect Determination of H, Hess s Law 7-8 Standard Enthalpies of Formation 7-9 Fuels as Sources of Energy General Chemistry: Chapter 7 Slide Getting Started: Some Terminology System Surroundings General Chemistry: Chapter 7 Slide 4

3 Terminology Energy, U The capacity to do work. Work Force acting through a distance. Kinetic Energy The energy of motion. General Chemistry: Chapter 7 Slide 5 Energy Kinetic Energy e k = 1 2 mv2 Units: kg m 2 s 2 = J Work w = Fd Units: kg m s 2 m = J General Chemistry: Chapter 7 Slide 6

4 Potential Energy Energy Energy due to condition, position, or composition. Associated with forces of attraction or repulsion between objects. Energy can change from potential to kinetic. General Chemistry: Chapter 7 Slide 7 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. General Chemistry: Chapter 7 Slide 8

5 Heat Energy transferred between a system and its surroundings as a result of a temperature difference. Heat flows from hotter to colder. Temperature may change. Phase may change (an isothermal process). General Chemistry: Chapter 7 Slide 9 The Zeroth Law The Law of Thermal Equilibrium If two systems are in thermal equilibrium with a third system, then they must be in thermal equilibrium with each other. This law allows us to define temperature scales and thermometers. Two objects in thermal equilibrium with each other are taken to be the same temperature. General Chemistry: Chapter 7 Slide 10

6 Units of Heat Calorie (cal) The quantity of heat required to change the temperature of one gram of water by one degree Celsius. Joule (J) (Named after James Joule, a brewer) SI unit for heat 1 cal = J General Chemistry: Chapter 7 Slide 11 Heat Capacity The quantity of heat required to change the temperature of a system by one degree. Molar heat capacity (units J mol -1 o C -1 ) System is one mole of substance. Specific heat, c (J g -1 o C -1 ) System is one gram of substance Heat capacity (J o C -1 ) Mass H specific heat. q = mc T q = C T General Chemistry: Chapter 7 Slide 12

7 Conservation of Energy In interactions between a system and its surroundings the total energy remains constant energy is neither created nor destroyed. q system + q surroundings = 0 q system = -q surroundings General Chemistry: Chapter 7 Slide 13 Determination of Specific Heat General Chemistry: Chapter 7 Slide 14

8 Example 7-2 Determining Specific Heat from Experimental Data. Use the data presented on the last slide to calculate the specific heat of lead. q lead = -q water q water = mc T = (50.0 g)(4.184 J/g C)( ) C q water = 1.4x10 3 J q lead = -1.4x10 3 J =mc T = (150.0 g)(c)( ) C c lead = 0.13 Jg -1 C -1 General Chemistry: Chapter 7 Slide 15 Specific Heat Capacities of Some Elements, Compounds, and Materials Substance Specific Heat Capacity (J/g*K) Substance Specific Heat Capacity (J/g*K) Elements Materials aluminum, Al wood 1.76 graphite,c cement 0.88 iron, Fe glass 0.84 copper, Cu granite 0.79 gold, Au steel 0.45 Compounds water, H 2 O(l) ethyl alcohol, C 2 H 5 OH(l) 2.46 ethylene glycol, (CH 2 OH) 2 (l) 2.42 carbon tetrachloride, CCl 4 (l) General Chemistry: Chapter 7 Slide 16

9 7-3 Heats of Reaction and Calorimetry Chemical energy. Contributes to the internal energy of a system. Heat of reaction, q rxn. The quantity of heat exchanged between a system and its surroundings when a chemical reaction occurs within the system, at constant temperature. General Chemistry: Chapter 7 Slide 17 Heats of Reaction Exothermic reactions. Produces heat, q rxn < 0. Endothermic reactions. Consumes heat, q rxn > 0. Calorimeter A device for measuring quantities of heat. General Chemistry: Chapter 7 Slide 18

10 A system transferring energy as heat only. General Chemistry: Chapter 7 Slide 19 Bomb Calorimeter q rxn = -q cal q cal = q bomb + q water + q wires + Define the heat capacity of the calorimeter: q cal = 3m i c i T = C T all i heat General Chemistry: Chapter 7 Slide 20

11 Example 7-3 Using Bomb Calorimetry Data to Determine a Heat of Reaction. The combustion of g sucrose, in a bomb calorimeter, causes the temperature to rise from to C. The heat capacity of the calorimeter assembly is 4.90 kj/ C. (a) What is the heat of combustion of sucrose, expressed in kj/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. Note that 1 food Calorie is actually 1 kcal, so 19 Calories is 19 kcal. General Chemistry: Chapter 7 Slide 21 Example 7-3 Calculate q calorimeter : q cal = C T = (4.90 kj/ C)( ) C = (4.90)(3.41) kj = 16.7 kj Calculate q rxn : q rxn = -q cal = kj per g General Chemistry: Chapter 7 Slide 22

12 Example 7-3 Calculate q rxn in the required units: kj q rxn = -q cal = = kj/g g q rxn g = kj/g 1.00 mol = H 10 3 kj/mol (a) Calculate q rxn for one teaspoon: = (-16.5 kj/g)( q rxn 4.8 g 1 tsp 1.00 kcal )( ) = -19 kcal/tsp kj (b) General Chemistry: Chapter 7 Slide 23 Coffee-cup calorimeter. General Chemistry: Chapter 7 Slide 24

13 Sample Problem: Determining the Heat of a Reaction PROBLEM: You place 50.0 ml of M NaOH in a coffee-cup calorimeter at C and carefully add 25.0 ml of M HCl, also at C. After stirring, the final temperature is C. Calculate q soln (in J). (Assume the total volume is the sum of the individual volumes and that the final solution has the same density and specfic heat capacity as water: d = 1.00 g/ml and c = 4.18 J/g*K) PLAN: Need to determine the limiting reactant from the net ionic equation. Moles of NaOH and HCl as well as the total volume can be calculated. From the volume we use density to find the mass of the water formed. At this point q soln can be calculated using the mass, c, and T. The heat divided by the M of water will give us the heat per mole of water formed. General Chemistry: Chapter 7 Slide 25 Sample Problem Determining the Heat of a Reaction continued SOLUTION: HCl(aq) + NaOH(aq) H + (aq) + OH - (aq) H 2 O(l) NaCl(aq) + H 2 O(l) For NaOH M x L = mol OH - For HCl M x L = mol H + HCl is the limiting reactant mol of H 2 O will form during the rxn. total volume after mixing = L L x10 3 ml/l x 1.00 g/ml = 75.0 g of water q = mass x specific heat x T = 75.0 g x 4.18 J/g* 0 C x ( ) 0 C = 693 J (693 J/ mol H 2 O)(kJ/10 3 J) = 55.4 kj/ mol H 2 O formed General Chemistry: Chapter 7 Slide 26

14 7-4 Work In addition to heat effects chemical reactions may also do work. Gas formed pushes against the atmosphere. Volume changes. Pressure-volume work. General Chemistry: Chapter 7 Slide 27 A system losing energy as work only. Zn(s) + 2H + (aq) + 2Cl - (aq) Energy, E E<0 work done on surroundings H 2 (g) + Zn 2+ (aq) + 2Cl - (aq) General Chemistry: Chapter 7 Slide 28

15 Pressure Volume Work w = F H d = (P H A) H h = P V w = -P ext V General Chemistry: Chapter 7 Slide 29 Example Calculating Pressure-Volume Work. Suppose the gas in the previous figure is mol He at 298 K. How much work, in Joules, is associated with its expansion at constant pressure. Assume an ideal gas and calculate the volume change: V i = nrt/p = (0.100 mol)( L atm mol -1 K -1 )(298K)/(2.40 atm) = 1.02 L V f = 1.88 L V = L = 0.86 L General Chemistry: Chapter 7 Slide 30

16 Example Calculate the work done by the system: w= -P V 101 J = -(1.30 atm)(0.86 L)( ) 1 L atm = -1.1 H 10 2 J Hint: If you use pressure in kpa you get Joules directly. Where did the conversion factor come from? Compare two versions of the gas constant and calculate J mol -1 K L atm mol -1 K J L -1 atm -1 General Chemistry: Chapter 7 Slide The First Law of Thermodynamics Internal Energy, U (some books use E). Total energy (potential and kinetic) in a system (at constant Volume!). Translational kinetic energy. Molecular rotation. Bond vibration. Intermolecular attractions. Chemical bonds. Electrons. General Chemistry: Chapter 7 Slide 32

17 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 (or E = q + w) Law of Conservation of Energy The energy of an isolated system is constant General Chemistry: Chapter 7 Slide 33 First Law of Thermodynamics Energy entering a system is positive If heat is absorbed by a system, q > 0 If work is done on a system, w > 0 Energy leaving a system is negative If heat is lost by a system, q < 0 If work is done by a system, w < 0 If more energy enters the system, U > 0 If more energy leaves the system, U < 0 General Chemistry: Chapter 7 Slide 34

18 State Functions Any property that has a unique value for a specified state of a system is said to be a State Function. Water at K and 1.00 atm is in a specified state. d = g/ml This density is a unique function of the state. It does not matter how the state was established. General Chemistry: Chapter 7 Slide 35 Functions of State U is a function of state. Not easily measured. U has a unique value between two states. Is easily measured. General Chemistry: Chapter 7 Slide 36

19 Path Dependent Functions Changes in heat and work are not functions of state. Remember example 7-5, w = -1.1 H 10 2 J in a one step expansion of gas: Consider 2.40 atm to 1.80 atm and finally to 1.30 atm. w = (-1.80 atm)( )l (1.30 atm)( )l = L atm 0.68 L atm = -1.3 L atm = (-1.3 L atm)(101 J / 1 L atm) = H 10 2 J General Chemistry: Chapter 7 Slide Heats of Reaction: U and H Reactants Products U i U f U = U f -U i U = q rxn + w In a system at constant volume: U = q rxn + 0 = q rxn = q v But we live in a constant pressure world! How does q p relate to q v? General Chemistry: Chapter 7 Slide 38

20 Heats of Reaction General Chemistry: Chapter 7 Slide 39 Heats of Reaction q V = q P + w We know that w = - P V and U = q V, therefore: U = q P -P V q P = U + P V These are all state functions, so define a new function, Enthalpy. Let H = U + PV Then H = H f H i = U + PV If we work at constant pressure and temperature: H = U + P V = q P General Chemistry: Chapter 7 Slide 40

21 Comparing Heats of Reaction q P = -566 kj/mol = H P V = P(V f V i ) = RT(n f n i ) = -2.5 kj U = H - P V = kj/mol = q V General Chemistry: Chapter 7 Slide 41 Comparing Heats of Reaction U H General Chemistry: Chapter 7 Slide 42

22 Comparing U and H Most chemical reactions involve little or no PV work. In these cases, most or all of the energy change occurs as a transfer of heat. 3 General cases: 1. Reactions that do not involve gases. 2KOH(aq) + H 2 SO 4 (aq) K 2 SO 4 (aq) + 2H 2 O(l) because liquids and solids undergo very small volume changes, V = 0 and P V = 0. Therefore, H U. 2. Reactions in which the amount (or moles) of gases do not change. N 2 (g) + O 2 (g) 2NO(g) when the total amount of gaseous products equals the total amount of gaseous reactants, V = 0 and P V = 0. Therefore, H = U General Chemistry: Chapter 7 Slide 43 Comparing U and H 3. Reactions in which the amount (moles) of gas does change. 2H 2 (g) + O 2 (g) 2H 2 O(g) in this case V is not zero since there are three moles of gas to start and only two at the end. At 25 o C, when two moles of hydrogen and one mole of oxygen produce 2 mol of water vapour, H = kj and P V = nrt = -1 mol x J K -1 mol -1 x 298 K = J = -2.5 kj U = H - P V = kj - (-2.5 kj) = kj P V is small, so U and H are still very close General Chemistry: Chapter 7 Slide 44

23 Changes of State of Matter Consider two endothermic processes: evaporation and melting Molar enthalpy of vaporization: H 2 O (l) H 2 O(g) H vap = 44.0 kj at 298 K Molar enthalpy of fusion: H 2 O (s) H 2 O(l) H fus = 6.01 kj at K If it is for one mole of a substance, then it is called Molar Enthalpy and units are kj mol -1 (needed in discussions of the 2 nd and 3 rd law of thermodynamics) General Chemistry: Chapter 7 Slide 45 Example Enthalpy Changes Accompanying Changes in States of Matter. Calculate H for the process in which 50.0 g of water is converted from liquid at 10.0 C to vapor at 25.0 C. Break the problem into two steps: Raise the temperature of the liquid first then completely vaporize it. The total enthalpy change is the sum of the changes in each step. Set up the equation and calculate: q P = mc H 2O T + n H vap = (50.0 g)(4.184 J/g C)( ) C + = 3.14 kj kj = 125 kj 50.0 g 18.0 g/mol 44.0 kj/mol General Chemistry: Chapter 7 Slide 46

24 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 bar and at the temperature of interest. General Chemistry: Chapter 7 Slide 47 Enthalpy Diagrams General Chemistry: Chapter 7 Slide 48

25 7-7 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) H = kj ½N 2 (g) + ½O 2 (g) NO(g) H = kj H changes sign when a process is reversed NO(g) ½N 2 (g) + ½O 2 (g) H = kj General Chemistry: Chapter 7 Slide 49 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) NO(g) + ½O 2 (g) NO 2 (g) H = kj H = kj ½N 2 (g) + O 2 (g) NO 2 (g) H = kj General Chemistry: Chapter 7 Slide 50

26 Hess s Law Schematically General Chemistry: Chapter 7 Slide 51 Using Hess s Law Two gaseous pollutants that are formed in auto exhaust are CO and NO. An environmental chemist is studying ways to convert them to less harmful gases through the following equation: CO(g) + NO(g) CO 2 (g) + ½N 2 (g) H rxn =? Given the following information, calculate H. CO(g) + 1/2O 2 (g) CO 2 (g) N 2 (g) + O 2 (g) 2NO(g) H = kj H = kj General Chemistry: Chapter 7 Slide 52

27 7-8 Standard Enthalpies of Formation H f 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. General Chemistry: Chapter 7 Slide 53 Standard Enthalpies of Formation General Chemistry: Chapter 7 Slide 54

28 Standard Enthalpies of Formation General Chemistry: Chapter 7 Slide 55 Standard Enthalpies of Reaction H overall = -2 H f NaHCO 3 + H f Na2CO3 + H f CO 2 + H f H2O General Chemistry: Chapter 7 Slide 56

29 Enthalpy of Reaction H rxn = 3 H f products - 3 H f reactants General Chemistry: Chapter 7 Slide 57 Table 7.3 Enthalpies of Formation of Ions in Aqueous Solutions General Chemistry: Chapter 7 Slide 58

30 Determining H o rxn from H o f Nitric Acid, whose worldwide production is nearly 10 billion kg, is used to make many products, including fertilizers, dyes, and explosives. The first step in the process is the oxidation of ammonia: 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) Calculate H o rxn from Ho f (from Enthalpy Tables) H o f [NO] = 90.3 kj/mol H o f [H 2O(g)] = kj/mol H o f [NH 3 ] = kj/mol H o f [O 2 ] = 0 kj/mol General Chemistry: Chapter 7 Slide Fuels as Sources of Energy Fossil fuels. Combustion is exothermic. Non-renewable resource. Environmental impact. General Chemistry: Chapter 7 Slide 60

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