Forms of Energy and Their Interconversion. all energy is either poten:al or kine:c energy

Transcription

1 Chapter 6. Thermochemistry Drawing by Karen Thurber To create a rabbit, the magician must summon up the energy of the rabbit as well as some additional energy, equal to PV, to push the atmosphere out of the way. The total enery mustered up is the enthalpy (H = U + PV) Energy: Work: 1 Forms of Energy and Their Interconversion all energy is either poten:al or kine:c energy an object has poten:al energy due to its posi:on and kine:c energy due to its mo:on 2 1

2 like macroscopic objects, atoms and molecules also contain kine:c and poten:al energies when energy is transferred from one object to another it appears as work and/or heat 3 chemical energy energy released increasing poten:al energy increasing stability energy released 4 2

3 Units of Energy the SI unit of energy is the joule (J) J = 1 kg m s = 1 N m 1 kj = 1000 J the amount of work done by a force of 1N moving one m in the direc:on of the force. unit of energy used by electrical u:li:es for billing is the kwh, the amount of power used in an hour 6 1 kwh = J = 3.6 MJ another common (but outdated) unit is the calorie 1 cal = J the calorie is defined as the amount of energy needed to raise the temperature of 1 g of water by 1 o C 5 Unit Amount Required to Raise the temperature of 1 g water by 1 o C Amount Required to Light 100- W Bulb for 1 hour Amount Used by 75 kg Human to Run 1 km (approx) Amount Used by Average Canadian CiOzen in 1 Day Joule (J) x x x10 7 calorie (cal) x x x10 7 Calorie (Cal) x10 5 KilowaX- hour (kwh) 1.16x

4 kinetic energy = 1 2 mv J = 1 kg m s What is the energy, in kj, associated with a half- ton truck (4500 kg) moving at 100 km h - 1? A. 1.7 x 10 6 kj B. 1.3 x 10 4 kj C. 7.9 x 10 3 kj D. 1.7 x 10 3 kj E. 1.3 x 10 2 kj 7 The energy associated with moving a half- ton truck was determined to be 1.7x10 3 kj. What is this energy in kcal? A. 7.1 x 10 3 kcal B. 4.1 x 10 3 kcal C. 7.1 x 10 6 kcal D. 4.1 x 10 6 kcal 8 4

5 The Zeroth Law of Thermodynamics 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. A If T A = T B and T B = T C B C Then T A = T C 9 First law of thermodynamics (law of conservaoon of energy): In interactions between a system and its surroundings the total energy remains constant energy is neither created nor destroyed. Redheffer s perpetual mo:on machine q system + q surroundings = 0 q system = -q surroundings h\p://en.wikipedia.org/wiki/ Charles_Redheffer 10 5

6 When gasoline burns in a car engine, gases and heat are produced. The heat produced causes the products CO 2 and H 2 O to expand, which pushes the pistons outward. Excess heat is removed by the car s cooling system. If the expanding gases do 451 J of work on the pistons and the engine loses 325 J to the surroundings as heat, what is the energy change in J with respect to the system (engine). A. 776 B. 126 C D A balloon is gently heated over a toaster, absorbing 150 J of heat. As it absorbs the heat, it expands doing about 35 J of work. What is the change in internal energy of the balloon? A. 185 J B. 115 J C J D J 12 6

7 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 x specific heat. q = mcδt q = CΔT 13 Determination of Specific Heat 14 7

8 Determining Specific Heat from Experimental Data. Use the data presented on the last slide to calculate the specific heat of lead. 15 A 32.5 g iron rod (c=0.449 J g - 1 o C - 1 ), ini:ally at 22.7 C, is submerged into an unknown mass of water at 63.2 C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 59.5 C. What is the mass of the water? (c w = J g - 1 o C - 1 ). A g B g C g D g E g 16 8

9 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) ethylene glycol, (CH 2 OH) 2 (l) carbon tetrachloride, CCl 4 (l) 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. Exothermic reactions. Produces heat, q rxn < 0. Endothermic reactions. Consumes heat, q rxn > 0. Calorimeter A device for measuring quantities of heat. 18 9

10 A system transferring energy as heat only. 19 Bomb Calorimeter q rxn = -q cal q cal = q bomb + q water + q wires + Define the heat capacity of the calorimeter: q cal = Σ m i c i ΔT = CΔT all i heat Slide 20 10

11 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. 21 Coffee-cup calorimeter

12 You place 50.0 ml of M NaOH in a coffee-cup calorimeter at o C and carefully add 25.0 ml of M HCl, also at C. After stirring, the final T is o 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 specific heat capacity as water: d = 1.00 g/ml and c = 4.18 J/g*K) 23 Work In addition to heat effects chemical reactions may also do work. Gas formed pushes against the atmosphere. Volume changes. Pressure- volume work. Zn(s) + 2H + (aq) + 2Cl - (aq) Energy, E ΔE<0 work done on surroundings H 2 (g) + Zn 2+ (aq) + 2Cl - (aq) Slide 24 12

13 Pressure Volume Work w = F d = (P A) h = PΔV w = -P ext ΔV 25 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

14 How much work is done when a balloon is expanded from 0 to 2.75 L under a constant pressure of bar? A J B J C. 254 J 27 First Law of Thermodynamics in Full A system contains only internal energy, U (some books use E) Total energy (potential and kinetic) in a system (at constant Volume!) A system does not contain heat or work. These only occur during a change in the system. Law of Conservation of Energy ΔU = q + w (or ΔE = q + w) 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, ΔU > 0 If more energy leaves the system, ΔU < 0 The energy of an isolated system is constant 28 14

15 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. U is a function of state. Not easily measured. ΔU has a unique value between two states. Is easily measured. 29 Path Dependent Functions Changes in heat and work are not functions of state. Remember the PV work example 7-5 (slide 26): w = -1.1 x 10 2 J in a one step expansion of gas Consider 2.40 atm to 1.80 atm and finally to 1.30 atm: 30 15

16 Heats of Reaction: ΔU and ΔH Reactants Products U i U f ΔU = U f - U i ΔU = q rxn + w In a system at constant volume: ΔU = q rxn + 0 = q rxn = q v But we live in a constant pressure world! How does q p relate to q v? 31 Heats of Reaction 32 16

17 Heats of Reaction q V = q P + w We know that w = - PΔV and ΔU = q V, therefore: ΔU = q P - PΔV q P = ΔU + PΔV These are all state functions, so define a new function, Enthalpy. Let Then H = U + PV ΔH = H f H i = ΔU + ΔPV If we work at constant pressure and temperature: ΔH = ΔU + PΔV = q P 33 Comparing Heats of Reaction T = 300 K q P = -566 kj/mol = ΔH PΔV = P(V f V i ) = RT(n f n i ) = -2.5 kj ΔU = ΔH - PΔV = kj/mol = q V 34 17

18 Comparing Heats of Reaction ΔU ΔH 35 Comparing ΔU and ΔH Most chemical reac:ons involve li\le or no PV work. In these cases, most or all of the energy change occurs as a transfer of heat. 3 General cases: 1. Reac2ons 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, ΔV = 0 and PΔV = 0. Therefore, ΔH ΔU. 2. Reac2ons 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, ΔV = 0 and PΔV = 0. Therefore, ΔH = ΔU 36 18

19 Comparing ΔU and ΔH 3. Reac2ons in which the amount (moles) of gas does change. 2H 2 (g) + O 2 (g) 2H 2 O(g) in this case ΔV 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, ΔH = kj and PΔV = ΔnRT = - 1 mol x J K - 1 mol - 1 x 298 K = J = kj ΔU = ΔH - PΔV = kj - (- 2.5 kj) = kj PΔV is small, so ΔU and ΔH are s:ll very close 37 Changes of State of Matter Consider two endothermic processes: evaporation and melting Molar enthalpy of vaporization: H 2 O (l) H 2 O(g) ΔH vap = 44.0 kj at 298 K Molar enthalpy of fusion: H 2 O (s) H 2 O(l) ΔH 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) 38 19

20 Enthalpy Changes Accompanying Changes in States of Matter. Calculate ΔH for the process in which 50.0 g of water is converted from liquid at 10.0 C to vapor at 25.0 C. 39 Standard States and Standard Enthalpy Changes Define a particular state as a standard state. Standard enthalpy of reaction, ΔH 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

21 The chemical reac:on represen:ng aerobic respira:on is, C H O (s) + 6O (g) 6CO (g) + 6H O(g) Δ H = 2500 kj mol r and is the exact reverse of photosynthesis. A small child exhales about 3 moles of CO 2 in a 24 hour period, what is the energy change required to convert this amount of CO 2 back to glucose? 1 A kj B kj C kj D kj E kj 41 Propane burns according to the thermochemical reac:on below. Determine the mass of CO 2 produced by burning enough carbon to produce x 10 3 kj of heat. (MM CO2 = g mol - 1 ) C H (g) + 5 O (g) 3 CO (g) + 4 H O Δ H = kj mol r 1 A g B g C g D g 42 21

22 Indirect Determination of ΔH: Hess s Law ΔH 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) ΔH = kj ½N 2 (g) + ½O 2 (g) NO(g) ΔH = kj ΔH changes sign when a process is reversed NO(g) ½N 2 (g) + ½O 2 (g) ΔH = kj Hess s law of constant heat summa:on: If a process occurs in stages or steps (even hypothe:cally), 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) ½N 2 (g) + O 2 (g) NO 2 (g) ΔH = kj ΔH = kj ΔH = kj 43 Hess s Law Schematically 44 22

23 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) ΔH rxn =? Given the following information, calculate ΔH. CO(g) + 1/2O 2 (g) CO 2 (g) ΔH = kj N 2 (g) + O 2 (g) 2NO(g) ΔH = kj 45 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 equa:on: CO(g) + NO(g) CO 2 (g) + ½N 2 (g) ΔH rxn =? Using Hess s Law Given the following informa:on, calculate ΔH. CO(g) + ½O 2 (g) CO 2 (g) ΔH = kj N 2 (g) + O 2 (g) 2NO(g) ΔH = kj Rank Responses Other 46 23

24 ΔH f Standard Enthalpies of Formation 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 Standard Enthalpies of Formation 48 24

25 Write the balanced chemical reac:on corresponding to the enthalpy of forma:on of liquid ethanol (CH 3 CH 2 OH)? A. CH3CH2OH(l) CH 3(g) + CH 2(g) + OH(g) B. CH CH OH(l) 2C(graph) + 3H (g) + O (g) C. 2C(graph) + 3H (l) + O (l) CH CH OH(l) D. 2C(graph) + 3H (g) + O (g) CH CH OH(l) E. 2C(graph) + 6H(g) + O(g) CH3CH2OH(l) 49 Standard Enthalpies of Reaction ΔH overall = -2ΔH f NaHCO 3 + ΔH f Na2CO3 + ΔH f CO2 + ΔH f H2O 50 25

26 Enthalpy of Reaction ΔH rxn = ΣΔH f products - Σ ΔH f reactants 51 Determining ΔH o rxn from ΔHo f Nitric Acid, whose worldwide produc:on is nearly 10 billion kg, is used to make many products, including fer:lizers, dyes, and explosives. The first step in the process is the oxida:on of ammonia: 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) Calculate ΔH o rxn from ΔHo f (from Enthalpy Tables) ΔH o f [NO] = 90.3 kj/mol ΔHo f [H 2O(g)] = kj/mol ΔH o f [NH 3 ] = kj/mol ΔHo f [O 2 ] = 0 kj/mol Rank Responses Other 26

27 During photosynthesis, plants use energy from sunlight to form glucose and oxygen from carbon dioxide and water according to the following balanced equa:on. Calculate the heat of reac:on from the enthalpies of forma:on. 6CO (g) + 6H O(l) C H O (s) + 6O (g) Rank Responses Other h\p:// _121700/121668/Irriga:on- photosynthesis.gif Δ H f Δ H f Δ H f o C6H12O 6(s) o CO 2 (g) o H2O(l) = 1271 kj mol = kj mol = kj mol