Chapter 5. Thermochemistry

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1 Chapter 5 Thermochemistry Dr. A. Al-Saadi 1 Preview Introduction to thermochemistry: Potential energy and kinetic energy. Chemical energy. Internal energy, work and heat. Exothermic vs. endothermic reactions. PV work. Enthalpy. Calorimetry. Hess s law. Standard enthalpies of formation. Dr. A. Al-Saadi 2 1

2 Chapter 5 Section 1 The Nature of Energy Energy, including all of its different classes, is very essential for our lives. Macroscopic scale: fossil fuels, water and winds, solar, heat (thermal), electricity, etc. Microscopic scale: chemical reactions in living cells, nuclear energy, etc. Energy can be converted form one form to another. Energy: is the capacity to do work or to transfer heat. Dr. A. Al-Saadi 3 Chapter 5 Section 1 The Transformation of Energy The law of conservation of energy. Energy can be converted from one form to another but can neither be created nor destroyed. E Universe = constant First law of thermodynamics; the law of conservation of energy All forms of energy are either potential or kinetic. Dr. A. Al-Saadi 4 2

3 Chapter 5 Section 1 Kinetic Energy Kinetic Energy (KE): due to the motion of an object. KE = ½ (mu 2 ) m is mass and u is velocity Thermal Energy - one form of kinetic energy associated with the random motion of atoms/ molecules. Thermal energy α Temperature. Dr. A. Al-Saadi 5 Chapter 5 Section 1 Potential Energy Potential Energy (PE): due to position or composition. Chemical energy - is stored within structural units of chemical substances. Electrostatic energy is energy resulting from the interaction of charged particles. Q = charge ; d = distance + E el : repulsive E el : attractive d Dr. A. Al-Saadi 6 3

4 Chapter 5 Section 1 Interconversion between KE and PE KE and PE are interconvertible. PE KE Macroscale Mi icrocale sc Water behind a dam Gasoline before it burns (chemical energy) Water when it falls down from the dam when gasoline burns up and gives power to the engine (thermal energy) Remember that interconversion between PE and KE is always governed by the law of conservation of energy. Dr. A. Al-Saadi 7 Chapter 5 Section 1 Energy Changes in Chemical Reactions System vs. surroundings System Reactants and products Surroundings Reaction container, room, everything else Dr. A. Al-Saadi 8 4

5 Chapter 5 Section 1 Energy Changes in Chemical Reactions Combustion of fossil fule: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) + thermal energy System Reactants and products exothermic Heat flow Surroundings Reaction container, room, everything else Energy lost from the system = Energy gained by the surroundings Dr. A. Al-Saadi 9 Chapter 5 Section 1 Energy Changes in Chemical Reactions Formation of nitric oxide: N 2 (g) + O 2 (g) + thermal energy (heat) 2NO(g) Thermochemistry: The study of the transfer of heat (thermal energy) in chemical reactions System Reactants and products endothermic Heat flow Surroundings Reaction container, room, everything else Energy gained by the system = Energy lost from the surroundings Dr. A. Al-Saadi 10 5

6 Another example: Another example: Chapter 5 Section 1 Exothermic and Endothermic Reactions In methane combustion the energy (heat) flows out of the system. It is an exothermic reaction. H 2 O(l) H 2 O(s) + energy (heat) Nitric oxide formation is an endothermic reaction. N 2 (g) + O 2 (g) + energy (heat) 2NO(g) Here the heat flows from the surroundings into the system. H 2 O(s) + energy (heat) H 2 O(l) Dr. A. Al-Saadi 11 Chapter 5 Section 1 Exothermic and Endothermic Reactions Dr. A. Al-Saadi 12 6

7 Chapter 5 Section 1 Units of Energy Joule (J) is the SI unit for energy. The amount of energy possessed by a 2 kg mass moving at a speed of 1 m/s. The amount of energy exerted when a force of 1 Newton (N) is exerted over 1 m distance. Dr. A. Al-Saadi 13 Chapter 5 Section 1 Units of Energy Calorie (cal) - commonly used on food labels. 1 cal = J 1000 cal = 1 Cal = 1 kcal Food calories (Cal) are really 1000 calories (cal). Dr. A. Al-Saadi 14 7

8 Chapter 5 Section 1 Exercise Calculate the kinetic energy of a neon atom moving at a speed of 98 m/s. E k 1 26 ( E k kg)(98m/s) J 2 Dr. A. Al-Saadi 15 Chapter 5 Section 2 Types of Systems Types of systems: open (exchange of mass and energy) closed (exchange of energy) isolated (no exchange) Dr. A. Al-Saadi 16 8

9 Chapter 5 Section 2 States and State Functions The change in the state function (property) depends only on the initial and final states of the system and not on how the change was carried out. Energy, pressure, volume, temperature are examples of state functions. Elevation is always the same (state function). Distances traveled by the two hikers differ. Efforts exercised by them are also not the same (not state functions) Dr. A. Al-Saadi 17 Chapter 5 Section 2 Internal Energy (U) S(s) + O 2 (g) SO 2 (g) We can not calculate the total internal energy with any certainty for a individual reactants/products. But we can calculate the change in energy of the system experimentally. U = U products or U final U (reactants) or U initial Dr. A. Al-Saadi 18 9

10 Chapter 5 Section 2 Internal Energy (U) in Exothermic Reactions U i S(s) + O 2 (g) Some PE stored in the chemical bonds is converted to KE (thermal energy) via heat. In exothermic reactions products have lower energy on average. U = ve U f SO 2 (g) Dr. A. Al-Saadi 19 Chapter 5 Section 2 Internal Energy (U) in Endothermic Reactions In endothermic reactions, products have higher PE and weaker bonds on average. U f 2NO(g) U = +ve U i N 2 (g) + O 2 (g) Dr. A. Al-Saadi 20 10

11 Chapter 5 Section 2 Change in U for System and Surroundings U system + U surroundings = 0 U system = U surroundings Exothermic Endothermic ΔU < 0 ΔU > 0 Dr. A. Al-Saadi 21 Chapter 5 Section 2 Heat and Work Energy is defined as the capacity to do work (w) or transfer heat (q). ΔU = q + w When a system absorbs or releases heat, its internal energy changes. When a system does work on the surrounding, or when the surrounding does work on the system, the internal energy of the system changes. Dr. A. Al-Saadi 22 11

12 Chapter 5 Section 2 Internal Energy (U), Heat (q) and Work (w) U can be changed by a flow of work, heat, or both: ΔU = q + w Exothermic Endothermic ΔU < 0 ΔU > 0 Heat is flowing out of the system (q is -) Work is done by the system (w is -) U is decreasing (ΔU < 0) Heat is going into the system (q is +) Work is done on the system (w is +) U is increasing (ΔU > 0) Dr. A. Al-Saadi 23 Chapter 5 Section 2 Internal Energy (U), Heat (q) and Work (w) Exercise: What is the ΔU for a system undergoing g a process in which 15.6 kj of heat flows into it and 1.4 kj of work is done on it? In order to calculate ΔU, we must know the values and signs of q and w. It is an endothermic process where: q = kj and w = +1.4 kj. ΔU = ( ) kj = kj 1 kilojoule (kj) = 1000 J Dr. A. Al-Saadi 24 12

13 Chapter 5 Section 3 The Pressure-Volume, or PV, Work It is work done by or on a gas sample work = force distance pressure = force / area Thus: work = pressure area distance w = P A Δh w = P ΔV w = P ΔV This gives how much work is being done by the system to expand a gas ΔV against a constant external pressure P. Dr. A. Al-Saadi 25 Chapter 5 Section 3 Decomposition of NaN 3 at Constant Volume At constant volume => V = 0 U = q P V Normally, P is referred to the external pressure. U = q V Dr. A. Al-Saadi 26 13

14 Chapter 5 Section 3 Decomposition of NaN 3 at Constant Pressure At constant pressure => U = q P V q P = U + P V Dr. A. Al-Saadi 27 Chapter 5 Section 3 Enthalpy (H) Enthalpy of a system is defined as: H = U + PV ΔH = ΔU + ΔPV Is H a state function? and way? Yes it is, because E, P and V are all state functions. For a system at constant P and its w is only PV work: ΔH =ΔU + PΔV = q P At constant P where only PV work is allowed: ΔH=q P Dr. A. Al-Saadi 28 14

15 Chapter 5 Section 3 Internal Energy (U) and Enthalpy (H) For a chemical or physical process taking place under: Constant Volume Constant Pressure q V the heat flow (q) to or from the system is a measure of the q P change in internal energy (ΔU) for that specific process. change in enthalpy (ΔH) for that specific process. Dr. A. Al-Saadi 29 Chapter 5 Section 3 Enthalpy Change (ΔH) for Chemical Reactions Most of the chemical reactions performed in the laboratory are constant-pressure processes. Thus, the heat exchanged between the system and the surroundings is equal to the change in enthalpy. Heat of a reaction and change in enthalpy are equivalent terms when chemical reactions are studied at P constant. ΔH rxn = H products H reactants For H products > H reactants ΔH rxn = +ve => Endothermic reaction. For H products < H reactants ΔH rxn = ve => Exothermic reaction. Dr. A. Al-Saadi 30 15

16 Chapter 5 Section 3 Thermochemical Equations Thermochemical equations represent both mass and enthalpy changes. H f H i H 2 O(s) H 2 O(l) H = kj/mol This is an endothermic process. It requires 6.01 kj to melt one mole of ice, H 2 O(s). When the process is reversed it becomes exothermic. The magnitude of ΔH doesn t change, but the sign becomes ve. The enthalpy value will change if the number of moles varies from the 1:1 reaction stoichiometry. For example, to melt 2 moles of ice, kj energy is required. Dr. A. Al-Saadi 31 Chapter 5 Section 3 Thermochemical Equations H f H i This is an exothermic process. It releases kj when one mole of methane, CH 4, reacts. When the process is reversed it becomes endothermic. The magnitude of ΔH doesn t change, but the sign becomes +ve. The enthalpy value will change if the number of moles varies from the 1:2:1:2 reaction stoichiometry. For example, kj energy is released when 2 moles of methane, CH 4, react. Dr. A. Al-Saadi 32 16

17 Chapter 5 Section 3 Exercise When 1 mole of CH 4 is burned at constant P, 890 kj of energy is released as heat. Calculate ΔH for a process in which a 5.8 g sample of CH 4 is burned at constant P. At constant P: CH 4 (g) + 2O 2 (g) CO 2 (g) +2H 2 O(g) ΔH = 890 kj Dr. A. Al-Saadi 33 Chapter 5 Section 3 Exercise Given the following equation: C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O(l) ΔH = 2803 kj/mol Calculate the energy released when g of glucose is burned in oxygen. 1mol C6H12O kj g C6H12O kj g C H O 1mol C H O Dr. A. Al-Saadi 34 17

18 Chapter 5 Section 4 Calorimetry Calorimetry is to measure heat by observing the change in T when a body absorbs or releases energy as heat. Some materials need too much energy to raise their T 1 C. Other materials need much less energy to raise their T 1 C. Heat Capacity, C, is the amount of heat required to raise T of an object by 1 C. C = heat absorbed increase in T J J = C = K C depends on the amount of the substance. extensive property Dr. A. Al-Saadi 35 Chapter 5 Section 4 Specific Heat Capacity It would be more useful when C is given per unit mass, such as grams. Specific Heat Capacity (s) of a substance is: the amount of heat required to raise T of 1 gram of that substance by 1 C. s does NOT depend on the amount of the substance. intensive property s for H 2 O = J/ o C g s = q mass ΔT q = mass s ΔT q = C ΔT where q is heat, m is mass, s is specific heat, C is heat capacity, and T is change in temp. ( T = T final T initial ) Dr. A. Al-Saadi 36 18

19 Chapter 5 Section 4 Specific Heat Capacity It takes much less energy to raise the energy of 1g of a metal than for 1g for water Dr. A. Al-Saadi 37 Chapter 5 Section 4 Specific Heat Capacity Exercise: Calculate the amount of energy required to heat 95.0 grams of water from 22.5 C to 95.5 C. q = (95.0 g) (4.184 J/g C) )(95.5 C 22.5 C) q = 2.90 x 10 4 J or 29.0 kj Dr. A. Al-Saadi 38 19

20 Chapter 5 Section 4 Constant-Pressure Calorimetry T f T i P which is the atmospheric pressure remains constant. No heat goes in or out from the calorimeter. Surrounding medium is water. So, the heat released by the system is absorbed by water. Wellinsulated walls A simple Styrofoam-cup calorimeter Dr. A. Al-Saadi 39 Chapter 5 Section 4 Constant-Pressure Calorimetry 31.9 C 25.0 C Consider the following process of mixing i a strong acid with a strong base: Wellinsulated walls A simple Styrofoam-cup calorimeter HCl 50.0 ml 1.0 M + NaOH 50.0 ml 1.0 M T goes up from 25.0 o C to 31.9 o C. Calculate the heat released by the above acid-base reaction. Dr. A. Al-Saadi 40 20

21 Chapter 5 Section 4 Constant-Pressure Calorimetry Net ionic equation is: H + (aq) + OH - (aq) H 2 O(l) HCl 50.0 ml 1.0 M Heat required to raise T of water 1 o C is 4.18 J/ o C g. + NaOH 50.0 ml 1.0 M q released by rxn = q absorbed by water = q sys = specific heat capacity for water (s) mass of solution (m) increase in T (ΔT) J = (4.18 ) ( ml)(1.0 g/ml)(6.9 o o C g C) = J V d = 2.9 kj (which is also the change of enthalpy of this reaction). Dr. A. Al-Saadi 41 Chapter 5 Section 4 Constant-Pressure Calorimetry Can you now find ΔH in kj/mol for the neutralization reaction. H + (aq) + OH - (aq) H 2 O(l) Dr. A. Al-Saadi 42 21

22 Chapter 5 Section 4 Constant-Pressure Calorimetry Dr. A. Al-Saadi 43 Chapter 5 Section 4 Constant-Volume Calorimetry Because V is constant, PΔV = 0 and there will be no work done (w = 0). At constant V calorimeter: ΔU=q V A bomb calorimeter It is also know as bomb calorimeter, where the sample is ignited. Heat of combustion is usually measured using bomb calorimeters. Dr. A. Al-Saadi 44 22

23 Chapter 5 Section 4 Bomb Calorimeter Dr. A. Al-Saadi 45 Chapter 5 Section 4 Bomb Calorimeter All heat released by the combustion reaction is absorbed by the calorimeter. C cal is the energy required to raise T of water and other parts of the bomb calorimeter by 1 C. C cal for any bomb calorimeter can be determined by burning a substance with a known heat of combustion. Dr. A. Al-Saadi 46 23

24 Chapter 5 Section 4 Bomb Calorimeter Calculation The combustion of 1.00 g of benzoic acid releases kj of heat. In one experiment, we wanted to measure the heat capacity of a bomb calorimeter by burning 1.00 g of benzoic acid. The temperature increased by C. What is heat capacity of that bomb calorimeter? Dr. A. Al-Saadi 47 Chapter 5 Section 4 Bomb Calorimeter Calculation A combustion of g of octane (C 8 H 18 ) in a bomb calorimeter results in an increase int by 2.25 o C. Calculate Δ ΔU for the reaction if C cal. = 11.3 kj/ C. The reaction is: excess C 8 H 18 (s) + 25/2 O 2 (g) 8CO 2 (g) + 9H 2 O(g) kj q released by rxn = ΔT C cal. = (2.25 o C)(11.3 kj/ o C) = 25.4 kj To get ΔU (in kj/mol) for the reaction: 1 mol octane g octane = mol octane g octane For 1 mole of octane: 25.4 kj / mol octane = kj/mol (Exothermic) Thus q V = ΔU = kj/mol Dr. A. Al-Saadi 48 24

25 H is a state function Chapter 5 Section 5 Hess s Law For the reaction: A + B C+ D ΔH is the same whether: The reaction takes place in one step. The reaction takes place in a series of steps (two or more). A + E C + F ΔH 1 B + F D + E ΔH 2 A + B C + D ΔH = ΔH 1 + ΔH 2 Dr. A. Al-Saadi 49 H is a state function H is an extensive property Chapter 5 Section 5 Hess s Law For the reaction: A + B C+ D ΔH is the same whether: The reaction takes place in one step. The reaction takes place in a series of steps (two or more). Also notice that: When: A + B C + D has ΔH = +x, then C + D A + B has ΔH = x and 2A + 2B 2C + 2D has ΔH = +2x Dr. A. Al-Saadi 50 25

26 Chapter 5 Section 5 Hess s Law for the Oxidation of N 2 to NO 2 N 2 (g) + 2O 2 (g) + energy (heat) 2NO 2 (g) Step 2 Step 1 Dr. A. Al-Saadi 51 ½ ( ½) ½ Chapter 5 Section 5 Practicing Hess s Law Given the following data: 2ClF(g) + O 2 (g) Cl 2 O(g) + F 2 O(g) ΔH 1 = kj 2ClF 3 (g) + 2O 2 (g) Cl 2 O(g) + 3F 2 O(g) ΔH 2 = kj 2F 2 (g) + O 2 (g) 2F 2 O(g) ΔH 3 = kj Calculate ΔH for the reaction: ClF(g) + F 2 (g) ClF 3 (g) ClF(g) + ½ O 2 (g) ½ Cl 2 O(g) + ½ F 2 O(g) ½ ΔH 1 3 ½ Cl 2 O(g) + ½ F 2 O(g) ClF 3 (g) + O 2 (g) ½ ΔH 2 F 2 (g) + ½ O 2 (g) F 2 O(g) ½ ΔH 3 ClF(g) + F 2 (g) ClF 3 (g) ΔH = kj Dr. A. Al-Saadi 52 26

27 Chapter 5 Section 6 Standard Enthalpy of Formation (ΔH f o ) It is defined as the heat change that accompanies the formation of 1 mole of a compound from its elements in their standard states. N 2 (g) + 2O 2 (g) 2NO 2 (g) ΔH = 68 kj ½ N 2 (g) + O 2 (g) NO 2 (g) ΔH = 34 kj There is no method to determine the absolute H value. We can only calculate the change in H (ΔH)from thermodynamics experiments. ΔH o f Enthalpy of formation of an element (Na, He) or a compound (O 2, H 2 ) in its standard state is zero Standard state means 1 atm, 25 C. Dr. A. Al-Saadi 53 Chapter 5 Section 6 Standard Enthalpy of Formation (ΔH f o ) ΔH o f is ΔH that accompanies the formation of 1 mole of a product from its elements with all substances in their standard states. N 2 (g) + 2O 2 (g) 2NO 2 (g) ΔH = 68 kj ½ N 2 (g) + O 2 (g) 1NO 2 (g) ΔH = 34 kj standard states ΔH o f = 0 f ΔH f o = 34 kj/mol C(s) + 2H 2 (g) + ½O 2 (g) CH 3 OH(l) ΔH o f = 239 kj/mol 3 2Fe(s) + ½O 2 (g) Fe 2 O 3 (s) ΔH o f = 826 kj/mol Dr. A. Al-Saadi 54 27

28 Chapter 5 Section 6 Standard Enthalpy of Formation (ΔH f o ) Can you predict the reactions of formation of the compounds listed in the table from reactants at their standard states? 1/2 N 2 (g) + 3/2 H 2 (g) 1/2 N 2 (g) + O 2 (g) H 2 (g) + 1/2 O 2 (g) 2Al( Al(s) +3/2O 2 (g) 2 Fe(s) + 3/2 O 2 (g) C graphite (s) + O 2 (g) C graphite (s) + 1/2 O 2 (g) + 2H 2 (g) 8 C graphite (s) + 9H 2 (g) Dr. A. Al-Saadi 55 Chapter 5 Section 6 Calculating H rxn from Standard Enthalpies of Formation CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) C(s) + 2H 2 (g) CH 4 (g) o ΔH f (a) = 75 kj O 2 (g) standard state o ΔH f (b) = 0 kj C(s) + O 2 (g) CO 2 (g) o ΔH f (c) = 394 kj H 2 (g) + ½ O 2 (g) H 2 O(l) o ΔH f (d) = 286 kj ΔH o rxn = n products ΔH f o (products) n reactants ΔH f o (reactants) ΔH o o o o o rxn = [ΔH f (a)] [ΔH f (b)] + [ΔH f (c)] + 2 [ΔH f (d)] = [ 75 kj] + [0 kj] + [ 394 kj] + 2[ 286 kj] = -891 kj Dr. A. Al-Saadi 56 28

29 Dr. A. Al-Saadi 57 29

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