2. COURSE NAME: CHEMISTRY 101 COURSE CODE: 402101-4 Chapter 5 Thermochemistry

3. Energy is the capacity to do work. Thermal energy is the energy associated with the random motion of atoms and molecules Chemical energy is the energy stored within the bonds of chemical substances Nuclear energy is the energy stored within the collection of neutrons and protons in the atom Potential energy is the energy available by virtue of an object’s position 1

4. 3 Heat is the transfer of thermal energy between two bodies that are at different temperatures. Energy Changes in Chemical Reactions Temperature is a measure of the thermal energy. Temperature = Thermal Energy 2

5. 4 Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. Endothermic process is any process in which heat has to be supplied to the system from the surroundings. 2H 2 (g) + O 2 (g) 2H 2 O (l) + energy H 2 O (g) H 2 O (l) + energy energy + 2HgO (s) 2Hg (l) + O 2 (g) energy + H 2 O (s) H 2 O (l) Thermochemistry is the study of heat change in chemical reactions. 3

6. 5 Schematic of Exothermic and Endothermic Processes 4

7. First Law of Thermodynamics First Law: Energy of the Universe is Constant E = q + w q = heat. Transferred between two bodies w = work. Force acting over a distance (F x d) 5

8. Definition of Enthalpy Thermodynamic Definition of Enthalpy (H): H = E + PV E = energy of the system P = pressure of the system V = volume of the system 6

9. Changes in Enthalpy Consider the following expression for a chemical process:  H = H products - H reactants If  H >0, then q p >0. The reaction is endothermic If  H <0, then q p <0. The reaction is exothermic 7

10. Thermodynamic State Functions Thermodynamic State Functions: Thermodynamic properties that are dependent on the state of the system only. (Example:  E and  H) Other variables will be dependent on pathway (Example: q and w). These are NOT state functions. The pathway from one state to the other must be defined. 8

11. Thermodynamics is the scientific study of the interconversion of heat and other kinds of energy. State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. energy, pressure, volume, temperature  E = E final - E initial  P = P final - P initial  V = V final - V initial  T = T final - T initial 9

12. 11 Thermochemical Equations CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l)  H = -890.4 kJ/mol System gives off heat Exothermic  H < 0 H 2 O (s) H 2 O (l)  H = 6.01 kJ/mol The stoichiometric coefficients always refer to the number of moles of a substance If you reverse a reaction, the sign of  H changes H 2 O (l) H 2 O (s)  H = - 6.01 kJ/mol If you multiply both sides of the equation by a factor n, then  H must change by the same factor n. 2H 2 O (s) 2H 2 O (l)  H = 2 x 6.01 = 12.0 kJ 10

13. 12 Standard enthalpy of formation (  H 0 ) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f The standard enthalpy of formation of any element in its most stable form is zero.  H 0 (O 2 ) = 0 f  H 0 (O 3 ) = 142 kJ/mol f  H 0 (C, graphite) = 0 f  H 0 (C, diamond) = 1.90 kJ/mol f 11

14. 13 The standard enthalpy of reaction (  H 0 ) is the enthalpy of a reaction carried out at 1 atm. rxn aA + bB cC + dD H0H0 rxn d  H 0 (D) f c  H 0 (C) f = [+] - b  H 0 (B) f a  H 0 (A) f [+] H0H0 rxn n  H 0 (products) f =  m  H 0 (reactants) f  - Hess’s Law: When reactants are converted to 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. It doesn’t matter how you get there, only where you start and end.) 12

15. Hess’ Law Defined Hess’ Law:  H for a process involving the transformation of reactants into products is not dependent on pathway. Therefore, we can pick any pathway to calculate  H for a reaction. 13

16. Hess’ Law: An Example 14

17. Using Hess’ Law When calculating  H for a chemical reaction as a single step, we can use combinations of reactions as “pathways” to determine  H for our “single step” reaction. 15

18. Example (cont.) Our reaction of interest is: N 2 (g) + 2O 2 (g) 2NO 2 (g)  H = 68 kJ This reaction can also be carried out in two steps: N 2 (g) + O 2 (g) 2NO(g)  H = 180 kJ 2NO (g) + O 2 (g) 2NO 2 (g)  H = -112 kJ 16

19. Example (cont.) If we take the previous two reactions and add them, we get the original reaction of interest: N 2 (g) + O 2 (g) 2NO(g)  H = 180 kJ 2NO (g) + O 2 (g) 2NO 2 (g)  H = -112 kJ N 2 (g) + 2O 2 (g) 2NO 2 (g)  H = 68 kJ 17

20. Changes in Enthalpy Consider the following expression for a chemical process:  H = H products - H reactants If  H >0, then q p >0. The reaction is endothermic If  H <0, then q p <0. The reaction is exothermic 18

21. Example (cont.) Note the important things about this example, the sum of  H for the two reaction steps is equal to the  H for the reaction of interest. We can combine reactions of known  H to determine the  H for the “combined” reaction. 19

22. Hess’ Law: Details Once can always reverse the direction of a reaction when making a combined reaction. When you do this, the sign of  H changes. N 2 (g) + 2O 2 (g) 2NO 2 (g)  H = 68 kJ 2NO 2 (g) N 2 (g) + 2O 2 (g)  H = -68 kJ 20

23. Details (cont.) The magnitude of  H is directly proportional to the quantities involved (it is an “extensive” quantity). As such, if the coefficients of a reaction are multiplied by a constant, the value of  H is also multiplied by the same integer. N 2 (g) + 2O 2 (g) 2NO 2 (g)  H = 68 kJ  N 2 (g) + 4O 2 (g) 4NO 2 (g)  H = 136 kJ 21

24. Using Hess’ Law When trying to combine reactions to form a reaction of interest, one usually works backwards from the reaction of interest. Example: What is  H for the following reaction? 3C (gr) + 4H 2 (g) C 3 H 8 (g) 22

25. Example (cont.) 3C (gr) + 4H 2 (g) C 3 H 8 (g)  H = ? You’re given the following reactions: C (gr) + O 2 (g) CO 2 (g)  H = -394 kJ C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O (l)  H = -2220 kJ H 2 (g) + 1/2O 2 (g) H 2 O (l)  H = -286 kJ 23

26. Example (cont.) Step 1. Only reaction 1 has C (gr). Therefore, we will multiply by 3 to get the correct amount of C (gr) with respect to our final equation. C (gr) + O 2 (g) CO 2 (g)  H = -394 kJ 3C (gr) + 3O 2 (g) 3CO 2 (g)  H = -1182 kJ Initial: Final: 24

27. Example (cont.) Step 2. To get C 3 H 8 on the product side of the reaction, we need to reverse reaction 2. 3CO 2 (g) + 4H 2 O (l) C 3 H 8 (g) + 5O 2 (g)  H = +2220 kJ C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O (l)  H = -2220 kJ Initial: Final: 25

28. Example (cont.) Step 3: Add two “new” reactions together to see what is left: 3C (gr) + 3O 2 (g) 3CO 2 (g)  H = -1182 kJ 3CO 2 (g) + 4H 2 O (l) C 3 H 8 (g) + 5O 2 (g)  H = +2220 kJ 2 3C (gr) + 4H 2 O (l) C 3 H 8 (g) + 2O 2  H = +1038 kJ 26

29. Question 1 An exothermic reaction causes the surroundings to: A) become basic B) decrease in temperature C) Condense D) increase in temperature E) decrease in pressure Question 2 How much heat is evolved when 320 g of SO2 is burned according to the chemical equation shown below? 2 SO2(g) + O2(g) ----> 2 SO3(g) ΔHorxn = -198 kJ A) 5.04 x 10-2 kJ B) 9.9 x 102 kJ C) 207 kJ D) 5.0 x 102 kJ E) None of the above Question 3 The specific heat of aluminum is 0.214 cal/goC. Determine the energy, in calories, necessary to raise the temperature of a 55.5 g piece of aluminum from 23.0 to 48.6oC. A) 109 cal B) 273 cal C) 577 cal D) 347 cal E) 304 cal Question 4 A 60.0 g sample of an alloy was heated to 96.00oC and then dropped into a beaker containing 87.0 g of water at a temperature of 24.10oC. The temperature of the water rose to a final temperature of 27.63oC. The specific heat of water is 4.184 J/goC. What is the specific heat of the alloy? A) 0.313 J/goC B) 2.16 J/goC C) 0.118 J/goC D) 1.72 J/goC E) None of the above 27

30. Question 5 When 1.535 g of methanol (CH3OH) was burned in a constant-volume bomb calorimeter, the water temperature rose from 20.27oC to 26.87oC. If the mass of water surrounding the calorimeter was exactly 1000 g and the heat capacity of the bomb calorimeter was 1.75 kJ/oC, calculate the molar heat of combustion of CH3OH. The specific heat of water is 4.184 J/goC. A) -8.17 x 105 kJ/mol B) -817 kJ/mol C) 1.88 kJ/mol D) 817 kJ/mol E) None of the above 28

31. Question 6 To which one of the following reactions, occurring at 25oC, does the symbol ΔHof [H2SO4(l)] refer? A) H2(g) + S(s) + 2 O2(g) ----> H2SO4(l) B) H2SO4(l) ----> H2(g) + S(s) + 2 O2(g) C) H2(g) + S(g) + 2 O2(g) ----> H2SO4(l) D) H2SO4(l) ----> 2 H(g) + S(s) + 4 O(g) E) 2 H(g) + S(g) + 4 O(g) ----> H2SO4(l) 29

32. Question 7 Given: SO2(g) + ½O2(g) ----> SO3(g) ΔHorxn = -99 kJ, what is the enthalpy change for the following reaction? 2 SO3(g) ----> O2(g) + 2 SO2(g) A) 99 kJ B) -99 kJ C) 49.5 kJ D) -198 kJ E) 198 kJ 30

33. Question 8 Find the standard enthalpy of formation of ethylene, C2H4(g), given the following data: C2H4(g) + 3 O2(g) ----> 2 CO2(g) + 2 H2O(l) ΔHof = -1411 kJ; C(s) + O2(g) ----> CO2(g) ΔHof = -393.5 kJ; H2(g) + ½O2(g) ----> H2O(l) ΔHof = -285.8 kJ A) 731 kJ B) 2.77 x 103 kJ C) 1.41 x 103 kJ D) 87 kJ E) 52 kJ 31

34. Question 9 Calculate ΔHorxn for the combustion reaction of CH4 shown below given the following: ΔHof CH4(g) = -74.8 kJ/mol; ΔHof CO2(g) = -393.5 kJ/mol; ΔHof H2O(l) = -285.5 kJ/mol. CH4(g) + 2 O2(g) ----> CO2(g) + 2 H2O(l) A) -604.2 kJ B) 889.7 kJ C) -997.7 kJ D) -889.7 kJ E) None of the above 32

35. Question 10 A 1.300 g sample of benzoic acid (C7H6O2) was burned in a bomb calorimeter. The heat capacity of the entire apparatus, including the bomb, pail, thermometer, and water, was found to be 11,145 J/K. As a result of the reaction, the temperature of the calorimeter and water increased 4.627 K. What is the molar heat of combustion of benzoic acid? A) 4.84 x 106 kJ/mol B) -2.96 kJ/mol C) -4844 kJ/mol D) 549.1 kJ/mol E) 51.57 kJ/mol 33

36. Question 11 Which of the following is incorrectly matched? A) Radiant energy; solar energy able to influence global climate patterns B) Thermal energy; related to temperature irrespective of the volume C) Energy; capacity to do work D) Chemical energy; potential energy 34

37. Question 12 Energy is the ability to do work and can be: A) converted to one form to another B) can be created and destroyed C) used within a system without consequences D) none of the above 35

38. Question 13 Standard enthalpy of reactions can be calculated from standard enthalpies of formation of reactants. A) True B) False Question 14 In the equation ΔE is equal to q+w, which sign is correctly associated? A)q; - exothermic B) q; + endothermic C) w; + by system on surrounding D) w; + on system by surroundings 36