Thermochemistry: Energy Flow and Chemical Reactions

Transcription

1 Thermochemistry: Energy Flow and Chemical Reactions

2 Enthalpy of Formation formation = f = enthalpy associated with the formation of a compound from its constituent elements Examples of formation equations: C (graphite) + O 2 (gas) CO 2 (gas) 2C (graphite) + 2 (gas) C 2 2 (gas) C (graphite) (gas) + 1/2 O 2 (gas) + N 2 (gas) C 4 N 2 O(urea) The reactants should be elements in their natural form

3 Enthalpy of Formation Write balanced formation equations at standard conditions for each of the following substances: a) CaCl 2 b) NaCO 3 c) CCl 4 d) NO 3 Answer: a) Ca (s) + Cl 2(g) CaCl 2(s) b) Na (s) + ½ 2(g) + C graph + 3/2 O 2(g) NaCO 3(s) c) C graph + 2Cl 2(g) CCl 4(l) d) ½ 2(g) + ½ N 2(g) + 3/2 O 2(g) NO 3(l)

4 Enthalpy of Formation = depends of various conditions such as T, P, state of matter, etc., of the reactants and products To compare enthalpies, it is convenient to define a set of conditions called a standard state at which the enthalpies are tabulated. Standard state = pure form of the substance at 1.0 atmospheric pressure and at 25 C (298 K) o = standard enthalpy = enthalpy at standard conditions of all species o f = standard enthalpy of formation = change in enthalpy of reaction that forms 1.0 mol of compound from its elements with all the substances in their standard states If the element exists in more than one form under standard conditions, the most stable form of that element is used for o f

5 Enthalpy of Formation 2C (graphite) + 32 (gas) + ½ O2 (gas) C25O o f = kj C (graphite) + O 2 (gas) CO 2 (gas) o f = kj 2C (graphite) (gas) C 2 6 (gas) o f = kj The standard enthalpy of formation, o f values are tabulated in books and can be consulted every time it is needed.

6 Table 6.5 Selected Standard eats of Formation at 25 0 C(298K) Formula D 0 f(kj/mol) Formula D 0 f(kj/mol) Formula D 0 f(kj/mol) calcium Ca(s) CaO(s) CaCO 3 (s) carbon C(graphite) C(diamond) CO(g) CO 2 (g) C 4 (g) C 3 O(l) CN(g) CS s (l) chlorine Cl(g) Cl 2 (g) Cl(g) hydrogen (g) 2 (g) nitrogen N 2 (g) N 3 (g) NO(g) oxygen O 2 (g) O 3 (g) 2 O(g) 2 O(l) silver Ag(s) AgCl(s) sodium Na(s) Na(g) NaCl(s) sulfur S 8 (rhombic) 0 S 8 (monoclinic) 2 SO 2 (g) SO 3 (g)

7 Enthalpy of Formation Using the tabulated values, we can calculate the enthalpy of reaction D 0 rxn = S md 0 f(products) - S nd 0 f(reactants) where m = number of moles of each product n = number of moles of each reactant

8 Sample Problem Calculate the enthalpy of reaction of the equation below using standard enthalpies of formation. C38 (g) + 5 O2 3 CO2 (g) + 4 2O (l) Solution: o reaction = 3 o f [CO2(g)] + 4 o f [2O (l)] - o f [C38 (g)] - 5 o f [O2 (g)] = 3 ( kj) + 4 ( kj) ( kj) 5( 0 kj) = kj

9 Sample Problem Calculate the enthalpy of reaction of the equation below using standard enthalpies of formation. C66 (g) + 15/2 O2 6 CO2 (g) + 3 2O (l) Solution: o reaction = 6 o f [CO2(g)] + 3 o f [2O (l)] - o f [C66 (g)] - 15/2 o f [O2 (g)] = 6 ( kj) + 3 ( kj) (49.0 kj) 15/2(0 kj) = kj

10 Sample Problem Acetylene burns in air according to the following equation: C /2 O 2 2CO O Δ = kj Given that Δ fo of CO 2 = kj/mol and Δ fo of 2 O = kj/mol, what is the Δ fo of acetylene? Solution: o reaction = 2 o f [CO2(g)] + o f [2O (l)] - o f [C22 (g)] - 5/2 o f [O2 (g)] o f [C22 (g)] = 227 kj/mol

11 Enthalpy, Enthalpy of Formation decomposition formation The general process for determining D 0 rxn from D 0 f values. Elements Reactants -D 0 f D 0 f initial D 0 rxn Products final D 0 rxn = S md 0 f(products) - S nd 0 f(reactants)

12 Worksheet # Calculate reaction for each of the following: a) 2 2 S (g) + 3O 2 (g) 2 SO 2 (g) O (g) b) C 4 (g) + Cl 2 (g) CCl 4 (l) + Cl (g) (unbalanced) c) C 2 6 (g) + O 2 (g) CO 2 (g) + 2 O (l) (unbalanced) 2. The standard enthalpy change for the reaction: CaCO 3 (s) CaO (s) + CO 2 (g) = kj Given that o f of CO 2 (g) = kj/mol and that o f of CaO (s) = kj/mol, what is the o f of CaCO 3 (s)?

13 Worksheet # 9 5: Answers 1. Calculate reaction for each of the following: a) 2 2 S (g) + 3O 2 (g) 2 SO 2 (g) O (g) = kj b) C 4 (g) + 4Cl 2 (g) CCl 4 (l) + 4Cl (g) = kj c) C 2 6 (g) + 7/2 O 2 (g) 2CO 2 (g) O (l) = kj 2. The standard enthalpy change for the reaction: CaCO 3 (s) CaO (s) + CO 2 (g) = kj Given that o f of CO 2 (g) = kj/mol and that o f of CaO (s) = kj/mol, what is the o f of CaCO 3 (s)? Answer = kj/mol

14 Bond Energy Examine closely the following reaction:

15 Bond Energy Bond Energy the energy required to break a bond or to form a bond

16 Bond Energy When is bond energy negative? When is bond energy positive?

17 Bond Energy It takes energy to break a bond. Thus, bond breakage is positive (endothermic). Energy is released when a bond is formed. Thus, bond formation is negative (exothermic). Bond energy for a particular bond has the same numerical value. The sign changes depending on the whether bond is broken or formed.

18 Bond Energy

19 Bond Energy What is the relationship of bond energy and bond strength? The stronger is the bond, the greater the energy required to break the bond and larger is the bond energy.

20 Bond Energy What is the relationship of bond energy and bond order (number of bonds)? The higher the number of bonds, the greater the energy required to break them, the larger is the bond energy!

21 Bond Energy What is the relationship of bond energy and bond length? The shorter the bond length, the greater the energy required to break the bond and the larger is the bond energy!

22 Enthalpy of Reaction and Bond Energies

23 Enthalpy of Reaction and Bond Energies Using bond energies to calculate D 0 rxn. D 0 rxn = ΣD 0 reactant bonds broken + Σ D 0 product bonds formed

24 Enthalpy, Enthalpy of Reaction and Bond Energies 2CO + O 2 2CO 2 =? BOND BREAKING D 0 1 = + sum of BE D 0 2 = - sum of BE BOND FORMATION D 0 rxn

25 Enthalpy of Reaction and Bond Energies 2CO + O 2 2CO 2 =? O O O O reaction = sum of bond energies reactants + sum of bond energies products = 2[BE(CΞO)] + [BE(O 2 )] + 4 [-BE(C = O)] = 2(1070) (-799) kj = kj The structures of each of the substances have to be known!

26 Enthalpy of Reaction and Bond Energies + 2 O O 2 O + O O reaction = 4 BE(C-) + 2 BE(O 2 ) + 4[-BE(O-)] + 2[-BE(C = O)] = 4 (413) + 2 (498) - 4(467) 2(799) kj D 0 rxn= - 818kJ

27 Enthalpy, Enthalpy of Reaction and Bond Energies BOND BREAKING 4BE(C-)= +1652kJ 2BE(O 2 )= + 996kJ D 0 (bond breaking) = +2648kJ BOND FORMATION 2[-BE(C O)]= -1598kJ 4[-BE(O-)]= -1868kJ D 0 (bond forming) = -3466kJ D 0 rxn= -818kJ

28 Sample Problem Use bond energies to calculate D 0 rxn for the following reaction: C Cl 2 C 2 4 Cl 2 + Cl Cl Cl Cl Answer: -168 kj

29 Sample Problem Use bond energies to calculate D 0 rxn for the following reaction: O + C Ξ O O O Answer: - 22 kj

30 Worksheet # Calculate the reaction (using bond energies) for the chlorination of methane to form chloroform? Cl Cl Cl Cl Cl Cl 2. Calculate the reaction (using bond energies) for the following reaction. + Br Br

31 Worksheet # 9 6: Answers 1. Calculate the reaction (using bond energies) for the chlorination of methane to form chloroform? Cl Cl Cl Cl Cl Cl Answer: -330 kj 2. Calculate the reaction (using bond energies) for the following reaction. + Br Br Answer: -59 kj

32 Table 9.5 eats of Combustion(D comb ) of Some Carbon Compounds Two-Carbon Compounds Ethane (C 2 6 ) Ethanol (C 2 6 O) One-Carbon Compounds Methane (C 4 ) Methanol (C 4 O) Structural Formulas C C C C O C C O Sum of C-C and C- Bonds Sum of C-O and O- Bonds D comb (kj/mol) D comb (kj/g)

33 Points to Ponder.. Why are fossil fuels (not other compounds) used as gasoline in vehicles? What is the difference if one uses bioethanol as a fuel?

34 Points to Ponder.. Why are carbohydrates the major source of energy of humans?

35 omework What is the difference between the enthalpy calculated using enthalpy of formation and bond energies? Calculate and compare the enthalpies of reaction (using enthalpy of formation) in the reactions used in worksheet #9-6.

36 Worksheet # Calculate the reaction for the following reaction using bond energies and standard enthalpy of formation. The equation is not balanced. O O + N O N N + O 2. Calculate the reaction for the following reaction using bond energies and standard enthalpy of formation. + Br Br Br Br

37 Worksheet # 9 7 N O N Urea -, f = kj/mol Br 1,2 - dibromoethane -, f = kj/mol Br

38 Worksheet # 9 7: Answers 1. Calculate the reaction for the following reaction using bond energies and standard enthalpy of formation. Ans: kj (from, formation) O O + 2 N O N N + O 2. Calculate the reaction for the following reaction using bond energies and standard enthalpy of formation. Br + Br Br Br Ans: - 92 kj (from bond energy) and 90.1 kj(from, formation)

39 Enthalpies of Reaction Which is a more accurate calculation? the one using the enthalpies of formation (state of substances indicated, position of atoms indicated, values are not averaged, etc..) bond energies are used to only estimate the enthalpy of reaction and if enthalpy of formation of some substances are unknown

40 ess Law ess s Law of eat Summation = the enthalpy change of an overall process is the sum of the enthalpy changes of its individual steps Steps in calculating an unknown Δ 1. Identify the target equation, the step whose Δ is unknown and note the number of moles of reactants and products. 2. Manipulate the equations with known Δ values so that the target numbers of moles of reactants and products are on the correct side. Remember: - change the sign of Δ when you reverse an equation - multiply numbers of moles and Δ by the same factor 3. Add the manipulated equations to obtain the target equation. Add their Δ values to obtain the unknown Δ.

41 Sample Problem Calculate Δ rxn for: Ca + ½ O 2 + CO 2 CaCO 3 Given the following set of reactions: Ca + ½ O 2 CaO Δ = kj CaCO 3 CaO + CO 2 Δ = kj Answer: kj

42 Sample Problem Calculate Δ rxn for: C F 2 2CF 4 + 4F Given the following set of reactions: 2 + F 2 2F Δ = kj C + 2F 2 CF 4 Δ = kj 2C C 2 4 Δ = kj Answer: kj

43 Worksheet # Given the following data 2ClF (g) + O 2 (g) Cl 2 O (g) + F 2 O (g) 2ClF 3 (g) + 2O 2 (g) Cl 2 O (g) + 3F 2 O (g) 2F 2 (g) + O 2 (g) 2F 2 O (g) Calculate the o reaction for the reaction ClF (g) + F 2 (g) ClF 3 (g) = kj = kj = kj 2. Calculate o reaction for 2NOCl (g) N 2 (g) + O 2 (g) + Cl 2 (g) Given the following set of reactions: ½ N2 (g) + ½ O2 (g) NO (g) NO (g) + ½ Cl2 (g) NOCl (g) = 90.3 kj = kj

44 Worksheet # 9 8: Answers 1. Given the following data 2ClF (g) + O 2 (g) Cl 2 O (g) + F 2 O (g) 2ClF 3 (g) + 2O 2 (g) Cl 2 O (g) + 3F 2 O (g) 2F 2 (g) + O 2 (g) 2F 2 O (g) Calculate the o reaction for the reaction ClF (g) + F 2 (g) + ClF 3 (g) = kj = kj = kj Ans: kj 2. Calculate o reaction for 2NOCl (g) N 2 (g) + O 2 (g) + Cl 2 (g) Given the following set of reactions: ½ N2 (g) + ½ O2 (g) NO (g) NO (g) + ½ Cl2 (g) NOCl (g) Ans: kj = 90.3 kj = kj

45 eats of Formation and ess s Law The common lead-acid car battery produces a large burst of currrent, even at low temperatures, and is rechargeable. The reaction that occurs while recharging a dead battery is: 2PbSO 4(s) O (l) Pb (s) + PbO 2(s) SO 4(l) a. Use Δ fo values to calculate Δ rxn. b. Use the following equations to check your answer in part (a). Pb(s) + PbO 2 (s) + 2SO 3(g) 2PbSO 4 (s) SO 3(g) + 2 O (l) 2 SO 4 (l) Δ = -768 kj Δ = -132 kj Answer: = 2( ) + (-276.6) + 0-2( ) 2( ) = kj

46 Energy from Foods eats of Combustion of Some Fats and Carbohydrates Substance D comb (kj/g) Fats vegetable oil margarine butter Carbohydrates table sugar (sucrose) brown rice maple syrup -10.4

47 Energy from Foods Which is a better source of energy for our body? Proteins Carbohydrates Fats

48 Energy from Foods Calorie determination of food labels: carbohydrate: (sucrose, a simple sugar) C12 22 O11 D formation = kj/mol C1222O O2 12 CO O = 12 ( kj) + 11 (-286 kj) kj = kj

49 Energy from Foods Calorie determination of food labels: protein: (glycine, a simple amino acid) C25O2N D formation = kj/mol 4C25O2N(s) + 9O2(g) -> 8CO2(g) + 102O(l) + 2N2(g) = 8( kj) + 10( kj) ( )4 kj = kj

50 Energy from Foods Calorie determination of food labels: Fats: (lauric acid) C12 24 O2 D formation = kj/mol C1224O2 + 12O2 12 CO O = 12 ( kj) + 11 (-286 kj) kj = kj/mol

51 Points to ponder. ow do cold packs and hot packs work?