1. Thermochemistry: Chemical Energy Chapter 8

2. 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 Electrical energy is the energy associated with the flow of electrons Potential energy is the energy available by virtue of an object’s position, or stored energy Kinetic energy is moving energy.  E = q rxn + w

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 90 0 C 40 0 C greater thermal energy

4. Thermochemistry is the study of heat change in chemical reactions. The system is the specific part of the universe that is of interest in the study. open mass & energyExchange: closed energy isolated nothing SYSTEMSURROUNDINGS

5. Calorimetry and Heat Capacity Measure the heat flow at constant pressure (  H).

6. Calorimetry and Heat Capacity Measure the heat flow at constant volume (  E).

7. First Law of Thermodynamics Energy is neither created nor destroyed. In other words, the total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa. Use Fig. 5.5

8. 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)

9. Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.  H = H (products) – H (reactants)  H = heat given off or absorbed during a reaction at constant pressure H products < H reactants  H < 0 H products > H reactants  H > 0

10. Enthalpy Changes01 Enthalpies of Physical Change:

11. Enthalpies of Physical and Chemical Change Enthalpy of Fusion (  H fusion ): The amount of heat necessary to melt a substance without changing its temperature. Enthalpy of Vaporization (  H vap ): The amount of heat required to vaporize a substance without changing its temperature. Enthalpy of Sublimation (  H subl ): The amount of heat required to convert a substance from a solid to a gas without going through a liquid phase.

12. Thermochemical Equations H 2 O (s) H 2 O (l)  H = 6.01 kJ Is  H negative or positive? System absorbs heat Endothermic  H > 0 6.01 kJ are absorbed for every 1 mole of ice that melts at 0 0 C and 1 atm.

13. Thermochemical Equations CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l)  H = -890.4 kJ Is  H negative or positive? System gives off heat Exothermic  H < 0 890.4 kJ are released for every 1 mole of methane that is combusted at 25 0 C and 1 atm.

14. H 2 O (s) H 2 O (l)  H = 6.01 kJ The stoichiometric coefficients always refer to the number of moles of a substance Thermochemical Equations If you reverse a reaction, the sign of  H changes H 2 O (l) H 2 O (s)  H = - 6.01 kJ 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

15. H 2 O (s) H 2 O (l)  H = 6.01 kJ The physical states of all reactants and products must be specified in thermochemical equations. Thermochemical Equations H 2 O (l) H 2 O (g)  H = 44.0 kJ How much heat is transfered when 266 g of white phosphorus (P 4 ) burn in air? P 4 (s) + 5O 2 (g) P 4 O 10 (s)  H = -3013 kJ 266 g P 4 1 mol P 4 123.9 g P 4 x -3013 kJ 1 mol P 4 x = -6470 kJ

16. The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius. C = ms Heat (q) absorbed or released: q = ms  t q = C  t  t = t final - t initial

17. Calorimetry and Heat Capacity Assuming that a can of soda has the same specific heat as water, calculate the amount of heat (in kilojoules) transferred when one can (about 350 g) is cooled from 25 °C to 3 °C. Specific heat ( Water ) = 4.18 g °C J Temperature change = 3 C-25 °C = - 22 °C Mass = 350 g q = (specific heat) x (mass of substance) x  T

18. Calorimetry and Heat Capacity Calculate the amount of heat transferred. = -32 000 J -22 °C 1000 J 1 kJ x 350 g x Heat evolved = = -32 kJ g °C 4.18 J x -32 000 J

19. Calorimetry and Heat Capacity

20. Bomb Calorimetry Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy,  E, not  H.  E = q rxn + w W = VΔP For most reactions, the difference is very small. ΔE almost = q rxn Constant-Volume Calorimetry

21. No heat enters or leaves! q sys = q water + q Cal + q rxn q sys = 0 q rxn = - (q water + q Cal ) q water = ms  t q Cal = C Cal  t

22. Constant-Pressure Calorimetry No heat enters or leaves! q sys = q water + q cal + q rxn q sys = 0 q rxn = - (q water + q cal ) q water = ms  t q cal = C cal  t Reaction at Constant P  H = q rxn

24. Hess’s Law Haber Process: Multiple-Step Process N2H4(g)N2H4(g)2H 2 (g) + N 2 (g) Hess’s Law: The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction. 2NH 3 (g)3H 2 (g) + N 2 (g)  H° = -92.2 kJ  H° 1 = ? 2NH 3 (g)N 2 H 4 (g) + H 2 (g)  H° 1+2 = -92.2 kJ 2NH 3 (g)3H 2 (g) + N 2 (g)  H° 2 = -187.6 kJ

25.  H° 1 =  H° 1+2 -  H° 2 Hess’s Law = -92.2 kJ - (-187.6 kJ) = 95.4 kJ  H° 1 +  H° 2 =  H° 1+2

26. Hess’s Law01 Hess’s Law: The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction. 3 H 2 (g) + N 2 (g)  2 NH 3 (g)  H° = –92.2 kJ

27. Hess’s Law The industrial degreasing solvent methylene chloride (CH 2 Cl 2, dichloromethane) is prepared from methane by reaction with chlorine: CH 4 (g) + 2 Cl 2 (g) CH 2 Cl 2 (g) + 2 HCl(g) Use the following data to calculate  H° (in kilojoules) for the above reaction : CH 4 (g) + Cl 2 (g) CH 3 Cl(g) + HCl(g)  H° = –98.3 kJ CH 3 Cl(g) + Cl 2 (g) CH 2 Cl 2 (g) + HCl(g)  H° = –104 kJ

28. Standard Heats of Formation H 2 (g) + 1 / 2 O 2 (g)  H 2 O(l)  H° f = –286 kJ/mol 3 / 2 H 2 (g) + 1 / 2 N 2 (g)  NH 3 (g)  H° f = –46 kJ/mol 2 C(s) + H 2 (g)  C 2 H 2 (g)  H° f = +227 kJ/mol *Standard state (°): P = 1atm, t = 25°C, Concentration = 1 mol/Lit

29. Standard Heats of Formation Standard Heats of Formation (  H° f ): The enthalpy change for the formation of 1 mole of substance in its standard state from its constituent elements in their standard states. The standard heat of formation for any element in its standard state is defined as being ZERO.  H° f = 0 for an element in its standard state

30. Standard Heats of Formation Calculating  H° for a reaction:  H° =  H° f (Products) –  H° f (Reactants) For a balanced equation, each heat of formation must be multiplied by the stoichiometric coefficient. aA + bB cC + dD  H° = [c  H° f (C) + d  H° f (D)] – [a  H° f (A) + b  H° f (B)]

31. Standard Heats of Formation -1131Na 2 CO 3 (s)49C6H6(l)C6H6(l)-92HCl(g) -127AgCl(s)-235C 2 H 5 OH(g)95.4N2H4(g)N2H4(g) -167Cl - (aq)-201CH 3 OH(g)-46NH 3 (g) -207NO 3 - (aq)-85C2H6(g)C2H6(g)-286H 2 O(l) -240Na + (aq)52C2H4(g)C2H4(g)-394CO 2 (g) 106Ag + (aq)227C2H2(g)C2H2(g)-111CO(g) Some Heats of Formation,  H f ° (kJ/mol)

32. Benzene (C 6 H 6 ) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol. 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) H0H0 rxn 6  H 0 (H 2 O) f 12  H 0 (CO 2 ) f = [+] - 2  H 0 (C 6 H 6 ) f [] H0H0 rxn = [ 12x(–393.5) + 6x( -286 ) – [ 2x(49.04) ] = -6542.0 kJ -6542.0 2 mol = - 3271.0 kJ/mol C 6 H 6 See Page 292 for Standard Heat of Formation

33. Standard Heats of Formation Using standard heats of formation, calculate the standard enthalpy of reaction for the photosynthesis of glucose (C 6 H 12 O 6 ) and O 2 from CO 2 and liquid H 2 O. C 6 H 12 O 6 (s) + 6O 2 (g)6CO 2 (g) + 6H 2 O(l)  H° = ?  H° = [  H° f (C 6 H 12 O 6 (s))] - [6  H° f (CO 2 (g)) + 6  H° f (H 2 O(l))] = 2816 kJ [(6 mol)(-393.5 kJ/mol) + (6 mol)(-285.8 kJ/mol)]  H° = [(1 mol)(-1260 kJ/mol)] -

34. Bond Dissociation Energy Bond Dissociation Energy: Can be used to determine an approximate value for  H° f.  H° f = D (Bonds Broken) – D (Bonds Formed) ΔH o rxn = D (Reactant bonds) ­ - D (Product bonds) For the reaction: H-H (g) + Cl-Cl (g)  2H-Cl (g)  H° rxn = (D(H–H) + D(Cl-Cl))-(2 X D(H–Cl) )  H° rxn = -185 kJ  H° f = -92.5 KJ/mol

35. Bond Dissociation Energy02

36. Calculate an approximate  H° (in kilojoules) for the synthesis of ethyl alcohol from ethylene: C 2 H 4 (g) + H 2 O(g)  C 2 H 5 OH(g)HHH H C C O-H C HH + H-O-H H C HH ΔH o rxn = D (Reactant bonds) ­- D (Product bonds) ΔH o rxn = (D C=C + 4 D C­H + 2 D O­H ) - ­ (D C­C + D C­O + 5 D C­H + D O­H )

37. ΔH o rxn = [(1 mol)(611 kJ/mol) + (4 mol)(410 kJ/mol) +(2 mol)(460 kJ/mol)]­ - [(1 mol)(350 kJ/mol) + (1 mol)( 350 kJ/mol) + (5 mol)(410 kJ/mol) + (1 mol)(460 kJ/mol)] ΔH o rxn = ­39 kJ ΔH o rxn = (D C=C + 4 D C­H + 2 D O­H ) ­- (D C­C + D C­O + 5 D C­H + D O­H )

38. Thermodynamics State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. energy, pressure, volume, temperature

39. Predicting spontaneity of a chemical Reaction Second Law of Thermodynamics: Reactions proceed in the direction that increases the entropy of the system plus surroundings. A spontaneous process is one that proceeds on its own without any continuous external influence. A nonspontaneous process takes place only in the presence of a continuous external influence.

40. Introduction to Entropy02 The measure of molecular disorder in a system is called the system’s entropy; this is denoted S. Entropy has units of J/K (Joules per Kelvin).  S = S final – S initial –Positive value of  S indicates increased disorder. –Negative value of  S indicates decreased disorder.

41. Introduction to Entropy03

42. Introduction to Entropy05 Predict whether  S° is likely to be positive or negative for each of the following reactions. a. 2 CO(g) + O 2 (g)  2 CO 2 (g) b. 2 NaHCO 3 (s)  Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g) c. C 2 H 4 (g) + Br 2 (g)  CH 2 BrCH 2 Br(l) d. 2 C 2 H 6 (g) + 7 O 2 (g)  4 CO 2 (g) + 6 H 2 O(g)

43. Introduction to Entropy04 To decide whether a process is spontaneous, both enthalpy and entropy changes must be considered: Spontaneous process: Decrease in enthalpy (–  H), Increase in entropy (+  S). Nonspontaneous process: Increase in enthalpy(+  H),Decrease in entropy (–  S).

44. Introduction to Free Energy01 Gibbs Free Energy Change (  G): Weighs the relative contributions of enthalpy and entropy to the overall spontaneity of a process.  G =  H – T  S  G < 0Process is spontaneous  G = 0Process is at equilibrium  G > 0Process is nonspontaneous

45. Which of the following reactions are spontaneous under standard conditions at 25°C? AgNO 3 (aq) + NaCl(aq)  AgCl(s) + NaNO 3 (aq)  G° = –55.7 kJ 2 C(s) + 2 H 2 (g)  C 2 H 4 (g)  G° = 68.1 kJ N 2 (g) + 3 H 2 (g)  2 NH 3 (g)  H° =-92.2 kJ  S° = -199 J/K ΔG o = ΔH o ­-TΔS o = (­-92.2 kJ) ­- (298 K)(-­0.199 kJ/K) = ­-32.9 kJ Because ΔG o is negative, the reaction is spontaneous.

46. Introduction to Free Energy04 Equilibrium (  G° = 0): Estimate the temperature at which the following reaction will be at equilibrium. Is the reaction spontaneous at room temperature? – N 2 (g) + 3 H 2 (g)  2 NH 3 (g)  H° = –92.0 kJ  S° = –199 J/K –Equilibrium is the point where  G° =  H° – T  S° = 0 –T = 462 K