1. Thermochemistry Chapter 6 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. What is Thermochemistry? Thermochemistry : study of heat change in chemical reactions.

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

4. Heat : transfer of thermal energy between two bodies that are at different temperatures. Temperature: measure of the thermal energy. Temperature = Thermal Energy 90 0 C 40 0 C greater thermal energy Temperature vs. Thermal Energy

5. System: specific part of the universe that is of interest in the study. open mass & energyExchange: closed energy isolated nothing SYSTEMSURROUNDINGS Effects on a system

6. Exothermic process: transfers thermal energy from the system to the surroundings. Endothermic process: transfers thermal energy from the surrounds to the system. 2H 2 (g) + O 2 (g) 2H 2 O (l) + energy energy + 2HgO (s) 2Hg (l) + O 2 (g)

7. Endo vs Exo during phase changes H 2 O (g) H 2 O (l) H 2 O (s) H 2 O (l) + energy energy + gas  liquid  solid loses energy! solid  liquid  gas gains energy!

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

9. How much heat is given off when an 869 g iron bar cools from 94 0 C to 5 0 C? c Fe = 0.444 J/g 0 C  t = t final – t initial = 5 0 C – 94 0 C = -89 0 C q = mc  t = 869 g x 0.444 J/g 0 C x –89 0 C= -34,000 J

10. Each line represents a different calculation and is numbered accordingly. Typical Heat Curve

11. If we have 85 grams of ice at -10 °C. How much energy is needed to convert it to water vapor at 120 o C? Step 1 – We are below 0°C Step 2 - We are at 0 °C Typical Heat Curve – Problem Solving

12. Step 3 – We are above 0 °C but below 100 °C Step 4 - We are at 100 °C Step 5 – We are above 100 °C Typical Heat Curve – Problem Solving; cont’d

13. Now, add up all the steps for total heat. Summary: – When the substance is heating up and is in a single phase, the formula used is : q=mc p ∆T – When a phase change is occurring all the energy is involved with the intermolecular forces and because of this we do not see a temperature change. To calculate we must multiply the moles by molar heat; q= mol∆H fusion OR q= mol∆H vap Typical Heat Curve – Problem Solving; cont’d Note: During the cooling process ΔH becomes negative

14. No heat enters or leaves! q sys = q water + q bomb + q rxn q sys = 0 q rxn = - (q water + q bomb ) q water = mc  t q bomb = C bomb  t Constant-Volume Calorimetry

15. Heat of a Calorimeter A 2.05 g piece of magnesium is burned in a bomb calorimeter with a heat capacity of 2100. J/ o C. The calorimeter contains 450. g of water and the temperature increases by 5.84 o C. Calculate the heat released by the Mg in kJ/mol. q Mg = -(q H2O + q bomb ) Note: the negative sign is just a reminder that heat is released

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

17. Basic Calorimetry Example A 3.75 g piece of hot iron is dropped into a water bath and the temperature of the iron decreases by 46.8 o C. How much heat is released? The specific heat of iron is 0.47J/g o C q=mc Δ T What is the sign of q?

18. System vs Surroundings Calcuations When 5.84g of HCl is added to water the temperature of the solution increases from 25.0 o C to 37.8 o C. What is the Δ H for this reaction? HCl  H + + Cl - Δ H= _____ J/mol q HCl = -q H2O

19. System vs Surroundings Calcuations A 14.2 g sample of an unknown metal is placed in a water bath and the temperature of the metal decreases from 67.9 o C to 34.5 o C. The temperature of the 40.0 g of water shows an increase of 12.0 o C.What is the specific heat of the unknown metal? q metal = - q H2O

20. Enthalpy (H): 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

21. 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.

22. 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.

23. 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 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 Thermochemical Equations

24. How much heat is evolved 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 Example Problem

25. This is used as a reference point for all enthalpy expressions. The standard enthalpy of formation of any element in its most stable form is zero. Standard enthalpy of formation (  H 0 ): heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f  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 Standard Enthalpy of Formation standard states heat of FORMATION change in enthalpy

26. You will always be given a table as a reference. There is a table in the back of your book!

27. standard enthalpy of reaction (  H 0 ): 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  - Enthalpy of a Reaction

28. 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 n  H 0 (products) f =  m  H 0 (reactants) f  - 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 = [ 12(-393.5) + 6(-187.6) ] – [ 2(49.04) + 0 ] = -5946 kJ -5946 kJ 2 mol = - 2973 kJ/mol C 6 H 6 Enthalpy of Reaction Example

29. Hess’s Law 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. Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. Examples of state functions: energy, pressure, volume, temperature

30. Hess’s Law Example Problem H 2 (g) + ½O 2 (g) → H 2 O(l) C(s) + O 2 (g) → CO 2 (g) CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(l) This reaction must be reversed! The ΔH is changed to positive! CO 2 (g) + 2H 2 O(l) → CH 4 (g) + 2O 2 (g) ∆ H 0 = +890.8 kJ 2 [ H 2 (g) + ½O 2 (g) → H 2 O(l) ] This reaction must x2! The ΔH is x2! Goal Reaction: C(s) + 2H 2 (g) → CH 4 (g)

31. 2H 2 (g) + O 2 (g) → 2H 2 O(l) C(s) + O 2 (g) → CO 2 (g) C(s) + 2H 2 (g) → CH 4 (g) CO 2 (g) + 2H 2 O(l) → CH 4 (g) + 2O 2 (g) ADD IT UP!! ∆ H 0 = +890.8 kJ

32. enthalpy of solution (  H soln ): heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent.  H soln = H soln - H components Which substance(s) could be used for melting ice? Which substance(s) could be used for a cold pack? Enthalpy of Solution

33. The Solution Process for NaCl  H soln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol

34. Entropy and Free Energy Chapter 18 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

35. A waterfall runs downhill A lump of sugar dissolves in a cup of coffee At 1 atm, water freezes below 0 0 C and ice melts above 0 0 C Heat flows from a hotter object to a colder object A gas expands in an evacuated bulb Iron exposed to oxygen and water forms rust spontaneous nonspontaneous Spontaneous Processes

36. spontaneous nonspontaneous X Gas Movement Spontaneity

37. Does a decrease in enthalpy mean a reaction proceeds spontaneously? CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l)  H 0 = -890.4 kJ H + (aq) + OH - (aq) H 2 O (l)  H 0 = -56.2 kJ H 2 O (s) H 2 O (l)  H 0 = 6.01 kJ NH 4 NO 3 (s) NH 4 + (aq) + NO 3 - (aq)  H 0 = 25 kJ H2OH2O Spontaneous reactions NO!

38. Entropy (S): measure of the randomness or disorder of a system. orderS S  S = S f - S i If the change from initial to final results in an increase in randomness S f > S i  S > 0 Entropy

39. Entropy and Phase Changes For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas state S solid < S liquid << S gas H 2 O (s) H 2 O (l)  S > 0

40. Processes that lead to an increase in entropy more disorder (  S > 0)

41. (a) Condensing water vapor Randomness decreases Entropy decreases (  S < 0) (b) Forming sucrose crystals from a supersaturated solution Randomness decreases Entropy decreases (  S < 0) (c) Heating hydrogen gas from 60 0 C to 80 0 C Randomness increases Entropy increases (  S > 0) (d) Subliming dry ice Randomness increases Entropy increases (  S > 0) How is Entropy affected?

42. First Law of Thermodynamics Energy can be converted from one form to another but energy cannot be created or destroyed. Second Law of Thermodynamics The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process.  S univ =  S sys +  S surr > 0 Spontaneous process:  S univ =  S sys +  S surr = 0 Equilibrium process: Laws of Thermodynamics

43. aA + bB cC + dD S0S0 rxn dS 0 (D) cS 0 (C) = [+] - bS 0 (B) aS 0 (A) [+] S0S0 rxn nS 0 (products) =  mS 0 (reactants)  - standard entropy of reaction (  S 0 ): entropy change for a reaction carried out at 1 atm and 25 0 C. rxn Standard Entropy of Reaction Look Familiar??

44. Standard Entropy Problem What is the standard entropy change for the following reaction at 25 0 C? 2CO (g) + O 2 (g) 2CO 2 (g) S 0 (CO) = 197.9 J/K mol S 0 (O 2 ) = 205.0 J/K mol S 0 (CO 2 ) = 213.6 J/K mol S0S0 rxn = 2 x S 0 (CO 2 ) – [2 x S 0 (CO) + S 0 (O 2 )] S0S0 rxn = 427.2 – [395.8 + 205.0] = -173.6 J/K mol JUST LIKE STANDARD ENTHALPY CALCULATIONS!

45. If a reaction produces more gas molecules than it consumes  becoming more disordered If the total number of gas molecules decreases  becoming less disordered If there is no net change in the total number of gas molecules, then  S 0 may be positive or negative BUT  S 0 will be a small number. What is the sign of the entropy change for the following reaction? 2Zn (s) + O 2 (g) 2ZnO (s)  S is negative Gases and Entropy  S 0 > 0 positive  S 0 < 0 negative

46. Exothermic Process  S surr > 0 Endothermic Process  S surr < 0 Heat and Entropy

47. Third Law of Thermodynamics The entropy of a perfect crystalline substance is zero at the absolute zero of temperature.

48. For a constant- temperature process  G =  H sys -T  S sys G = Gibbs free energy H = Enthalpy T = Temperature (K) S = Entropy  G < 0 The reaction is spontaneous in the forward direction.  G > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction.  G = 0 The reaction is at equilibrium. Gibbs Free Energy

49. aA + bB cC + dD G0G0 rxn d  G 0 (D) f c  G 0 (C) f = [+] - b  G 0 (B) f a  G 0 (A) f [+] G0G0 rxn n  G 0 (products) f =  m  G 0 (reactants) f  - standard free-energy of reaction (  G 0 ): free-energy change for a reaction when it occurs under standard-state conditions. rxn Standard free energy of formation (  G 0 ): free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. f  G 0 of any element in its stable form is zero. f Gibbs Free Energy Calculation

50. 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) G0G0 rxn n  G 0 (products) f =  m  G 0 (reactants) f  - What is the standard free-energy change for the following reaction at 25 0 C? G0G0 rxn 6  G 0 (H 2 O) f 12  G 0 (CO 2 ) f = [+] - 2  G 0 (C 6 H 6 ) f [] G0G0 rxn = [ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ Is the reaction spontaneous at 25 0 C? YES!  G 0 = -6405 kJ < 0 spontaneous Gibbs Energy of Formation Problem

51.  G =  H - T  S What if…