1. ENERGY OF REACTIONS Entropy, Enthalpy, and Gibb’s Free Energy

2. ENERGY SO FAR  Review:  Energy is the ability to do work or produce heat  There is potential and kinetic energy  Energy can neither be created or destroyed  Every compound needs energy to increase temperature or to change from one state of matter to another

3. ENERGY SO FAR  Units of energy:  Joule (J)  Calorie (cal)  1 cal = 4.184J

4. OTHER ASPECTS OF ENERGY  Energy is also an important component of chemical reactions  Example:  4Fe (s) + 3O 2(g)  2Fe 2 O 3(s) + 1625kJ  In this example, you combine iron and oxygen to produce iron (III) oxide. This reaction also produces 1625kJ of energy/heat.

5. ENERGY AND REACTIONS  To better explain the energy changes in reactions, chemists have come up with:  Enthalpy (H) : the heat content of substances under constant pressure  For a chemical reaction, we describe the change in enthalpy. This is called:  Enthalpy (or heat) of reaction (δH rxn ): the change in heat or energy in a chemical reaction

6. HOW DO WE MEASURE δH rxn  To measure the heat produced or used by a reaction, scientists again use calorimeters. (REVIEW: What is a calorimeter?)  To calculate, we have the following formula:  δH rxn = H products - H reactants

7. PREVIOUS EXAMPLE  Previously, we looked at the following:  4Fe (s) + 3O 2(g)  2Fe 2 O 3(s) + 1625kJ  According to this equation, we produced (or lost) 1625kJ of energy to the environment (you feel this as hot) - EXOTHERMIC  Therefore H reactants > H products  You would have to add 1625kJ of energy to the products to equal the reactants

8. WHAT THIS MEANS  For the reaction:  The reactants have 1625kJ more than the products – you have to add energy to the product side  δH rxn = H reactants - H products  δH rxn = -1625kJ  Therefore, exothermic reactions are always a negative “-” δH rxn

9. ENDOTHERMIC  Since we know that exothermic reactions have a negative heat of reaction:  WHAT IS THE SIGN FOR AN ENDOTHERMIC REACTION?

10. WHAT THIS MEANS  Endothermic reactions are always a positive “+” δH rxn  You have to add energy to the reactant side to equal the products  Therefore we always have to add energy to an endothermic reaction:  Example:  27kJ + NH 4 NO 3 (s)  NH 4 + (aq) + NO 3 - (aq)  δH rxn = H products - H reactants  δH rxn = +27kJ

11. CALCULATING HEATS OF REACTION  Chemists have measured different heats of reaction for combustion reactions.  In each case, they do this under a condition of standard pressure and temperature.  They call these heats of combustion:  δH comb

12. REVIEW  What is the general format for a combustion reaction?

13. COMBUSTION REACTION  Molecule + O 2  CO 2 + H 2 O  Therefore, these are values for the combustion of 1 MOLE of the molecule  Example:  We will look at the combustion of 1 mole of glucose (sugar)  C 6 H 12 O 6 + 6O 2  6CO 2 + 6H 2 O + 2808kJ  δH comb = -2808kJ/mole

14. CALCULATING HEAT OF COMBUSTION  Example: You start with 3.55x10 3 g of glucose (C 6 H 12 O 6 ). How much energy is released when glucose goes through a complete combustion reaction?  STEPS:  Convert to moles  Use the δH comb for glucose to calculate energy

15. ANSWER  Molar mass of glucose: 180g/mole  δH comb = -2808kJ/mole  3.55x10 3 g | 1 mole | -2808kJ | 180g | 1 mole  -5.54x10 4 kJ

16. TRY THESE 1. The heat of combustion for octane (C 8 H 18 ) is -5471kJ/mole. If you start with 1550g of octane, how much energy is released? 2. Sucrose (C 12 H 22 O 11 ), or table sugar, has a heat of combustion of -5644kJ/mole. If you add 2.5g of sucrose to your cereal, how much energy will be added to your cereal?

17. ANSWERS 1. -7.44X10 4 kJ 2. -41 kJ

18. TRY THIS  You have a cup filled with 125mL of glucose (C 6 H 12 O 6 ). If the density of glucose is 1.54g/mL, how many moles of glucose do you have?  If the heat of combustion for glucose is -2805 kJ/mole, what is the heat produced from the cup of glucose?

19. ANSWER  1.07 moles glucose  -3.00 x10 3 kJ

20. TRY THIS  A combustion reaction with octane (C 8 H 18 ) releases a total of -5.55x10 4 kJ of energy. If the heat of combustion for octane is -250kJ/mole, how many grams of octane did you start with?

21. ANSWER 2.53X10 4 g

22. HEATS OF REACTION  As we said earlier, most heats of reaction or heats of combustion are measured using a calorimeter. Sometimes (because some reactions are toxic or unstable), we cannot use a calorimeter and must find an alternative way to measure heats of reaction.

23. HEATS OF REACTION  Hess’s law: If you add two or more thermochemical equations to produce a final equation for a reaction, then the sum of the enthalpy changes for the individual reactions is the enthalpy change for the final reaction

24. EXAMPLE: FIND THE HEAT OF REACTION  2S + 3O 2  2SO 3  This reaction can occurs in 2 steps:  S + O 2  SO 2 ; δH = -297kJ  2SO 3  2SO 2 + O 2 ; δH = 198kJ  How do these reactions go together to form the final reaction?

25. 2S + 3O 2  2SO 3  Step 1: reverse the second reaction because SO 3 is a product. Therefore, you have to change the sign for the heat of reaction  S + O 2  SO 2 ; δH = -297kJ  2SO 2 + O 2  2SO 3 ; δH = -198kJ  NOTE: Now the second reaction is exothermic

26. 2S + 3O 2  2SO 3  Step 2: there are 2 moles of S in the first reaction. This means we have to double everything in the first reaction (including the heat of reaction)  2(S + O 2  SO 2 ; δH = -297kJ)  2S + 2O 2  2SO 2 ; δH = -594kJ  2SO 2 + O 2  2SO 3 ; δH =-198kJ

27. 2S + 3O 2  2SO 3  Step 3: you add the two reactions together to get the final product  2S + 2O 2  2SO 2 ; δH = -594kJ  2SO 2 + O 2  2SO 3 ; δH = -198kJ__  2S + 3O 2  2SO 3 ; δH =-792kJ__  NOTE: If a compound is on opposite sides of a reaction, they cancel out

28. SUMMARY 1. You need to have known chemical reactions that you can combine to form the final chemical reaction 2. You need the δH for each reaction 3. If you need to reverse a reaction, you must also change the sign of δH 4. If you need to multiply a reaction by a number, you must also multiply the δH 5. When all reactions are completed, you then add them together to get the final δH

29. TRY THE FOLLOWING  Combine the two reactions to find the heat of reaction when hydrogen peroxide (H 2 O 2 ) breaks apart to form water and oxygen)  2H 2 O 2  2H 2 O + O 2 ;ΔH = ?  2H 2 + O 2  2H 2 O;δH = -572kJ  H 2 + O 2  H 2 O 2 ; δH = -188kJ

30. 2H 2 O 2  2H 2 O + O 2  Step 1: We need to reverse the second reaction  2H 2 + O 2  2H 2 O; δH = -572kJ  H 2 O 2  H 2 + O 2 ; δH = 188kJ  Step 2: You need 2 moles of H 2 O 2  2H 2 + O 2  2H 2 O; δH = -572kJ  2H 2 O 2  2H 2 + 2O 2 ; δH = 376kJ

31. 2H 2 O 2  2H 2 O + O 2  Step 3: Combine the reactions  2H 2 + O 2  2H 2 O; δH = -572kJ  2H 2 O 2  2H 2 + 2O 2 ; δH = 376kJ  2H 2 O 2  2H 2 O + O 2 δH = -196kJ  NOTE: When you cancel out reactants and products, 2H 2 cancels out 2H 2, but only 1O 2 is cancelled out in the products

32. TRY THIS 2 Ca + 2C + 3O 2  2CaCO 3 1. Ca + 2C  CaC 2 ;dH rxn = -62.8kJ 2. CO 2  C + O 2 ; dH rxn = -393kJ 3. 2CaCO 3 + 2CO 2  2CaC 2 + 5 O 2 ; dH rxn = +1538kJ

33. OTHER PRACTICE  4 NH 3 + 5 O 2  4 NO + 6 H 2 O  N 2 + O 2  2 NO  ∆H = -180.5 kJ  N 2 + 3 H 2  2 NH 3  ∆H = -91.8 kJ  2 H 2 + O 2  2 H 2 O  ∆H = -483.6 kJ

34. PREDICTING SPONTANEOUS CHEMICAL REACTIONS  Energy is an essential ingredient for a chemical reaction  Not all chemical reactions happen spontaneously  To determine if a reaction occurs spontaneously, we need to look at another concept: ENTROPY

35. ENTROPY  Entropy (S): a measure of the number of possible ways that the energy of a system can be distributed  In other words, entropy is the tendency for molecules to spread out as far as possible from each other  Since molecules spreading out is dependent on temperature, the unit for entropy is (J/K) – Joules/Kelvin

36. KELVIN TEMPERATURE  Before we go any further, we need to review Kelvins:  Kelvin (K)  A temperature scale based on absolute 0 (the coldest possible temperature)  The Celsius temperature converts to Kelvin:  K = °C + 273

37. THERMODYNAMICS  Thermodynamics: The laws of thermodynamics, in principle, describe the specifics for the transport of heat and work.  The only law we are interested in is the second law of thermodynamics

38. SECOND LAW OF THERMODYNAMICS  Second Law of Thermodynamics: spontaneous processes always proceed in such a way that the entropy of the universe increases.

39. PREDICTING CHANGES IN ENTROPY  Reminder:  δH rxn = H reactants – H products  We have a similar equation for entropy  δS rxn = S products - S reactants  If the entropy increases during a reaction, then S products > S reactants and δS rxn is positive  If the entropy decreases during a reaction, then S products < S reactants and δS rxn is negative

40. PREDICTING CHANGES IN ENTROPY  Changing states of matter: entropy changes when you go between solid, liquid or gas  H 2 O(l)  H 2 O (g); δS rxn > 0  Dissolving a gas in a solid or liquid always decreases entropy  When you increase the number of gas particles in a reaction, entropy tends to increase  Zn(s) + HCl(aq)  ZnCl 2 (aq) + H 2 (g) ; δS rxn > 0

41. PREDICTING CHANGES IN ENTROPY  With some exceptions, entropy increases when a solid dissolves in a liquid  The solid tends to break apart  Random motion of particles increases as the temperature increases  As you increase temperature, the entropy increases

42. PREDICT THE FOLLOWING  Try to determine whether entropy increases or decreases for the following reactions: 1. CF(g) + F 2 (g)  CF 3 (g) 2. NH 3 (g)  NH 3 (aq) 3. C 10 H 8 (s)  C 10 H 8 (l) 4. H 2 O (l)  H 2 (g) + O 2 (g)

43. ANSWERS 1. Entropy decreases: You go from 2 molecules of gas to 1 2. Entropy decreases: You are dissolving a gas into a liquid 3. Entropy increases: You have gone from a solid to a liquid 4. Entropy increases: You are going from no gas to 2 molecules of gas

44. TRY THIS CH 4 + NH 3  HCN + 3 H 2  N 2 + 3 H 2  2 NH 3 ; ∆H = -91.8 kJ  C + 2 H 2  CH 4 ;∆H = -74.9 kJ  H 2 + 2 C + N 2  2 HCN; ∆H = +270.3 kJ

45. TRY THIS N 2 H 4 + H 2 → 2NH 3 N 2 H 4 + CH 4 O  CH 2 O + N 2 + 3H 2 ΔH = -37 kJ N 2 + 3H 2 → 2NH 3 ΔH = -46 kJ CH 4 O → CH 2 O + H 2 ΔH = -65 kJ

46. TRY THIS 2C + 2H 2 O  CH 4 + CO 2  C + H 2 O  CO + H 2 ; ∆H = 131.3kJ  CO + H 2 O  CO 2 + H 2 ; ∆H = -41.2kJ  CH 4 + H 2 O  3H 2 + CO; ∆H = 206.1kJ

47. PUTTING IT ALL TOGETHER  Generally, exothermic reactions are spontaneous (you do not need to add energy to make them work)  Generally, spontaneous reactions will increase in entropy  Yet, there is a way to know for sure if a reaction will occur spontaneously or not

48. GIBB’S FREE ENERGY  In 1878, William Gibbs (an American) combined enthalpy and entropy to determine if a reaction was spontaneous  He called his equation Gibbs Free Energy equation  Free energy: the energy that is available to do work (δG)

49. GIBB’S FREE ENERGY  Gibbs Free Energy Equation:  δG rxn = δH rxn – TδS rxn  δG (kJ): represents free energy  δH (kJ): represents change in enthalpy  T (K): temperature in kelvins  δS (J/K): represents change in entropy

50. SPONTANEOUS?  Like entropy or enthalpy, Gibb’s free energy can be positive or negative  If positive (+): if the Gibb’s free energy is positive, the reaction is not spontaneous (you would need to add energy to make the reaction work)  If negative (-): if the Gibb’s free energy is negative, the reaction is spontaneous

51. EXAMPLE  N 2 (g) + 3H 2 (g)  2NH 3 (g)  δH rxn = -91.8kJ  δS rxn = -197J/K  T = 25°C 1. Why does the entropy decrease in this reaction? 2. What is the Gibb’s free energy? 3. Is this reaction spontaneous?

52. STEP 1: WRITE WHAT YOU KNOW  δG = ?  δH = -91.8kJ  T = 25°C +273 = 298K  δS = -197J/K = -0.197kJ/K  NOTE: Temperature must be in Kelvin and entropy is converted to kJ so that it matches the units for δG and δH.

53. STEP 2: PUT THE NUMBERS IN THE FORMULA  δG rxn = δH rxn – TδS rxn  δG rxn = -91.8kJ – (298K)(-0.197kJ/K)  δG rxn = -91.8kJ – (-58.7kJ)  δG rxn = -91.8kJ + 58.7kJ  δG rxn = -33.1kJ

54. STEP 3: DETERMINE IF IT IS SPONTANEOUS 1. The reason why the entropy of this reaction decreased is that you are going from 4 moles of gas to 2 moles of gas. The amount of gas is reduced and lowers the entropy. 2. δG rxn = -33.1kJ 3. Since the Gibb’s free energy is negative, this is a spontaneous reaction.

55. TRY THIS  A Chemical reaction has the following information:  δH rxn = 145 kJ  δS rxn = 322 J/K  T = 109°C 1. What is the Gibb’s free energy? 2. Is this reaction spontaneous?

56. ANSWER 1. δG = +22kJ 2. Since the Gibb’s free energy is positive, this reaction is NOT spontaneous  Challenge : At what temperature would this reaction become spontaneous?

57. ANSWER  δH rxn = 145 kJ  δS rxn = 322 J/K  T = 109°C  We need to find what temperature δG<0  0kJ = 145kJ – T(0.322kJ/K)  T(0.322kJ/K) = 145kJ  T = 450K  the reaction has to be at least 450K to be spontaneous

58. WHEN ARE REACTIONS SPONTANEOUS?  Reactions are spontaneous when:  δH rxn 0  δH rxn < 0; δS rxn < 0 (spontaneous at lower temperatures only)  δH rxn > 0; δS rxn > 0 (spontaneous at higher temperatures only  Reactions are not spontaneous when:  δH rxn > 0; δS rxn < 0