Homework Ch. 17 thermodynamics

General Questions:

You Should write #1-15 on your paper and fill them in as we go along in class.

1. What is the symbol for Enthalpy? What does it mean?

2. What is the symbol for Entropy? What does it mean?

3. What is an exothermic reaction?

4. What is an endothermic reaction?

5. Look at the following phase changes and say whether they are endo or exothermic reactions

a. Liquid to gas

b. Solid to liquid

c. Gas to liquid

d. Liquid to Solid

6. True or False: Enthalpy is negative for exothermic reactions.

7. True or False: The more disordered a system is, the more negative the Entropy.

8. True or False: If a reaction is spontaneous, it will go to completion very quickly.

9. True or False: Phase changes deal with intermolecular forces while molecular composition or decomposition deals with intramolecular forces.

10. How many kCal/gram are in Carbohydrates? Fats? Proteins?

11. True or False: If the specific heat of a substance is high, then it is easy to heat up that substance and that substance cools down very fast.

12. True or False: Heat deals with temperature and mass. Temperature is independent of the amount of mass.

13. True or False: When water freezes, heat is released.

14. True or False: When ice melts, heat is released.

15. True or False: The more negative DG, the more spontaneous the reaction will be.

Math Stuff:

Finding Specific Heat of a Substance: (q = C_p x m x DT). You are going to solve for s (specific heat).

16. P. 548 #16 Answer: 3562 J

17. P. 548 #17 Answer: 569K

18. 10 g of a metal at 100^oC is placed in a calorimeter with 20 g of water. The water’s original temperature was 10^oC and its final temperature was 15^oC. What is the specific heat of the metal?

19. Use the exact same set up as in problem #18 (same masses, same temps..etc), only the final temperature of the water will be 40^oC (instead of 15^oC). What is the specific heat of this metal? Which metal had the higher specific heat - #18 or #19?

20. What does your answers to #18and #19 tell you about the value of specific heat on raising the temperature of water? (In other words, if you were going to heat up a swimming pool with a large metal block, would you use one with high specific heat, or low specific heat?)

Calculating Heat of phase changes on PHYSICAL reactions

(q = C_p x m x DT) or (q = m x heat of vap.) or (q = m x heat of fus.)

21. Assume you have a 10 g sample of water. Draw a temp. vs. time curve. Begin at the bottom of the curve (where water is frozen at –10^oC) and calculate the heat (q) at each step of the way. I want you to end when water completely boils at 100^oC. DO NOT GO ON AFTER THIS POINT.

Here’s some hints: You will need to use all three of the equations for calculating heat. Heat of fusion for water is: 334 J/g. Heat of vaporization of water is 2260 J/g.

The specific heat for liquid water is 4.18 J/g^oC. The specific heat for solid water is 2.03 J/g^oC.

Your final answer will look like this….: You will have a q value for solid ice reaching the melting point. You will have a q value for solid ice melting. You will have a q value for liquid water rising up to the boiling point. You will have a q value for liquid water boiling away. That is a total of 4 q values.

22. Are the q values for #21 going to be positive or negative? Remember that heat is positive if it is endothermic and negative if it is exothermic. And in order to heat ice to melt and boil it, you must add heat. So, is the system taking in heat or releasing it?

23. Do the same thing you did for #21, only this time, do it with copper. Assume you have 10 g of solid copper and you want to find out the heat required to boil it. Draw a heating curve with temperature along the y-axis and time along the x-axis (don’t worry about the amount of time it takes, just write the word “time”). Start at a temperature of 100^oC and end at the boiling point of 2927^oC.

Here are some vital statistics on copper: Copper’s Heat of fusion is 205 J/g Copper’s Heat of Vaporization is 4726 J/g. Specific heat of the solid of copper is 0.387 J/g^oC. Specific heat of the liquid form is 0.695 J/g^oC. Melting point: 1084^oC. Boiling point: 2927^oC.

24. Do the same thing you did for #21 and 23, only this time, do it with gold. Assume you have 10 g of liquid gold and you want to find out the heat required to FREEZE it. Draw a heating curve. Start at a temp of 2856^oC (gold_(l)) and end at the sub-freezing point of 100^oC. Will q be positive or negative?

Here are some vital statistics on gold: Gold’s heat of fusion is 64.5 J/g. Gold’s heat of Vaporization is 1578 J/g Specific heat of solid gold is 0.129 J/g^oC. Specific heat of liquid gold is 0.345 J/g^oC). Melting point: 1064^oC. Boiling point: 2856^oC

Finding Enthalpy Changes in CHEMICAL reactions (DH = DH_products - DH_reactants)

Hint: Remember that DH_f for any element in standard state is 0. Also remember to multiply your DH value by the number of moles in the balanced equation.

25. Use the table on p. 902 to find the Enthalpy Change (DH) for the combustion reaction of sucrose with oxygen to form carbon dioxide and LIQUID water. REMEMBER TO BALANCE IT FIRST! DH for Sucrose is –2222 KJ/mole

26. Find the enthalpy change for the combustion reaction of Methane (again, it will be LIQUID water).

27. #2 p. 522

Nutrition and Enthalpy Changes (remember that calories is Enthalpy in nutrition terms) (Also remember that Enthalpy is listed as kcal/gram instead of kJ/mole when you are talking about food)

28. What is the energy released when a quarter-pound hamburger is digested? The mass of the hamburger is 114 g. It is 41% water, 16% protein, 14% fat and 29% carbohydrate. (HINT: Multiply the percents by 114 g of burger (eg. 0.16 x 114 g) Remember how many kcals are in each gram of fat, protein and carbohydrate – see your notes if you don’t remember).

29. Look at a package of food (any food, candy bar, cereal box…etc) and find the grams of carbohydrate, protein and fat (combine saturated and unsaturated fats). Determine the amount of calories from each one of these parts of food. YOU MUST WRITE YOUR DATA FROM THE BOX IN YOUR ANSWER.

30. Look at that same package of food. Divide the total calories listed on the package by the three individual parts of food to determine the % of calories from fat, the % of calories from protein, the % of calories from Carbohydrate.

31. Compare the energy (Kcalories) available in a 12oz glass of Milk to a 12oz can of soda. 12oz of liquid has a mass of 360 g. Assume that the Milk is 87.6% water, 3.3% protein, 2.8% fat and 4.7% Carbohydrate. The Soda is 89% water and 11% carbohydrate. (HINT: To find the grams of protein, fat and carbohydrate in each drink, multiply 360 grams by the individual percents (eg. 3.3 percent = 0.033)).

Entropy and Gibbs Free Energy (DG = DH - TDS) and (DH = DH_products - DH_reactants) and

(DS = DS_products - DS_reactants)

32. What is the DG for making LIQUID water out of hydrogen and oxygen gas? You will need to know that the DH_f of hydrogen and oxygen is ZERO. But the DS of hydrogen is 0.131 kJ/K mole and DS of oxygen is 0.205 kJ/K mole. You can find the DH_f of LIQUID water on p. 902, and the DS of liquid water is 0.07 kJ/K mole. Assume the temperature is 273K. Is this reaction spontaneous? Is the reaction spontaneous the other way (meaning, can water spontaneously turn into hydrogen and oxygen?)

33. What is the DG for the following balanced equation: 2FeO -----à 2Fe + O₂?

You will need to know the following: DH_f of FeO is –272 kJ/mole. (it is zero for Fe and O₂).

DS for FeO is 0.061 kJ/K mole, for Fe is 0.027 kJ/K mole, and for O₂ is 0.205 kJ/K mole. Assume a temperature of 298K

When you have figured out DG, is the reaction spontaneous? Is it spontaneous the other way? Do you know what this reaction is?

34. What is the DG for the following balanced equation: CO₂ + 2H₂O ----à CH₄ + 2O₂?

DH_f of CO₂ is –393.5 kJ/mole, of H₂O is –285.8 kJ/mole of CH₄ is –75 kJ/mole

DS of CO₂ is 0.214 kJ/K mole, of H₂O is 0.070 kJ/K mole of CH₄ is 0.186 kJ/K mole of O₂ is 0.205 kJ/K mole. Assume the temperature is 298 K.

35. Figure out what temperature would be necessary (at a minimum) to make the reaction in problem #34 be spontaneous. (in other words, a DG of Zero!)

BIG TEST HINT: I am going to give you the DG = DH - TDS equation and ask you if something is spontaneous with: Large positive DH value and a large negative DS value…etc. Think about it.

Answer Key

18. q_water=418J

Cp_metal= 0.492 J/g^oC

19. q_H2O= 2508J

S_metal= 4.18J/g^oC

21. a. 203J

b. 3340J

c. 4180J

d. 22600J

23. a. 3808J

b. 2050J

c. 12809J

d. 47260J

24. a. –15780J

b. –6182J

c. –645J

d. –1244J

25. DH= -5644 KJ

26. DH= -890.3 KJ

28. Prot. 72.96 Kcal

Fat 143.64 Kcal

Carb. 132.24 Kcal

S= 349 Kcal

31. Milk

Prot. 47.52 Kcal

Fat 90.72 Kcal

Carb. 67.68 Kcal

S= 205.92 Kcal

31 continued. Soda

Prot. 0 Kcal

Fat 0 Kcal

Carb. 158 Kcal

S=158 Kcal

32. DG= -482 KJ/mol

33. DH= 544 KJ/mol DS= +0.137 KJ/(mol)(K) DG= + 503 KJ/mol

34. DH=+890.1KJ/mol DS=+0.242 KJ/(mol)(K) DG= +817 KJ/mol

35. A minimum temp. of 3678K would be needed.