Student Worksheet for Thermochemistry

Transcription

1 Student Worksheet for Attempt to work the following practice problems after working through the sample problems in the videos. Answers are given on the last page(s). Relevant Equations/Information Specific Heat: Q=mc T T= Final Temperature - Initial Temperature Where Q= Energy in J, m= mass in grams, c= Specific Heat in J/g C, and T= C Hess Law: H = of H (energy) absorbed and released by all reactions/bonds broken or made - Can also be calculated as H = Hproducts Hreactants Influence of H on Reaction Type (see energy diagram below) - Negative H= Exothermic Reaction (releases energy into the surroundings) - Positive H= Endothermic Reaction (requires energy to be put in before it can proceed) Endothermic Reaction Exothermic Reaction Hints for using H to determine endothermic versus exothermic reactions. The interpretation of the H value depends on how it is used. For examples: a) In your video, you are only looking at the H value of the bonds. Thus, if the reactants have more energy than the products (a + H), energy was added to the system (left graph above) and is endothermic. b) When using Hess law, you are looking at the amount of energy released and absorbed for the entire molecule (not individual bonds). This is also known as Hf. Breaking a bond absorbs energy, and making a bond releases energy. Consider the following equation and scenario: A + B C 2017 Supercharged Science 1

2 If 10 kj (total) of energy are absorbed to break the bonds in s A & B, the 10 kj are now available for the formation of molecule C. Remember that energy is conserved. When C is formed, it uses 5 kj of energy. This leaves 5 kj of energy (such as heat) still in the surroundings. Using Hess Law to determine if the reaction is endothermic or exothermic, - H = Hproducts Hreactants - H = 5 kj 10 kj = -5 kj of energy are released (making it exothermic) like the right graph above. c) Atoms in their elemental form (O2, H2, C, etc) have an Enthalpy of formation equal to 0 kj/mol Supercharged Science 2

3 1. Gold has a specific heat of J/g C. How many joules of heat energy are required to raise the temperature of 40 grams of gold from 20 C to 70 C? 2. An unknown substance with an unknown mass absorbs 1500J while undergoing a temperature increase of 30 C. It s specific heat is 0.49 J/g C. What is the mass of the substance? 3. If the temperature of 50.1 g of ethanol increases from 35 C to 87.8 C, how much heat has been absorbed by the ethanol? The specific heat of ethanol is 2.44J/g C. 4. If 425 g of water at 75 C loses 6570 J of heat, what is the final temperature of the water? Liquid water has a specific heat of 4.18 J/g C. 5. Tea or coffee is drank in cold weather because it warms us. The beverage is 40 C and the body is normally 37.2 C, but warms to 39 C after consumption. If you drink a cup of tea with a mass of 300g of water, how much heat energy was transferred to the body? The specific heat of water is 4.184J/g C Supercharged Science 3

4 6. When testing to see if a solid is pure Titanium, the temperature is raised from 25 C to 57 C after the absorption of J of heat. If the mass of the solid is 2.7 g and the specific heat of Titanium is known to be 0.54 J/g C. What is the specific heat of the solid and is it pure Titanium? For questions 7-11, use the bond energies provided to classify them as Endothermic or Exothermic. 7. Bond Energy per Bond (kj per mole) H-H 432 O-O 204 O-H H2 + O2 2 H2O 8. Bond Energy per Bond (kj per mole) C-H 413 N-H 391 C-N 305 H-H 432 CH4 + NH3 CH3NH2 + H Supercharged Science 4

5 9. Bond Energy per Bond (kj per mole) H-Cl 427 S-Cl 271 H-S 347 Cl-Cl HCl + SCl2 H2S + 2 Cl2 10. Bond Energy per Bond (kj per mole) N-N 160 H-H 432 N-H 391 N2 + 3 H2 2 NH3 11. Bond Energy per Bond (kj per mole) C-Cl 339 O-H 467 H-Cl 427 C-O 358 CHCl3 + H2O HCl + CHCl2OH 2017 Supercharged Science 5

6 For questions 12-15, use the Enthalpy formations provided and Hess Law to classify them as Endothermic or Exothermic. 12. H2O -286 CO -111 H2 0 CO2-394 H2O (l) + CO (g) H2 (g) + CO2 (g) 13. NH3-368 N2O4 27 N 2 0 H2O NH3 (g) + 3 N2O4 (g) 7 N2 (g) + 12 H2O (g) 2017 Supercharged Science 6

7 14. Fe 0 O2 0 Fe2O Fe (s) + 3 O2 (g) 2 Fe2O3 (s) 15. CaO -635 CO2-394 CaCO3-1,207 CaO (s) + CO2 (g) CaCO3 (s) 2017 Supercharged Science 7

8 1. Gold has a specific heat of J/g C. How many joules of heat energy are required to raise the temperature of 40 grams of gold from 20 C to 70 C? Q = mc T You are solving for Q = 40*0.129*(70-20) = 40*0.129*50 = 258 J of energy was absorbed 2. An unknown substance with an unknown mass absorbs 1500J while undergoing a temperature increase of 30 C. It s specific heat is 0.49 J/g C. What is the mass of the substance? Q = mc T You are solving for m. m = Q c T = 1500 ( ) = 102 grams 3. If the temperature of 50.1 g of ethanol increases from 35 C to 87.8 C, how much heat energy has been absorbed by the ethanol? The specific heat of ethanol is 2.44J/g C. Q = mc T You are solving for Q = 50.1*2.44*( ) = 50.1*2.44*52.8 = 6,454.5 J or ~6.5 kj of energy was absorbed 4. If 425 g of water at 75 C loses 6570 J of heat, what is the final temperature of the water? Liquid water has a specific heat of 4.18 J/g C. This one is a little different because you are not solving for an entire variable of the specific heat equation. Instead, you are solving for the final temperature in the T equation. NOTE: The water is losing (has less) heat, so a negative sign goes in front of the Q value. There are two ways to approach this problem, depending on which method you are most comfortable with. Option A: Q = mc T T = Q mc = 6570 ( ) 2017 Supercharged Science 8

9 = = C The problem tells you that the initial temperature was 75 C. To solve for the final temperature, T = Final Temperature-Initial Temperature C = Final Temperature 75 C (Add 75 C to both sides) C= Final Temperature Option B: Q = mc(final T- Initial T) Final T- Initial T= Q mc Minor manipulation of the equation Add Initial T to both sides Final T= Q mc + Initial T = 6570 ( ) + 75 C = C + 75 C = C 5. Tea or coffee is drank in cold weather because it warms us. The beverage is 40 C and the body is normally 37.2 C, but warms to 39 C after consumption. If you drink a cup of tea with a mass of 300g of water, how much heat energy was transferred to the body? The specific heat of water is 4.184J/g C. Q = mc T You are solving for Q Q = 300*4.184*( ) = 300*4.184*1.8 = 2,259.4 J or 2.26 kj of energy was absorbed 6. When testing to see if a solid is pure Titanium, the temperature is raised from 25 C to 57 C after the absorption of J of heat. If the mass of the solid is 2.7 g and the specific heat of Titanium is known to be 0.54 J/g C. What is the specific heat of the solid and is it pure Titanium? Q = mc T You are solving for c c= Q = = = 1.78 J/g C m T (2.7 32) 86.4 Because Titanium has a heat capacity of 0.54 and the solid has a c of 1.78, it is NOT Titanium Supercharged Science 9

10 For questions 7-11, use the bond energies provided to classify them as Endothermic or Exothermic. 7. Bond Energy per Bond (kj per mole) H-H 432 O-O 204 O-H H2 + O2 2 H2O H O Total 1,068 kj OH ,868 Total 1,868 kj 1,068 kj 1,868 kj= -800 kj = Exothermic Note: You should recognize this problem from the end of the video. In the video, the reaction proceeded in the reverse direction. It is important to recognize that if a reaction is endothermic in one direction, it is always exothermic in the opposite direction. 8. Bond Energy per Bond (kj per mole) C-H 413 N-H 391 C-N 305 H-H 432 CH4 + NH3 CH3NH2 + H2 CH Supercharged Science 10

11 NH Total 804 CN H Total kj- 737 kj= + 67 kj = Endothermic Note: In this example, there are 4 CH bonds, but when reviewing the equation, only 1 participates in the reaction. Hence, you could have accounted for them all (as well as all 3 NH bonds), but the answer would be the same regardless. You only need to account for bonds that participate in the reaction. 9. Bond Energy per Bond (kj per Bond) H-Cl 427 S-Cl 271 H-S 347 Cl-Cl HCl + SCl2 H2S + 2 Cl2 HCl SCl Total 1396 HS Cl Total ,396 kj 1,180 kj = kj = Endothermic 2017 Supercharged Science 11

12 10. Bond Energy per Bond (kj per mole) N-N 160 H-H 432 N-H 391 N2 + 3 H2 2 NH3 N H Total 1456 NH Total kj 2346 kj = -890 kj = Exothermic 11. Bond Energy per Bond (kj per mole) C-Cl 339 O-H 467 H-Cl 427 C-O 358 CHCl3 + H2O HCl + CHCl2OH CCl OH Total 806 HCl Supercharged Science 12

13 CO Total kj 785 kj = + 21 kj = Endothermic Use the Enthalpy of Formation Table on Page 2 and Hess Law to classify each of the reactions in as Endothermic or Exothermic. 12. H2O -286 CO -111 H2 0 CO2-394 H2O (l) + CO (g) H2 (g) + CO2 (g) Hformation Hformation H2O -286 H2 0 CO -111 CO2-394 Total/Sum -397 Total/Sum -394 H = Hproducts Hreactants = (-397) = = +3 kj = Endothermic 2017 Supercharged Science 13

14 13. NH3-368 N2O4 27 N2 0 H2O NH3 (g) + 3 N2O4 (g) 7 N2 (g) + 12 H2O (g) Hformation Hformation NH3-46 * 8 = -368 N2 0 * 7 = 0 N 2 O 4 +9 * 3 = 27 H 2 O -242 * 12 = -2,904 Total/Sum -341 Total/Sum -2,904 H = Hproducts Hreactants = -2,904 (-341) = -2, = -2,563 kj = Exothermic For this question (in comparison to # 12), there are different values for H2O. Remember to make note of the phase of the molecule. The enthalpy of formation values are different because #12 addresses water in liquid phase whereas #13 addresses water in gas phase. 14. Fe 0 O2 0 Fe2O Fe (s) + 3 O2 (g) 2 Fe2O3 (s) Hformation Hformation Fe 0 * 4 = 0 Fe2O3 2 * = -1,648 O2 0 * 3 = 0 Total/Sum 0 Total/Sum -1, Supercharged Science 14

15 H = Hproducts Hreactants = -1,648 0 = -1,648 kj = Exothermic 15. CaO -635 CO2-394 CaCO3-1,207 CaO (s) + CO2 (g) CaCO3 (s) Hformation Hformation CaO -635 CaCO3-1,207 CO2-394 Total/Sum -1,029 Total/Sum -1,207 H = Hproducts Hreactants = -1,207 (-1,029) = -1, ,029 = kj = Exothermic 2017 Supercharged Science 15