1. OB: Students will develop a mastery of all thermochem math problems Calculators, reference tables, and thinking caps (again).

2. Calculate the specific heat capacity constant for aluminum if your hunk of pure aluminum has mass of 147.2 g and the temperature changes from 279 K to 365 K when you impart 11,355 Joules of energy onto it.

3. Calculate the specific heat capacity constant for aluminum if your hunk of pure aluminum has mass of 147.2 g and the temperature changes from 279 K to 365 K when you impart 11,393 Joules of energy onto it. q = mC ΔT 11,355 J = (147.2g)(C)(86.0K) 11,355 J (147.2g)(86.0K) 0.897 J/g ·K = C = C NOTE: metals have a fairly low specific heat capacity, water’s is 4.18 J/g·K

4. Water is a common enough substance in your life that the regents will not remind you what phase changes it might go through in any problem. You will have to determine that from the temperatures in the problem. Ice can be colder than 273 K (0 °C) and can be made colder or warmer while staying a solid. Water phase changes at 273 K. When ever there is a temperature change (ice can change temp) we use the basic heat formula: q = mCΔT If water is liquid, we use the 4.18 J/g·K constant. When the H 2 O is solid ice, it has a different constant, C ICE = 2.10 J/g·K When you are given information like this, you must use it in your problems, like the next slide will do. Write the ice constant near table B in your reference table now.

5. For fun, you obtain a block of ice (2550 g) at -5.00°C and sit on it in a tub until it melts to body temperature of 36.0°C (do math on next slide) q=mCΔT C = 2.10 J/g·K q = mH F q=mCΔT C = 4.18 J/g·K 268 K 309 K 0 °C -5.00 °C 36.0 °C

6. For fun, you obtain a block of ice (2550 g) at -5.00°C and sit on it in a tub until it melts to body temperature of 36.0°C This is a 3 step thermochem problem, ice warms, ice melts, water warms. Make note to use the proper “C” value for your H 2 O in the proper place. Add up the three sets of Joules for one total answer.

7. For fun, you obtain a block of ice (2550 g) at -5.00°C and sit on it in a tub until it melts to 36.0°C q=mCΔT = (2550g)(2.10J/g·K)(5.00 K) = 26,775 J q = mH F = (2550g)(334J/g) = 851,700 J q=mCΔT = (2550g)(4.18J/g·K)(36.0K) = 383,724 J 1,262,199 J 1,260,000 J with 3 SF Thermochem is like eating out in a fancy restaurant, multi courses with ONE total bill. Add up the parts to get the total sum.

8. Compare how much energy it takes to condense15.0 g steam into water with the amount of energy it takes to vaporize 15.0 g water into steam. vaporize q = mH V condense q = mH V

9. Compare how much energy it takes to condense15.0 g steam into water with the amount of energy it takes to vaporize 15.0 g water into steam. vaporize q = mH V q = (15.0g)(2260 J/g) q = 33,900 J condense q = mH V q = (15.0g)(2260 J/g) q = 33,900 J

10. When 56.0 grams of carbon and sufficient hydrogen synthesize completely into ethane (C 2 H 6 ) gas, how much energy is released (or absorbed) in JOULES???

11. When 56.0 grams of carbon and sufficient hydrogen synthesize completely into ethane (C 2 H 6 ) gas, how much energy is released (or absorbed)? NOTE: ΔH is a negative number, that means energy is a product, which we can write this way… 2C + 3H 2 C 2 H 6 + 84.0 kJ 56.0 g C 1 X 1 mole C 12 g C = 4.67 moles C MR carbon energy 1 84.0 4.67 x 2X = 392.28 kJ X = 196 kJ

12. DRAW this Cooling curve for Chromium metal Questions on next slide. A B C D E F Heat removed at a constant rate (time) KELVINKELVIN 1 2

14. What temps are 1 + 2? What’s PE doing BC and CD? What’s KE doing AB and DE Why is BC longer than DE? Which thermochem formula do you use for BC? How about for EF? A B C D E Heat removed at a constant rate (time) KELVINKELVIN 1 2 Cooling curve for Chromium metal F

15. Why is BC longer than DE? BC represents the condensation phase change for chromium gas to liquid, which is a “bigger” energy event that the freezing of chromium. A B C D E F Heat removed at a constant rate (time) KELVINKELVIN 2944 2180 Cooling curve for Chromium metal PE KE q = mH V q = mC ΔT

16. Converting balanced chemical equations into balanced thermochemical equations. Look at table I, choose the second equation, propane combusts. Write the balanced chemical equation with the ΔH, then, write a balanced thermochemical equation below it (properly). Then, write the most endothermic equation of all, with the ΔH, then below that one, the balanced thermochemical equation properly.

17. Converting balanced chemical equations into balanced thermochemical equations. Look at table I, choose the second equation, propane combusts. Write the balanced chemical equation with the ΔH, then, write a balanced thermochemical equation below it (properly). C 3 H 8 + 5O 2 3CO 2 + 4H 2 0 ΔH= -2219.2 kJ C 3 H 8 + 5O 2 3CO 2 + 4H 2 0 + 2219.2 kJ Energy is a PRODUCT in an exothermic reaction Then, write the most endothermic equation of all, with the ΔH, then below that one, the balanced thermochemical equation properly. 2C + H 2 C 2 H 2 ΔH = +227.4 kJ 2C + H 2 + 227.4 kJ C 2 H 2 Energy is a REACTANT in an endothermic reaction

18. DUE TOMORROW nothing Hand in Friday: Dorito’s Lab