1. 1 10 pt 15 pt 20 pt 25 pt 5 pt 10 pt 15 pt 20 pt 25 pt 5 pt 10 pt 15 pt 20 pt 25 pt 5 pt 10 pt 15 pt 20 pt 25 pt 5 pt 10 pt 15 pt 20 pt 25 pt 5 pt 7.1 Endo Exo 7.2 Calorimetry/He at Capacity 7.3 State Changes 7.4 Enthalpy of Reactions 7.5 Hess’s’s’ss Law (rusty puzzles)

2. 2 myard Explain System vs. Surroundings ?

3. 3 System is the specific molecule(s) reacting while the surroundings is the everything else stuff

4. 4 myard Explain endothermic and exothermic ?

5. 5 Endothermic takes energy from a reaction, exothermic releases energy into a reaction

6. 6 myard You hold a beaker in your hand and it feels hot in your hand. A. is the reaction endo or exo B. what is the system and what is the surroundings

7. 7 A. It is exo B. The system is the beaker and the surroundings is your hand C. let go of the damn cup your gonna burn yourself.

8. 8 myard A thermometer is put in a cup. The temperature then decreases. Is this endo or exo and explain were the energy is coming and going.

9. 9 The reaction is endothermic and the energy is voming form the thermonmeter to the stuff in the cup

10. 10 myard Which of these is an exothermic process? I. Melting ice II. Boiling water III. Freezing water A. I and II B. II only C. III only

11. 11 I DON’T KNOW! THESE QUESTIONS AREN’T ON THE AP EXAM HAHAHAHAHAA! Its C melting and boiling are endothermic.

12. 12 myard Convert 5896 joules to calories

13. 13 5896/4.18 = 1410 calories

14. 14 myard Convert.015209 Calories to joules

15. 15.015209∙1000 = 15.209 cal ∙ 4.18 = 63.57 Joules

16. 16 myard How many joules are needed to make 33g of 9 o water to 55 o

17. 17 MCAT ∆H = 33 ∙ 4.18 ∙ (55-9) ∆H = 6345.24 joules

18. 18 myard Define specific heat capacity in your own words

19. 19 If it was something along the lines of the amount of energy needed to change 1 gram of substance one C o yadayadayada your good

20. 20 myard If it took 4500 joules to heat 5 g of a substance with a heat capacity of 5.6 J/goC from 15o, what is the final temp.

21. 21 MCA(t2 –t1) 4500 = 5 ∙ 5.6 ∙ (x – 15) 4500 = 28x – 420hehe 4080 = 28x x = 145.7 o C

22. 22 myard True or False: temperature changes in a change of state

23. 23 False. The molecules absorb/release the energy to change the state

24. 24 myard Why on the heating curve of water are there stops in the rise of temp?

25. 25 Because those are the places at which state changes are being made and those don’t rise no temp

26. 26 myard How much energy does it take to boil 40 grams of water

27. 27 ∆H =MCA ∆H = 40 ∙ 547.2cal/g o C ∆H = 21888 cal

28. 28 myard If it takes 456 cal to melt 78 grams of a substance, what is the enthalpy of fusion

29. 29 456 = 78 ∙ H f H f = 5.84

30. 30 myard How much energy does it take to heat 10g of water from -2 o C to 100 o C?

31. 31 ∆H1 = 10 ∙.51 ∙ 2 = 10.2 cal ∆H2 = 10 ∙80.87 = 818.7 cal ∆H3 = 10 ∙ 1 ∙ 100 = 1000 cal ∆H1 + ∆H2 + ∆H3 = 1828.9 cal

32. 32 myard Explain enthalpy of formation

33. 33 The enthalpy change when a compound is formed from the elemental state

34. 34 myard Why is it products minus reactants

35. 35 Because the reactants take energy to break so that energy takes away from the energy made by the products being created.

36. 36 myard What do you have to do to this before you find the enthalpy of reaction? CH 3 OH + O 2 = CO 2 +H 2 O

37. 37 Balance it! You need the coefficients to multiply the enthalpy of formations by.

38. 38 myard 4NH 3 + 5O 2 = 4NO + 6H 2 O what is the heat of formation of this reaction

39. 39 H rxn = (4 ∙ 90.3 + 6 ∙ -241.8)- (4 ∙ -46.11 + 0) H rxn = 905 kj

40. 40 myard 2CH 2 OH + 3O 2 = 2CO 2 + 4H 2 O H rxn = -1454 kj how much energy is released if 985.7 g of water are made

41. 41 985.7/18.02 = 54.7 54.7moles/2moles ∙ -1454kj H = 39797.14

42. 42 myard Find the heat of reaction for the equation N 2 + 2O 2 = 2NO 2 A. N 2 + O 2 = 2NO 180kj B. 2NO 2 = 2NO + O 2 112kj

43. 43 N 2 + O 2 = 2NO 2NO + O 2 = 2NO 2 N 2 +2O 2 = 2NO 2 A + -B 180 -112 = 68kj

44. 44 myard Find the enthalpy of reaction for this reaction H 2 CO 3 = H 2 O + CO 2 A. H 2 CO + O 2 = H 2 CO 3 H=-140KJ B. H 2 CO + O 2 = H 2 O + CO 2 H=-62.5KJ

45. 45 H 2 CO 3 = H 2 CO + O 2 H 2 CO + O 2 = H 2 O + CO 2 H 2 CO 3 = H 2 O + CO 2 -A + B 140 – 62.5 = 77.5 KJ

46. 46 myard What is the value for ΔH for the following reaction? CS 2 (l) + 2 O 2 (g) → CO 2 (g) + 2 SO 2 (g) A. C(s) + O 2 (g) → CO 2 (g) ΔH f = -393.5 kJ/mol B. S(s) + O 2 (g) → SO 2 (g) ΔH f = -296.8 kJ/mol C. C(s) + 2 S(s) → CS 2 (l) ΔH f = 87.9 kJ/mol

47. 47 C(s) + O 2 (g) → CO 2 (g) 2 S(s) + 2 O 2 (g) → 2 SO 2 (g) CS 2 (l) → C(s) + 2 S(s) CS 2 (l) + 2 O 2 (g) → CO 2 (g) + 2 SO 2 (g) A +2C + -B -393.5 - 593.6 - 87.9 = -1075.0 kJ/mol

48. 48 myard Find the enthalpy of reaction for this reaction 4PCl 5 (g) = P 4 (s) + 10Cl 2 (g) A. 4PCl 3 (g) = P 4 (s) + 6Cl 2 (g) H=1084KJ B. PCl 3 (g) + Cl 2 (g)=>PCl 5 (g) H=-111KJ

49. 49 4PCl 3 (g) = P 4 (s) + 6Cl 2 (g) 4PCl 5 (g) = 4PCl 3 (g) + 4Cl 2 (g) 4PCl 5 (g) = P 4 (s) + 10Cl 2 (g) A + -4B 1084 + 444 = 1528KJ

50. 50 myard Calculate the enthalpy for this reaction: 2C(s) + H 2 (g) ---> C 2 H 2 (g)ΔH° =kJ A. C 2 H 2 (g) + (5/2)O 2 (g) ---> 2CO 2 (g) + H2O(l)ΔH° = -1299.5 kJ B. C(s) + O 2 (g) ---> CO 2 (g)ΔH° = - 393.5 kJ C. H 2 (g) + (1/2)O 2 (g) ---> H2O(l)ΔH° = -285.8 kJ

51. 51 2CO 2 (g) + H2O(l) ---> C 2 H 2 (g) + (5/2)O 2 (g) 2C(s) + 2O 2 (g) ---> 2CO 2 (g) H2(g) + (1/2)O 2 (g) ---> H2O(l) 2C(s) + H 2 (g) ---> C 2 H 2 (g) -A + 2B + C 1299 + -787 + -285.8 = 226.7