Energy

1. Energy • The ability to do work or produce heat • Potential- Stored energy • Energy stored in chemical bonds • Kinetic- Energy of movement • Energy of moving molecules

2. Law of Conservation of Energy • Energy can not be created or destroyed, it can only be converted from one form to another.

3. Chemical Potential Energy • Chemical bonds of a compound contain energy • That energy can be converted to do work or is converted to heat

4. Heat • Heat flows from warm objects to cooler objects. • The two objects will reach equilibrium and record the same temperature

5. Measuring Heat • Calorie- The amount of energy required to raise 1 g of water 1oC • 1 food Calorie = 1000 calories • Joule- The unit of measurement for heat energy • 1 Joule = .2390 calories

6. Problem • How Joules of heat would be obtained from a 364 nutritional Calorie hamburger? • See chart on p. 491 for conversions

7. Answer • 64 Cal x 1000 cal x 1 J = 1523012 J 1 Cal .2390 cal

8. Specific Heat • The amount of energy required to raise 1 gram of any substance 1oC • Specific heat determines how fast or slow that substances gain or lose heat.

9. Specific Heat and Water • Water has a high specific heat • It gains and loses heat energy very slowly • This is why bodies of water stay cool long into spring and warm long into fall

10. Calculating Heat Change • q = c x m x T • q = Heat gain or loss- J • c = Specific heat - J/(g . oC) • m = mass g • T = Change in temperature- oC

11. Calculating Heat Change • How much heat is required to raise 2.3 kg of iron 23 oC? • q = c x m x T • q = .449 J/(g.oC) x 2300g x 23oC • q = 23752.1 J

12. Calculating Heat Change • If there is more than one substance you will need to calculate the heat change separately • Ex. Water in an iron pot would require calculations for both iron and water

13. Determining Specific Heat • 1. Heat the unknown substance to a set temperature • 2. Place the substance in a measured mass of water • 3. Measure the temperature change when equilibrium is reached

14. Determining Specific Heat • 4. Calculate the amount of energy gained by the water. This equals the amount of energy lost by the substance • Use the heat value to solve for the specific heat of the substance

15. Calorimeter Lab • Calorimeter- A device used to measure the amount of heat gained or lost by a substance • See handout for instructions

16. Thermochemistry • The study of heat changes involved in chemical reactions • Reactions are either exothermic (release energy) or endothermic (absorb energy)

17. The Universe • A system plus its surroundings Surroundings The System

18. Enthalpy (H) • The heat content of a substance at a constant pressure • There is no way to measure the total enthalpy of a substance but you can calculate the change in enthalpy for a reaction

19. Enthalpy Changes • To calculate enthalpy change use the following equation • Hrxn = Hproduct - Hreactants • If Hrxn is positive the reaction is endothermic, if negative the reaction is exothermic

20. Enthalpy change • Exothermic reaction E n t h a l p y 4Fe(s) + 3O2(g) Reactants H= -1625 Fe2O3(s) Product

21. Enthalpy change • Endothermic reaction E n t h a l p y NH4+(aq) + NO3-(aq) Product H= 27 kJ NH4NO3(s) Reactant

22. Heat Gain and Enthalpy • q = c x m x T • H = Hproduct - Hreactants • q = H since heat gain or loss is the same as enthalpy change at constant pressure

23. Writing Thermochemical Equations • 1. Write the equation • 2. Write the H value to the right of the equation

24. Writing Thermochemical Equations • 3. If the reaction is exothermic the H value is negative, if endothermic, positive

25. Enthalpy of Combustion • The amount of energy released when one mole of a substance is burned • Hocomb = Heat of combustion

26. Changes of State • Hvap = Heat of vaporization • Hcond = Heat of condensation • Hfus = Heat of fusion • Hsolid = Heat of solid

27. Changes of state • Changing phase from a solid to a liquid, liquid to gas or vice versa requires a change in energy

28. Changes of state • The amount of energy required to convert one mole of a substance from a liquid to a gas or from a liquid to a solid

29. Equations for phase changes • H2O(l) --> H2O(g) Hvap=40.7 kJ • H2O(g) --> H2O(l) Hvap=-40.7 kJ • H2O(s) --> H2O(l) Hfus = 6.01 kJ • H2O(l) --> H2O(s) Hvap=-6.01 kJ

30. Assignment • P. 500 14, 15, 16, 17 • p. 504 20-22

31. Lab- Energy Change • P. 503 • Use Microsoft Excel to generate graph

32. Hess’s Law • States that if you can add 2 or more equations to produce a final equation for a reaction, then the sum of the enthalpy changes for the individual reactions = the enthalpy for the final reaction.

33. Hess’s Law • What does this mean? • You can predict enthalpy changes for reactions that can not be observed directly. • Ex. A reaction that takes several thousand years. Converting carbon to a diamond.

34. Applying Hess’s Law • What is the enthalpy change for the following reaction? • 2S(s) + 3O2(g)  2SO3(g)

35. Applying Hess’s Law • We know the following equations • S(s) + O2(g) SO2(g) • H=-297 kJ • 2SO3(g) 2SO2(g) + O2(g) H=198 kJ

36. Applying Hess’s Law • 1. You need to change the coefficients in the two equations to match the molar amounts in the original equation

37. Applying Hess’s Law • 2S(s) + 3O2(g)  2SO3(g) • S(s) + O2(g) SO2(g) • H=-297 kJ • Since there are 2 mol of S in the original equation the second equation must by multiplied by a factor of 2.

38. Applying Hess’s Law • 2(S(s) + O2(g) SO2(g)H=-297 kJ) • 2S(s) + 2O2(g) 2SO2(g) • H=2(-297 kJ) = -594 kJ • Now we have the amount of heat required for the reaction of 2 moles of S

39. Applying Hess’s Law • Since SO3 is the product in the original reaction we have to reverse the second reaction from 2SO3(g) 2SO2(g) + O2(g) H=198 kJ to • 2SO2(g) + O2(g)  2SO3(g) H= -198 kJ

40. Applying Hess’s Law • 2S(s) + 2O2(g)  2SO2(g) H= -594 kJ • 2SO2(g) + O2(g) 2SO3(g) H= -198 kJ • 2S+2O2+2SO2+ O2  2SO2+2SO3 H= -792kJ • 2S + 3O2 2SO3H= -792 kJ • So the original reaction evolves 792 kJ

41. Assignment • P. 508 # 28 and 29

42. Enthalpy of formation • Elements are free atoms. It takes no heat to form them. They are in the lowest energy state possible. • Elements are assigned a value of 0 kJ for Hf since there is no energy needed to create them because they are the simplest substances on Earth

43. Enthalpy of formation • Compounds are formed by combining elements. Energy is required to combine elements, therefore the H has a value greater than 0 if elements form a compound. • This is an endothermic reaction.

44. Enthalpy of formation • Hrxn = Hf(product) - Hf(reactant) • Add up all of the Hf products and subtract the sum of the Hf reactants • This equals the net H of the reaction

45. Reaction Spontaneity • Spontaneous Process- a physical or chemical change that occurs with no outside intervention • Most exothermic reactions are spontaneous and most endothermic reactions are not spontaneous.

46. Reaction Spontaneity • Not all exothermic reactions are spontaneous and not all endothermic reactions are not spontaneous • Ice melting is spontaneous but requires energy to occur. This makes in endothermic and spontaneous

47. Reaction Spontaneity • What accounts for an endothermic process being spontaneous? • Entropy

48. Entropy • The measure of the randomness and disorder of a system • Substances are more likely to exist in a high state of randomness than in a low state of randomness

49. Entropy • Substances are more likely to have high entropy than low entropy • Spontaneous processes always proceed in such a way that the entropy of the universe increases

50. Entropy • Think of a deck of cards • What are the odds that well shuffled cards will come out in order?