1. States of Matter & Thermochemistry

2. 10.1 Kinetic Theory Kinetic theory of gases- valid only at extremely low density 1.A gas is composed of particles, usually molecules or atoms Hard spheres Insignificant volume Far from each other 2. The particles in a gas move rapidly in constant motion. 3. All collisions are perfectly elastic.

3. Kinetic Energy (KE) When a gas is heated, it absorbs thermal energy. Some of this is converted to KE to increase the motion of particles. As a substance moves from solid to gas the KE increases Average KE of a gas is proportional to the Kelvin temperature. Particles at 200K have twice the KE of particles at 100K. The Kelvin temperature scale is used because 0K (absolute zero) is the temperature at which all motion stops.

4. Gas Pressure Atmospheric pressure- –results from the collisions of air molecules with objects. Are atmospheric pressures higher or lower? Why? Lower because higher elevation. Higher because lower elevation. More air is above it.

5. Barometer- instrument used to measure atmospheric pressure. SI unit of pressure pascal Standard atmospheric pressure = 101.3 kilopascals (kPa) Standard atmospheric pressure = 1 atm 1 atm = 101.3 kPa 1 atm 101.3 kPa STP = 1 atm and 0 o C

6. 10.2 Nature of Liquids Liquids and solids are condensed states of matter Liquids and gases flow. Particles of a liquid are held together by weak attractive forces.

7. Vaporization process by which a liquid changes to a gas or vapor.

8. Evaporation Evaporation- when gas molecules escape from the surface of a liquid. Vaporization in open container As temperature increases more molecules achieve enough energy to escape Evaporation is a cooling process (ex. Sweating) Highest energy (hottest) particles escape, leaving the cooler ones behind

9. Molecules at the surface are attracted to fewer other molecules than are molecules in the interior. Remember: must have enough energy to break the attractive forces for the liquid to vaporize. So, molecules with fewer attractive forces will vaporize into a gas first

10. Vapor Pressure (VP) Pressure produced in a closed container by vapor particles colliding with the walls As temperature increases, vp increases.

11. Dynamic Equilibrium  Rate of evaporation = rate of condensation

12. Boiling Point (BP) The temperature at which the vp of the liquid equals the external or atmospheric pressure. Normal bp = bp at 1 atm Mountains = low bp (Denver’s bp is 95ºC) Pressure cooker = high bp (bp above 100ºC) The temperature of a liquid never exceeds its bp

13. 10.3 Nature of Solids Melting Point (MP) –The temperature at which a solid changes into a liquid –Vibrations are strong enough to overcome attractions forces –Melting point = freezing point –Ionic solids = high melting point (strong attractive forces) –Molecular solids = low melting point (weak attractive forces)

14. Crystal Most solids are crystalline Atoms, ions, or molecules are arranged in an orderly, repeating, 3-D pattern called a crystal lattice. Unit cell- smallest group of particles within a crystal that retains the geometric shape of the crystal Allotropes- are two or more different forms of the same element in the same physical state. Carbon has four allotropes.

15. These 4 Carbon allotropes are: Diamond –Tightly packed –Dense –Hard –Crystalline Graphite –Loosely packed –Low density –Soft –Crystalline Fullerenes –Includes Buckminsterfullerene (Buckyballs), a 60 carbon sphere Soot –Amorphous carbon (non-crystalline form) Other examples of Amorphous solids : rubber, plastic, asphalt, glass

16. 10.4 Phase Changes Change of physical state Melting, freezing, evaporation, condensation, sublimation, deposition The temperature of a substance does not change during a phase change. Sublimation- Change of a solid to a gas without going through the liquid phase EX: ice cube “shrinkage” freeze drying, freezer burn Deposition is the opposite

18. 11.1 Thermochemistry Thermochemistry- is concerned with the heat changes that occur during chemical reactions. Energy – The capacity for doing work. It comes in many forms. Kinetic Energy: motion of particles Potential Energy: stored energy EXAMPLE: When gasoline burns, the potential energy stored in its chemical bonds is released as KE to do work, such as moving car.

19. Heat (q)- energy that transfers from one object to another because of a temperature difference between them. Heat always flows from a warmer object to a cooler object. System – the part of the universe on which you focus your attention Surroundings – everything else in the universe Law of Conservation of Energy – in any chemical or physical process, energy is neither created nor destroyed. EXAMPLE: Heat may be lost by the system, but it is not destroyed. It is transferred to the surroundings.

20. Temp (ºC) 1. Solid 3. Liquid 5. Gas 2. Heat of Fusion 4. Heat of Vaporization Melting Point Boiling Point E n d o t h e r m i c ( h e a t i n g ) E x o t h e r m i c ( C o o l i n g )

21. Endothermic process- system absorbs heat from the surroundings Exothermic process – system releases heat to the surroundings Endothermic or Exothermic process? –Evaporating alcohol: –Leaves burning: –Boiling water: –Water cooling: –Melting ice: –Freezing water:

22. Energy units: calorie (cal), Joule (J) calorie – quantity of heat needed to raise the temperature of 1 g of water 1ºC A food Calorie is used in nutrition and is capitalized. 1 Calorie = 1000 cal = 1 kcal Example: If a label on a candy bar indicated it contains 180 Calories, that is really 180 kcal, or 180,000 calories! If “burned” the sugar and fat in the candy bar release 180,000 cal of energy.

23. Heat capacity- amount of heat needed to increase the temperature of an object 1 o C. It depends on mass and composition. EXAMPLE: It takes more heat to increase the temperature of a large pot of water than a small cup of water. It takes more heat to raise the temperature of water than metal.

24. Specific Heat (C)- the amount of heat it takes to raise the temperature of 1g of substance 1 o C. It can be calculated: C = q m∆T Specific heat units = J/g o C or cal/g o C A low specific heat is matter that loses or gains heat quickly (Tiles on space shuttle) A high specific heat is matter that loses or gains heat slowly (water) Water has a uniquely high specific heat compared to other substances.

25. To calculate the heat energy required for a temperature change, use the following formula: q = mC∆T –q = energy –m = mass –C = specific heat –∆T = change in temperature EXAMPLE: How much energy is required to heat an iron nail with a mass of 7.0g from 25 o C until it becomes red hot at 750 o C? equation: q = mC∆T First, write out your givens.

26. EXAMPLE: How much energy is required to heat an iron nail with a mass of 7.0g from 25 o C until it becomes red hot at 750 o C? (C iron = 0.40 J/g o C) q = mC∆T q = x m = 7.0 g C = 0.40 J/g o C (found in specific heat table) ∆T = 750 o C - 25 o C = 725ºC q = (7.0 g)(0.40 J/g o C)(725ºC) q = 2030J

27. If 5750 J of energy is added to a 455g piece of glass at 24.0ºC, what is the final temperature of the glass? (c glass = 0.50 J/gºC) q = mC∆T q = 5750 J m = 455 g C = 0.50 J/gºC (found in specific heat table) T i = 24.0 ºCT f = x ∆T = x – 24 o C 5750 J = (0.50 J/gºC)(455 g)(x-24 o C) 5750 = 227.5 (x – 24) 25.27 = x – 24 x = 49.3 ºC

28. A 30.0g sample of an unknown metal is heated from 22.0 ºC to 59.2 ºC. During the process, 1.00 KJ of energy is absorbed by the metal. What is the specific heat of the metal? q = mC∆T q = 1.00 KJ (must change to J… 1 KJ = 1000 J) m = 30.0 g C = x T i = 22.0 ºC T f = 59.2 ºC ∆T = 59.2 ºC - 22.0 ºC = 37.2 ºC 1000 J = (30.0 g)(x)(37.2 ºC) x = 0.896 J/gºC

29. If it takes 3590 calories to heat up a sample of water by 12.2 ºC, what is the mass of the water? (C water = 1.00 cal/gºC) q = mC∆T q = 3590 calories m = x C = 1.00 cal/gºC ∆T = 12.2 ºC 3590 cal = (x)(1.00 cal/gºC)(12.2 ºC) x = 294 g Notice, we used “cal” instead of “J” to match the units of q.

30. 11.2 Calorimetry is the measurement of heat change for chemical and physical processes. Heat released by system = heat absorbed by surroundings Calorimeter- insulated device used to measure the absorption or release heat in a chemical or physical process.

31. EXAMPLE: A 25g sample of a metal at 75.0 ºC is placed in a calorimeter containing 25g of H 2 O at 20 ºC. The temperature stopped changing at 29.4 ºC. What is the specific heat of the metal? 2 Steps to solve: 1. First, solve for q for the water. Then, use the value of q for the metal. 2. Solve for specific heat of the metal. You will have 2 sets of givens. Remember the final temperature and energy for both will be the same.

32. A 25g sample of a metal at 75.0 ºC is placed in a calorimeter containing 25g of H 2 O at 20 ºC. The temperature stopped changing at 29.4 ºC. What is the specific heat of the metal? H 2 O metal q = m = 25 g C = 1.00 cal/gºC C = T i = 20 ºC T i = 75.0 ºC T f = 29.4 ºC ∆T = 9.4 ºC ∆T = 45.6 ºC H 2 O: q = (25g)(1.00 cal/gºC)(9.4ºC) = 235 cal Metal: 235 cal = (25g)(x)(45.6ºC) x = 0.21 cal/gºC 1. First, solve for q for the water. Then, use the value of q for the metal. 2. Solve for specific heat of the metal using same q.

33. Enthalphy (H) Enthalpy (H) is the amount of heat a system contains. Enthalpy Change: (same as energy) ∆H = q= C x m x ∆T Exothermic reaction has - ∆H Endothermic reaction has + ∆H Thermochemical equation includes heat changes. Physical state must be included!!!

34. EXAMPLE –CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) + 890KJ ∆H= -890 KJ (energy is released because it is a product) Exothermic –2H 2 0 + 241.8 KJ  2H 2 (g) + O 2 (g) ∆H= 241.8 KJ (energy is required because it is a reactant) Endothermic Exothermic – energy is a product Endothermic – energy is a reactant ∆H is also called the heat of reaction.

35. 11.3 Heat in Change of State The specific heat of water is 1.0cal/gºC The specific heat ice and steam is 0.5cal/gºC q = mc∆T Heat of fusion - heat required to melt 1 gram of solid Heat of solidification - heat released as 1 gram of liquid freezes Heat of fusion of water = 80 cal/g = heat of solidification q = mH f

36. Heat of vaporization- heat required to vaporize 1 gram of a liquid Heat of condensation- heat released as 1 gram of a gas condenses to a liquid. Heat of vaporization for water = 540 cal/g = Heat of Condensation q = mH v q= mass x heat of phase change *Temperature is constant during a phase change

37. Heating/Cooling Curve shows energy changes. All matter follows curve when energy is added or lost.

38. Temp (ºC) 1. Solid 3. Liquid 5. Gas 2. Heat of Fusion 4. Heat of Vaporization Melting Point Boiling Point E n d o t h e r m i c ( h e a t i n g ) E x o t h e r m i c ( C o o l i n g )

39. -H-H-H-Horizontal portions of curve indicate a physical state change. Notice temperature remains constant. However, there is a change in particle position resulting in a change in potential energy. S- SS- Slope portions show temperature change which indicates a change in kinetic energy as well.

40. EXAMPLE: How much heat, in calories, is needed to melt 150g of ice at 0 ºC? (H f =80cal/g) q = mH f q = ? m = 150 g H f = 80 cal/g q = (150 g)(80 cal/g) q = 12,000 cal 0ºC 100ºC 1

41. EXAMPLE: How much heat, is calories, is needed to heat the liquid water in the above problem to 20. ºC? (C = 1.00 cal/gºC) q = mC∆T q = ? m = 150. g C = 1.00 cal/gºC T i = 0 ºC T f = 20. ºC ∆T = 20. ºC q = (150. g)( 1.00 cal/gºC )( 20. ºC) q = 3000 cal 0ºC 100ºC 1

42. EXAMPLE: A 50g sample of ice is held at -10 ºC. Will 270 cal of heat be sufficient to raise the temperature of the ice to 0 ºC ? q = mC∆T q = ? (compare to 270 cal) m = 50. g C = 0.50 cal/gºC T i = -10 ºC T f = 0. ºC ∆T = 10. ºC q = (50. g)(0.50 cal/gºC )(1 0. ºC) q = 250 cal 270 cal enough? YES 0ºC 100ºC 1 b/c ICE

43. EXAMPLE: How many calories are released when 36g of steam at 100 ºC condenses to water at 100 ºC? q = mH v q = ? m = 36 g H v = 540 cal/g q = (36 g)(540 cal/g) q = 19,000 cal 0ºC 100ºC 1

44. EXAMPLE: How many calories are needed to convert 5.0g of ice at -15ºC to steam at 130ºC? (five problems) 1. q 1 = mC∆T q 1 = ? m = 5.0 g C = 0.50 cal/gºC T i = -15ºC T f = 0.ºC ∆T = 15.ºC q 1 =(5.0g)(0.50 cal/gºC)(15ºC) q 1 =37.5 cal 2.q 2 = mH f q 2 = ? m = 5.0 g H f = 80 cal/g q 2 = (5.0 g)(80 cal/g) q 2 = 400 cal 0ºC 100ºC 21 3 4 5 2 out of 5 steps completed…

45. EXAMPLE: How many calories are needed to convert 5.0g of ice at -15ºC to steam at 130ºC? (five problems) 3. q 3 = mC∆T q 3 = ? m = 5.0 g C = 1.0 cal/gºC T = 100ºC - 0.ºC = 100ºC q 3 = (5.0g)(1.0 cal/gºC)(100ºC) q 3 = 500 cal 4. q 4 = mH v q 4 = ? m = 5.0 g H v = 540 cal/g q 4 = (5.0 g)(540 cal/g) 0ºC 100ºC 2 1 3 4 5 q 4 = 2700 cal ALMOST THERE!!!

46. 5. q 5 = mC∆T q 5 = ? m = 5.0 g C = 0.50 cal/gºC T i = 100ºC T f = 130ºC ∆T = 30.ºC q 5 =(5.0g)(0.50 cal/gºC)(30.ºC) q 5 =75 cal Last step: ADD ALL CALCULATIONS TOGETHER. q 1 + q 2 + q 3 + q 4 + q 5 = q total 37.5 cal + 400 cal + 500 cal + 2700 cal + 75 cal = 3713 cal 100ºC 0ºC 1 4 5 3 2

47. 11.4 Standard heat of formation of a compound (∆H f º) ∆H f º of a free element in its standard state is zero. This is another way to calculate ∆H for a reaction. ∆H = ∆H prod - ∆H react -use table of “Standard Heats of Formation”.

48. Calculate ∆H for the following reaction: CaCO 3 (s)  CaO(s) + CO 2 (g) *First, make sure equations are balanced. *You will multiply the ∆H by the coefficient. (all coefficients for this problem are 1, so do not need to worry about that) ∆H = ∆H prod - ∆H react [ (∆H CaO ) + (∆H CO2 ) ] – [∆H CaCO3 ] [(-393.5 KJ) + (-635.1 KJ)] – [-1207.0 KJ]= 178.4 KJ, endothermic b/c ∆H > 0

49. Calculate the heat of reaction for the following reaction: 2H 2 (g) + O 2 (g)  2H 2 O (g) ∆H = ∆H prod - ∆H react [2(∆H H2O )] – [ 2(∆H H2 ) + (∆H O2 )] [2(-241.8)] – [ 0 + 0] = -483.6 KJ; exothermic b/c ∆H < 0 2 is from the coefficient 0 b/c elements in standard state (even diatomics) always have ∆H = 0