Kinetic Theory & Thermochemistry Explained

1. Ch. 10 & 11 States of Matter & Thermochemistry

2. 10.1 Kinetic Theory Kinetic theory of gases- valid only at extremely low density • 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. 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.

5. 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.

6. Endothermic (heating) 4. Heat of Vaporization Boiling Point 5. Gas Temp (ºC) 2. Heat of Fusion 3. Liquid Exothermic (Cooling) Melting Point 1. Solid Heating Curve

7. 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:

8. 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.

9. Heat capacity- amount of heat needed to increase the temperature of an object 1oC. 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.

10. Specific Heat (C)- the amount of heat it takes to raise the temperature of 1g of substance 1oC. It can be calculated: • C = q m∆T • Specific heat units = J/goC or cal/goC • 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.

11. 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 25oC until it becomes red hot at 750oC? equation: q = mC∆T First, write out your givens.

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

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

14. 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 Ti = 22.0 ºC Tf = 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

15. Notice, we used “cal” instead of “J” to match the units of q. If it takes 3590 calories to heat up a sample of water by 12.2 ºC, what is the mass of the water? (Cwater = 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

16. 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.

17. EXAMPLE: A 25g sample of a metal at 75.0 ºC is placed in a calorimeter containing 25g of H2O 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.

18. A 25g sample of a metal at 75.0 ºC is placed in a calorimeter containing 25g of H2O at 20 ºC. The temperature stopped changing at 29.4 ºC. What is the specific heat of the metal? H2O metal q = q = m = 25 g m = 25 g C = 1.00 cal/gºC C = Ti = 20 ºC Ti = 75.0 ºC Tf = 29.4 ºC Tf = 29.4 ºC ∆T = 9.4 ºC ∆T = 45.6 ºC 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. H2O: q = (25g)(1.00 cal/gºC)(9.4ºC) = 235 cal Metal: 235 cal = (25g)(x)(45.6ºC) x = 0.21 cal/gºC

19. 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!!!

20. EXAMPLE • CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) + 890KJ • ∆H= -890 KJ (energy is released because it is a product) • Exothermic • 2H20 + 241.8 KJ  2H2(g) + O2(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.

21. Endothermic (heating) 4. Heat of Vaporization Boiling Point 5. Gas Temp (ºC) 2. Heat of Fusion 3. Liquid Exothermic (Cooling) Melting Point 1. Solid Heating Curve

22. 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. • Slope portions show temperature change which indicates a change in kinetic energy as well.

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

24. 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 Ti = 0 ºC Tf = 20. ºC ∆T = 20. ºC q = (150. g)(1.00 cal/gºC )(20. ºC) q = 3000 cal 100ºC 1 0ºC

25. 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 Ti = -10 ºC Tf = 0. ºC ∆T = 10. ºC q = (50. g)(0.50 cal/gºC )(10. ºC) q = 250 cal 270 cal enough? YES 100ºC b/c ICE 1 0ºC

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

27. EXAMPLE: How many calories are needed to convert 5.0g of ice at -15ºC to steam at 130ºC? (five problems) 1. q1 = mC∆T q1 = ? m = 5.0 g C = 0.50 cal/gºC Ti = -15ºC Tf = 0.ºC ∆T = 15.ºC q1 =(5.0g)(0.50 cal/gºC)(15ºC) q1 =37.5 cal • q2 = mHf q2 = ? m = 5.0 g Hf = 80 cal/g q2 = (5.0 g)(80 cal/g) q2 = 400 cal 5 4 100ºC 3 1 2 0ºC 2 out of 5 steps completed…

28. 5 EXAMPLE: How many calories are needed to convert 5.0g of ice at -15ºC to steam at 130ºC? (five problems) 3. q3 = mC∆T q3 = ? m = 5.0 g C = 1.0 cal/gºC T = 100ºC - 0.ºC = 100ºC q3 = (5.0g)(1.0 cal/gºC)(100ºC) q3 = 500 cal 4. q4 = mHv q4 = ? m = 5.0 g Hv = 540 cal/g q4 = (5.0 g)(540 cal/g) 4 100ºC 3 2 1 0ºC ALMOST THERE!!! q4 = 2700 cal

29. 5 4 100ºC 5. q5 = mC∆T q5 = ? m = 5.0 g C = 0.50 cal/gºC Ti = 100ºC Tf = 130ºC ∆T = 30.ºC q5 =(5.0g)(0.50 cal/gºC)(30.ºC) q5 =75 cal Last step: ADD ALL CALCULATIONS TOGETHER. q1 + q2 + q3 + q4 + q5 = qtotal 37.5 cal + 400 cal + 500 cal + 2700 cal + 75 cal = 3713 cal 3 2 1 0ºC

30. 11.4 • Standard heat of formation of a compound (∆Hfº) • ∆Hfº of a free element in its standard state is zero. • This is another way to calculate ∆H for a reaction. • ∆H = ∆Hprod - ∆Hreact -use table of “Standard Heats of Formation”.

31. Calculate ∆H for the following reaction: CaCO3(s)  CaO(s) + CO2(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 = ∆Hprod - ∆Hreact [ (∆HCaO) + (∆HCO2) ] – [∆HCaCO3] [(-393.5 KJ) + (-635.1 KJ)] – [-1207.0 KJ]= 178.4 KJ, endothermic b/c ∆H > 0

32. Calculate the heat of reaction for the following reaction: 2H2(g) + O2(g)  2H2O (g) ∆H = ∆Hprod - ∆Hreact [2(∆HH2O)] – [ 2(∆HH2) + (∆HO2)] [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