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1. Chemistry I Honors—Unit 7:Thermodynamics/Kinetics/Equilibrium

2. Objective #1:Temperature vs. Heat & Heat Calculations I. Temperature vs. Heat: • Temperature is a measure of the averagekinetic energy in a sample of matter • Heat is energy transferred between samples of matter because of differences in their temperature

3. Objectives #1:Temperature vs. Heat & Heat Calculations II. Heat Calculations • Converting between Celsius and Kelvin temperature scales: K = oC + 273 • Review of formula and units: Q = (m) (c) ∆(t) “Q” = heat in Joules or calories “m” = mass in grams “c” = specific heat in J/g.K or J/g.C “∆t” = change in temperature Remember that a change in temp on Celsius scale EQUALS a change on the Kelvin scale!!!!

4. Objectives #1:Temperature vs. Heat & Heat Calculations • Examples—Its PLUG and CHUG time!!!  Q = (m) (c) ∆(t)

5. Example #1 Calculate the temperature change that will occur when 5.00 grams of water absorbs 500. J of heat energy, given that the specific heat capacity of water is 4.18 J/gK or oC. Q = mc∆t ∆t = Q / mc = 500. J / (5.00 g) (4.18 J/g.oC) = 23.9 oC

6. Calculate the temperature change using the above parameters but substituting copper for water. The specific heat capacity of copper is .385 J/g.oC. Q = mc∆t ∆t = Q / mc = 500. J / (5.00 g) (.385 J/g.oC) = 260. oC What do these answers tell us?

7. Example #2 The temperature of a 74 g sample of material increases from 15oC to 45oC when it absorbs 2.0 kJ of energy as heat. What is the specific heat capacity of this material? Q = mc∆t c = Q / m∆t = 2.0 X 103 J / (74 g) (30oC) = .901 J / g.oC

8. Example #3 If 1000 kJ of heat are added to 50 ml of liquid mercury at 300. K, what will the final temperature of the mercury be? The density of mercury is 13.6 g / ml and the specific heat capacity of mercury is 140 J/g.K. Q = mc∆t ∆t = Q / mc = 1 X 106 J / (13.6 g/ml) (50 ml) (140 J/g.K) = 10.5 K Tf = Ti + ∆t 300. K + 10.5 K = 310.5 K

9. Objective #2:Heating & Cooling Curves, Energy Changes & Phase Changes (see curve in lecture guide) III. Heating / Cooling Curve Terminology • Specific Heat – energy required to change temperature of material 1o C • Energy Changes During Phase Changes • Heat of Fusion – energy required to melt substance at melting point 2. Heat of Solidification – energy released to freeze substance at freezing point

10. Objectives #2: Heating and Cooling Curves, Energy Changes & Phase Changes • Heat of Vaporization – energy required to boil substance at its BP • Heat of Condensation – energy released to condense substance at its condensation point • Kinetic Energy – energy of motion; increases or decreases where temp. changes; remains the same during phase change • Potential Energy – stored energy; increases or decreases during phase changes; remains the same where temp. changes

11. Objectives #2:Heating/Cooling Curves, Energy Changes & Phase Changes C. Key Temperature Points 1. Melting Point 2. Freezing Point 3. Boiling Point 4. Condensation Point D. Phase Changes 1. Melting – solid to liquid 2. Freezing – liquid to solid 3. Boiling – liquid to gas 4. Condensing – gas to liquid Occur at same temp ! Occur at same temp !

12. Objectives #2:Heating/Cooling Curves, Energy Changes & Phase Changes • E. Example Problems: • To change Temperature, use Q = (m)(c)(Dt) • To change Phase, use DH = (# moles)(Molar heat) • A mole is a standard measure of the amount of a chemical need to have 6.02 x 1023 particles • Mole is used to compare different chemicals • Molar mass = # grams per mole; calculated from the formula of the compound Time to Plug & Chug—

13. 1. Calculate the energy required to melt 8.6 g of ice at its melting point. The molar heat of fusion of ice is 6.009 kJ/mole. ∆H = # moles of material X molar heat = (8.6 g X 1 mole / 18.0 g) X 6.009 kJ/mole = 2.9 kJ

14. 2. Calculate the mass of liquid water required to absorb 5.23 X 104 kJ of energy upon boiling. The molar heat of vaporization for water is 40.67 kJ/mole. ∆H = # moles of material X molar heat # moles of material = ∆H / molar heat = 5.23 X 104 kJ / 40.67 kJ / mole = 1290 moles 1290 moles H2O X 18.0 g / 1 mole = 23220 g

15. 3. Calculate the energy required to convert 15.0 g of liquid water at 25oC to steam at 100oC. The specific heat of liquid water is 4.18 J/g.C and the heat of vaporization of water is 40.67 kJ/mole. *warm water from 25oC to 100oC: Q = mc∆t = (15.0 g) (4.18 J/g.C) (75oC) = 4702.5 J = 4.7025 kJ

16. *boil ∆H = moles X molar heat = (15.0 g X 1 mole / 18.0 g) X 40.67 kJ/mole = 33.9 kJ Total = 4.0725 kJ + 33.9 kJ = 37.98 kJ

17. 4. Given the information below (see lecture guide), calculate the amount of energy required to convert 5.00 g of water at -25oC to steam at 110oC. (Remember a change in Celsius temp=an equivalent change in Kelvin temperature) *heat water from -25oC to 0oC: Q = mc∆t = (5.00 g) (2.09 J/g K) (25 K) = 261 J = .261 kJ

18. *melting: ∆H = moles X molar heat = (5.00 g X 1 mole / 18.0 g) X 6.01 kJ/mole = 1.67 kJ *warm water from 0oC to 100oC: Q = mc∆t = (5.00 g) (4.18 J g K) (100 K) = 2090 J = 2.090 kJ

19. *boil: ∆H = moles X molar heat = (5.00 g X 1 mole / 18.0 g) X 40.67 kJ/mole = 11.30 kJ *warm from 100oC to 125oC: Q = mc∆t = (5.00 g) (1.84 J /g K) (25 K) = 230 J = .230 kJ Total = .261 kJ + 1.67 kJ + 2.090 kJ + 11.30 kJ + .230 kJ = 15.55 kJ

20. Objectives #3-8:Energy, Entropy, and Reaction Rate • Chemical Reactions and Bond Energy: • All chemical reactions involve a change in energy • Energy can be “lost” to the environment by the system (the chemical reaction) and vice-versa • Bond energy is the energy required to break a bond which equals the energy released when a bond is formed Example: H – H + I-I → 2 HI 436 kJ 151 kJ 598 kJ

21. Objectives #3-8: Energy, Entropy, and Reaction Rate • Endothermic reactions absorbenergy while exothermic reactions release energy • Enthalpy and Entropy: • Enthalpy (∆H) is the energy difference between reactants and products in a chemical reaction • If the energy of the reactants is higher than the energy of the products, an exothermicreaction occurs • If the energy of the products is higher than the energy of the reactants, an endothermicreaction occurs

22. Objectives #3-8: Energy, Entropy, and Reaction Rate • Entropy (∆S) is a measure of the disorder of a system—there is more entropy if more molecules are formed or particles dissociate • Nature tends to favor processes that release energy and increase entropy Example: H2O(l) --› H2(g) + O2(g) the formation of gas is a more disordered state

23. Objectives #3-8: Energy, Entropy, and Reaction Rate • Collision theory, the Activated Complex, and Activation Energy: • according to the collision theory, chemical • reactions only occur if the reacting particles • have sufficient energy and a favorable • orientation

24. Objectives #3-8: Energy, Entropy, and Reaction Rate • The minimum energy required for the interacting particles to react is called the activation energy • If a collision is successful, an intermediate product forms, the activated complex, for a brief interval of time and then the final stable product forms • Example diagram: (see lecture guide) • Interpreting energy profile diagrams:

25. Factors affecting reaction rate:

26. Objective #9 Equilibrium and LeChatelier’s Principle *reversible reactions: products of a chemical reaction can re-form reactants *equilibrium – a dynamic state: rate of a forward reaction equals the rate of the reverse reaction *writing equilibrium expressions: EA + FB → GC + HD K = [C]G [D]H/[A]E[B]F only gases and aqueous ions are subject to equilibrium changes and therefore only these substances can be part of the expression (examples)

27. Objective #9 Equilibrium and LeChatelier’s Principle *the meaning of the equilibrium constant “K” a small K indicates that the reactants are favored a large K indicates that the products are favored *factors affecting equilibrium: pressure (only affects gases) concentration (does not affect liquids and solids) Temperature (effect depends on whether reaction is endothermic or exothermic) Common Ion Effect (requires the addition of a substance containing one of the ions in the reactant)

28. Objective #9 Equilibrium and LeChatelier’s Principle *LeChatelier’s Principle when a reaction at equilibrium is stressed, the reaction will respond in such a way to relieve the stress (examples)