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

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. II. Bomb Calorimeter Calculations A special type of calorimeter called a bomb calorimeter is used to determine the amount of heat released during a combustion reaction. A small sample of the substance to be combusted is placed in a small sealed cup containing oxygen. The cup is then placed in a “bomb” containing water and the temperature change of the water is determined

10. q = -CcalX ∆t Example 4: When 4.00 g of CH6N2 is combusted in a bomb calorimeter (in other words it goes BOOM!), the temperature of the calorimeter increases from 25.00oC to 39.50oC. If the heat capacity of the calorimeter is 7.794 kJ/oC, determine the heat of reaction in kJ/mole.

11. qrxn = -CcalΔt qrxn = -7.794 kJ/oC X14.5oC = -113 kJ • A mole is a standard measure of the amount of a chemical needed 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

12. moles = 4.00 g CH6N2 X 1 mole / 46.0 g = .0870 moles Heat of Reaction in kJ / mole = -113 kJ / .0870 moles = -1.30 X 103 kJ/mole

13. Objective #2:Heating & Cooling Curves, Energy Changes & Phase Changes (see curve in lecture guide) Phase Diagrams 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

14. 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

15. 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 !

16. 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) Time to Plug & Chug—

17. 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

18. 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 = 23200 g

19. 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

20. *boil ∆H = moles X molar heat = (15.0 g X 1 mole / 18.0 g) X 40.67 kJ/mole = 33.9 kJ Total = 4.7025 kJ + 33.9 kJ = 38.6 kJ

21. 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) = 26 J = .26 kJ

22. *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

23. *boil: ∆H = moles X molar heat = (5.00 g X 1 mole / 18.0 g) X 40.67 kJ/mole = 11.3 kJ *warm from 100oC to 110oC: Q = mc∆t = (5.00 g) (1.84 J /g K) (10 K) = 92 J = .092 kJ Total = .261 kJ + 1.67 kJ + 2.090 kJ + 11.30 kJ + .092 kJ = 15.41 kJ

24. III. Interpreting a Phase Diagram

26. *a phase diagram allows one to determine the phase that a substance is in at a given temperature and pressure *the phase diagram only shows one substance in its various phases *the boundaries between different phase regions represent areas of equilibrium in which the two phase changes are occurring at the same rate; for example at the liquid-gas boundary, molecules of gaseous vapor are moving into the liquid phase while molecules of liquid are moving into the gaseous phase

27. *if a point on the diagram does not fall on any line, only one phase is present Examples: 1. What phase is present at a temperature of 50oC and a pressure of 1 atm? liquid 2. What phase change occurs at 200oC if the pressure is increased? condensation 3. What phase change occurs at a constant pressure of 10 atm if the temperature is increased from -50oC to 50oC? melting 4. What happens to the melting point of this substance as pressure increases? decreases

28. 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 ΔH = -598 kJ + 587 kJ = -11 kJ (net change)

29. 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

30. 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

31. 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

32. 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 and next slide)

33. Energy Diagram Showing the Formation of an Activated Complex

34. Interpreting Energy Profile Diagrams • (on Smart file) • A spontaneous reaction occurs when entropy increases (ΔS is positive) and enthalpy decreases (ΔH is negative)

35. Reaction Rate *Reaction rate involves the relationship between the decrease in the amounts of reactants in a reaction and the increase in the amounts of products produced in a chemical reaction *The amounts in a reaction are often expressed in terms of molarity. Molarity is the number of moles of solute dissolved in one liter of solution. *For example, the rate of increase of oxygen gas produced in the following reaction:

36. 2N2O5 -- 4NO2 + O2 can be expressed as ∆[O2] / ∆t

37. Concentration as a Function of Time at 55oC

38. Example I: Calculate the rate of formation of oxygen gas from 300 s to 400 s. Δ[O2] / Δt = .0049 M - .0040 M / 400 s – 300 s = .0009 M / 100 s= 9 X 10-6 M/s

39. Example II: Calculate the rate of decomposition of dintrogenpentoxide from 300 s to 400 s Δ[NO2] / Δt = .0101 M - .0120 M / 400 s – 300 s = -.0019 / 100 s = 1.9 X 10-5 M/s

40. *If enough data is supplied, the rate at a specific point in time can be calculated instead of the average as is shown above. This more specific rate is referred to as the instantaneous rate. *The rate of a chemical reaction not only depends on the initial concentration of the reactants and products but also on the coefficients in the reaction *The coefficients in a chemical reaction refer the mole ratios of the chemicals involved.

41. *For example in the reaction above: 2 moles of N2O5 decompose to form 4 moles of NO2 and 1 mole O2 *The general rate of the reaction is therefore equal to the rate of consumption of a reactant (expressed as a negative value since its concentration is decreasing) or formation of a product (expressed as a positive value since its concentration is increasing) divided by its coefficient in the balanced equation.

42. Example III: Write the General Rate of Reaction for the above reaction -1/2 (Δ[N2O5] / Δt) = 1/4 (Δ[NO2] / Δt) = (Δ[O2 / Δt)

43. Write the General Rate of Reaction for the reaction below: 4NH3 + 5O2 -- 4NO + 6H2O -1/4 (Δ[NH3] / Δt) = -1/5 (∆[O2] / ∆t) = 1/4 (Δ[NO] / Δt) = 1/6 (Δ[H2O / Δt)

44. Factors affecting reaction rate:

45. Objective #9-11 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)

46. Write the equilibrium expression for the following reaction: 4NH3 + 5O2 -- 4NO + 6H2O (assume all components are gases) K = [NO]4 [H2O]6 / [NH3]4 [O2]5

47. Objective #9-11 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 *calculation of Kc: (based on molarity concentrations)

48. Example I: 2N2O5(g) ---- 4NO2(g) + O2(g) The following concentrations were measured for an equilibrium mixture at 500K. [N2O5] = .0300 M [NO2] = .037 M [O2] = .016 M Calculate the value of Kc Kc = [NO2]4 [O2] / [N2O5]2 Kc = [.037]4 [.016] / [.0300]2 Kc = [1.9 X 10-6] [.016] / 9.00 X 10-4 Kc = 3.4 X 10-5

49. *calculation of Kp: (based on partial pressures of gases in atmospheres) Example II: CH4(g) + 2H2S(g) --- CS2(g) + 4H2(g) What is the value of Kp at 1000 K if the partial pressures in an equilibrium mixture at 1000 K are .20 atm CH4, .25 atm. H2S, .52 atm. CS2, and .10 atm. of H2? Kp = [CS2] [H2]4 / [CH4] [H2S]2 = [.52] [.10]4 / [.20] [.25]2 = [.52] [.00010] / [.20] [.0625] = 4.2 X 10-3

50. Determining the Direction of an Equilibrium Reaction *in order to determine the direction that an equilibrium reaction is expected to go, one must compare the reaction quotient (Qc) which is based on concentration values that are not necessarily at equilibrium to the equilibrium constant (Kc) which is based on concentrations that are at equilibrium *once the quotient and constant are determined, the following generalizations can be made concerning the direction of the reaction: