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

Chemistry I Honors—Unit 7: Thermodynamics/Kinetics/Equilibrium. Objective #1: Temperature vs. Heat & Heat Calculations. I. Temperature vs. Heat: T emperature is a measure of the average kinetic energy in a sample of matter

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

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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 oC= K-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:Compare the following problems… 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/g K or oC. Q = (m)(c)(∆t) ∆t = Q / (m)(c) = 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 = (m)(c)(∆t) ∆t = Q / (m)(c) = 500. J / (5.00 g) (.385 J/g.oC) = 260. oC What do these answers tell us? Each substance has a unique value for specific heat. Specific heat is an intensive property!

  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 = (m)(c)(∆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 = (m)(c)(∆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. This apparatus is often used to determine the calorie content of food. 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. Changing the temp by 1o C is equivalent to 1 calorie.

  10. Bomb Calorimeter 1 cal = 4.18 J

  11. Bomb Calorimeter Formula: qrxn = (-Ccal) (∆t) Ccal: Calorimeter Constant Example 4: When 4.00 g of CH6N2 (methylhydrazine) 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.

  12. qrxn = (-Ccal) (Δt) qrxn = (-7.794 kJ/oC)(14.5oC) = -113 kJ But,the question asked for kJ/mole?? WTH?? • A mole is a standard measure of the amount of a chemical needed to have 6.02 x 1023 particles • The unit “mole” is the “common language” used to compare different chemicals • Molar mass = # grams of a specific chemical per mole; calculated from the formula of the compound

  13. SO…To find kJ/mole: moles = (4.00 g CH6N2) (1 mole CH6N2) 46.0 g CH6N2 = .0870 moles Heat of Reaction in kJ/mole = -113 kJ/.0870 moles = -1.30 X 103 kJ/mole

  14. Objective #2:Heating & Cooling Curves, Energy Changes & Phase Changes, Phase Diagrams

  15. Objective #2:Heating & Cooling Curves, Energy Changes & Phase Changes Phase Diagrams

  16. Objective #2:Heating & Cooling Curves, Energy Changes & Phase Changes; Phase Diagrams I. Heating / Cooling Curve Terminology • Specific Heat – energy required to change temperature of material 1o C • Energy Changes During Phase Changes 1. Heat of Fusion – energy required to melt substance at melting point 2. Heat of Solidification – energy released to freeze substance at freezing point 3. Heat of Vaporization – energy required to boil substance at its BP

  17. Objectives #2: Heating and Cooling Curves, Energy Changes & Phase Changes • 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

  18. 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 !

  19. Objectives #2:Heating/Cooling Curves, Energy Changes & Phase Changes Hints to remember: To change Temperature, use Q = (m)(c)(Dt) To change Phase, use DH = (# moles)(Molar heat) E. Example Problems: Time to Plug & Chug—

  20. Example #5: 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. “melt ice”…change of state, so… ∆H = (# moles of material) (molar heat) = (8.6 g) 1 mole X 6.009 kJ/mole 18.0 g = 2.9 kJ

  21. Example #6: 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. “ boiling”…change of state, so… ∆H = (# moles of material)(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 = 23200 g 1 mole

  22. Example #7: 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 oC and the heat of vaporization of water is 40.67 kJ/mole. Heat water from 25oC to 100oC…temp change…Q: Q = (m)(c)(∆t) = (15.0 g) (4.18 J/g.C) (75oC) = 4702.5 J = 4.7025 kJ

  23. Boil…phase change… ∆H = (moles) (molar heat) = (15.0 g x 1 mole ) (40.67 kJ/mole) 18.0 g = 33.9 kJ Total both parts = 4.7025 kJ + 33.9 kJ = 38.6 kJ

  24. Example #8: Given the information in the packet, 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) Step 1: Heat water from -25oC to 0oC: Q = (m)(c)(∆t) = (5.00 g) (2.09 J/g K) (25 K) = 26 J = .26 kJ

  25. Step 2: Melting… ∆H = moles X molar heat = (5.00 g X 1 mole ) X 6.01 kJ/mole 18 g = 1.67 kJ Step 3: Heat water from 0oC to 100oC: Q = (m)(c)(∆t) = (5.00 g) (4.18 J/g K) (100 K) = 2090 J = 2.090 kJ

  26. Step 4: Boil… ∆H = moles X molar heat = (5.00 g X 1 mole ) X 40.67 kJ/mole 18 g = 11.3 kJ Step 5: Heat from 100oC to 110oC Q = (m)(c)(∆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

  27. II. Interpreting a Phase Diagram

  28. A phase diagram allows one to determine the phase that a substance is in at a given temperatureand 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 twophase 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. • If a point on the diagram does not fall on any line, only one phase is present .

  29. Interpretation Practice: 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

  30. States of Matter & Phase Changes!!

  31. Objectives #4-8:Energy, Entropy, and Reaction Rate • Chemical Reactions and Bond Energy: • All chemical reactions involve a change in energy • Energy can be “lost” to the environmentby the system (the chemical reaction) and vice-versa • Bond energy is the energy required to breaka bond which equalsthe energy released when a bond is formed Example #9: H – H + I-I → 2 HI 436 kJ 151 kJ 598 kJ ΔH = -598 kJ + 587 kJ = -11 kJ (net change)

  32. Objectives #4-8: Energy, Entropy, and Reaction Rate • Endothermic reactions absorbenergy while exothermic reactions releaseenergy • 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 (∆H) is positive • If the energy of the products is higher than the energy of the reactants, an endothermicreaction occurs (∆H) is negative

  33. Objectives #4-8: 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 energyand increase entropy Example #10: H2O(l) H2(g) + O2(g) the formation of gas is a more disordered state

  34. Objectives #4-8: 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

  35. Objectives #4-8: 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 • “Switch Ball” Demo

  36. Example #11: Energy Diagram Showing the Formation of an Activated Complex

  37. Interpreting Energy Profile Diagrams: • A spontaneous reaction occurs when entropy increases (ΔS is positive) and enthalpy decreases (ΔH is negative) Exo- Endo-

  38. Factors Affecting Reaction Rate:

  39. 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 becomes… K = [C]G [D]H [A]E[B]F Remember: Only gases and aqueous ions are subject to equilibrium changes, and therefore only these substances can be part of the expression Chemical Products Reactants Coefficient = # of moles 1 mole = 6.02 x 1023 [ ] = concentration of chemical

  40. Example #17: Write the equilibrium expression for the following reaction: 4 NH3 + 5 O2 4 NO + 6 H2O (assume all components are gases) K = [NO]4 [H2O]6 [NH3]4 [O2]5

  41. Objective #9-11:Equilibrium and LeChatelier’s Principle • The meaning of the equilibrium constant “K”… • Asmallvalue for K indicates that the formation of the reactantsis favored • Alargevalue for K indicates that the formation of theproductsis favored • Calculation of Kc: • based on molarityconcentrations (M)… • # of particles within the aqueous/gas system…unit is moles/liter

  42. Example #18: 2 N2O5(g) 4 NO2(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] = [.037]4 [.016] [N2O5]2 [.0300]2 Kc = [1.9 X 10-6] [.016] 9.00 X 10-4 Kc = 3.4 X 10-5 *use Sci notation to eliminate zeros…

  43. Kp Calculations… • Based on partial pressures of gases; pressure units of atmospheres Example #19:--Qc CH4(g) + 2 H2S(g) CS2(g) + 4 H2(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] Kp = 4.2 X 10-3

  44. Determining the Direction of an Equilibrium Reaction • In order to determine the direction that an equilibrium reaction is expected to go, compare the reaction _________ (Qc) to the equilibrium _________ (Kc). Note that ( Qc) is based on concentration values that are ______ necessarily at equilibrium and (Kc) is based on concentrations that _________ at equilibrium. • Once the quotient and constant are determined, the following generalizations can be made concerning the direction of the reaction: quotient constant not are

  45. Rxn. Quotient v. Equilib. Constant If Qc < Kc Net reaction goes from reactants to products (Forward Reaction) If Qc > Kc Net reaction goes from products to reactants (Reverse Reaction) If Qc = KcNonet reaction occurs

  46. Example #20: A mixture of 1.57 mole N2O5, 1.92 mole NO2, and 8.13 mole O2 is introduced into a 20.0 L reaction vessel at 500 K. At this temperature, the equilibrium constant Kc is 170. Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction? 2 N2O5(g) 4 NO2(g) + O2(g) Qc = [1.92 / 20.0]4 [8.13 / 20.0] [1.57 / 20.0]2 = [.0000849] [.4065] = 0.0056 .00616 The Qc < Kc so the net direction of the reaction is reactants to products

  47. Example #21: The Kc of the reaction below is 57.0 at 700 K: H2(g) + I2(g) 2 HI(g) Determine the direction of the reaction if the components are in the following concentrations: [HI] = .10 M [I2] = .20 M [H2] = .10 M Qc = (.10)2= .50 (.10) (.20) Qc < Kc so reaction goes towards products

  48. 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)

  49. Objective #9-11 : 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)

  50. Example #22: N2(g) + 3 H2(g) 2 NH3(g) + 92 kJ an increase in the concentration of nitrogen gas will shift reaction towards the products side a decrease in the concentration of ammonia gas will shift reaction towards the products side cooling the reaction will shift the reaction towards the products side

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