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Wednesday, Nov. 6 th : “A” Day Thursday, Nov. 7 th : “B” Day (11:45 release) Agenda

Wednesday, Nov. 6 th : “A” Day Thursday, Nov. 7 th : “B” Day (11:45 release) Agenda. Lab: “ Calorimetry and Hess’s Law” Complete Calculations/Analysis/Hand In Start Ch. 10 Review Concept Review Work Time Chapter 10 Test/Concept Review Due: “A” day: Thursday, Nov. 14 th

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Wednesday, Nov. 6 th : “A” Day Thursday, Nov. 7 th : “B” Day (11:45 release) Agenda

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  1. Wednesday, Nov. 6th: “A” DayThursday, Nov. 7th: “B” Day (11:45 release)Agenda • Lab: “Calorimetry and Hess’s Law” • Complete Calculations/Analysis/Hand In • Start Ch. 10 Review • Concept Review Work Time Chapter 10 Test/Concept Review Due: “A” day: Thursday, Nov. 14th “B” day: Friday, Nov. 15th

  2. Lab: “Calorimetry and Hess’s Law” • We will work through the calculations, etc. together. • Make sure this lab is added to your table of contents before turning it in. • Make sure all of your data is labeled and has the proper units! • Don’t forget the reflection statement!

  3. Lab: “Calorimetry and Hess’s Law” Analysis 1. Organizing Data • Write a balanced chemical equation for each of the 3 reactions. #1: NaOH(s) + H2O(l) → NaOH(aq) + H2O(l) #2: HCl(aq) + NaOH(aq)→ NaCl(aq) + H2O(l) #3: HCl(aq) + NaOH(s)→ NaCl(aq) + H2O(l)

  4. Lab: “Calorimetry and Hess’s Law” 2. Analyzing Results • Add the first 2 equations from question #1 to get the equation for reaction #3: #1: NaOH(s) + H2O(l) → NaOH(aq) + H2O(l) + #2: HCl (aq) + NaOH(aq)→ NaCl(aq) + H2O(l) #3 NaOH(s) + HCl(aq)→ NaCl(aq) + H2O(l)

  5. Lab: “Calorimetry and Hess’s Law” 3. Explaining Events • Why does a plastic-foam cup make a better calorimeter than a paper cup? • A good calorimeter must insulate and not transfer (lose) heat. Plastic-foam cups are better insulators than paper cups and therefore make a better calorimeter.

  6. Lab: “Calorimetry and Hess’s Law” 4. Organizing Data • Calculate the change in temperature (ΔT) for each of the reactions. ΔT = Tfinal – Tinitial Example: ΔT1 =26.5°C – 21.5°C = 5.0°C ΔT2 = ΔT3 =

  7. Lab: “Calorimetry and Hess’s Law” 5. Organizing Data • Assuming that the density of the water and the solutions is 1.00 g/mL, calculate the mass,m, of liquid present for each of the 3 reactions. Example: #1 100.0 mL solution X 1.00 g = 100 g H2O (from data table) 1 mL

  8. Lab: “Calorimetry and Hess’s Law” 6. Analyzing Results • Use the calorimetry equation, q = mcpΔT, to calculate the heat released by each reaction. (cp water = 4.180 J/g·°C) Example: q = mcpΔT q1 = (100 g) (4.180 J/g·°C) (5.0°C) = 2,090 J = 2.09 kJ q2 = q3 =

  9. Lab: “Calorimetry and Hess’s Law” 7. Organizing Data • Calculate the moles of NaOH used in each of the 3 reactions. Example for reaction #1: 2.00 g NaOH X 1 mol NaOH = .05 mol NaOH (from table) 40 g NaOH Example for reaction #2: 50.0 mL NaOH X 1L X 1.0 mol NaOH = .05 mol NaOH 1,000 mL 1L NaOH

  10. Lab: “Calorimetry and Hess’s Law” 8. Analyzing Results • Calculate the ΔH values in kJ/molof NaOH for each of the 3 reactions. • Since the reactions release heat (exothermic), ΔH will be negative. • The heat released by the reactions was transferred to the water, so ΔH = -q Example reaction #1: ΔH1 = - 2.09 kJ(from #6) = - 41.8 kJ/mol .05 molNaOH(from #7)

  11. Lab: “Calorimetry and Hess’s Law” 9. Analyzing Results • Based on what you know about Hess’s Law, how should the enthalpies for the 3 reactions be mathematically related? ΔH1 + ΔH2 = ΔH3

  12. Lab: “Calorimetry and Hess’s Law” 10. Analyzing Results • Which types of heat of reaction apply to the enthalpies calculated in item 8. #1: heat of solution (NaOH dissolving) #2: heat of reaction (NaOH + HCl reaction) #3: heat of solution AND heat of reaction (both)

  13. Lab: “Calorimetry and Hess’s Law” Conclusions 11. Evaluating Methods • Find ΔH for the reaction of solid NaOH with HCl solution by direct measurement and by indirect calculation. Direct measurement: ΔH3 = -91.96 kJ/mol (from #8) Indirect Calculation: ΔH3 = ΔH1 + ΔH2 - 41.8 kJ/mol + (- 51 kJ/mol) = -92.8 kJ/mol

  14. Lab: “Calorimetry and Hess’s Law” 12. Drawing Conclusions • Could a mixture hot enough to cause burns result from mixing NaOH and HCl? There are 2 different reactions happening in the container that generate heat: 1. NaOH dissolving in water (heat of dissolution) 2. The reaction of the NaOH with the HCl (heat of reaction) • First, calculate the heat generated when NaOH dissolves: Moles NaOH: 55g NaOH X 1 mol NaOH = 1.4 mol NaOH (in container) 40 g NaOH Reaction #1: 1.4 mol NaOH X 41.8 kJ = 58.5 kJ 1 mol NaOH

  15. Lab: “Calorimetry and Hess’s Law” Next, use the mole ratio from the balanced reaction between NaOH and HCl to convert moles HCl in the container moles NaOH: NaOH + HClNaCl + H2O 1.35 moles HCl = 1.35 moles NaOH Reaction #2: 1.35 mol NaOH X 51 kJ = 68.9 kJ 1 mol NaOH Total heat of reaction: 58.5 kJ + 68.9 kJ = 127.4 kJ OR 127,400 J

  16. Lab: “Calorimetry and Hess’s Law” Finally, use the calorimetry equation, q = mcpΔT to find ΔT: 127,400 J = (450 g) (4.180 J/g·°C) ΔT ΔT = 67.7°C Initial temp = 25°C + 67.7°C = 92.7°C Water hotter than 60°C can cause 3rd degree burns, so YES, a mixture hot enough to cause burns could have resulted from mixing NaOH with HCl.

  17. Lab: “Calorimetry and Hess’s Law” 13. Applying Conclusions Which chemical is limiting? How many moles of the other reactant remained unreacted? • HCL is limiting (1.35 moles HCl vs. 1.4 moles NaOH) • .05 moles of NaOH left over after reaction (1.4 mol – 1.35 mol)

  18. Lab: “Calorimetry and Hess’s Law” 14. Evaluating Results • When chemists make solutions from NaOH pellets, they often keep the solution in an ice bath. Why? • The heat of solution for NaOH pellets is high enough to make the solution dangerously hot.

  19. Lab: “Calorimetry and Hess’s Law” 15. Evaluating Methods • Could the same type of procedure be used to determine ΔT for endothermic reactions? How would the procedure stay the same? What would change? • Yes, the procedure would work with endothermic reactions as well. The temperature of the water would decrease and ΔH would be positive.

  20. Lab: “Calorimetry and Hess’s Law” 16. Drawing Conclusions • Which is more stable, solid NaOH or NaOH solution? • NaOH solution is more stable because solid NaOH absorbs water from the atmosphere.

  21. Lab: “Calorimetry and Hess’s Law” Extensions 1. Applying Conclusions • Explain why adding an acid or a base to neutralize a spill is not a good idea. • The heat of reaction for a neutralization could cause a burn in addition to the burn caused by the acid or base itself.

  22. Lab: “Calorimetry and Hess’s Law” 2. Designing Experiments • How would you design a package to ship NaOH pellets to a very humid place? • The NaOH pellets could be packaged in an inert environment (Ar), in a foam container to contain any spills or leaks, and moisture-absorbing materials could be added to the packaging.

  23. Chapter Review/Concept Review Work Time • Use the rest of the time to work on the following: • Ch. 10 review, pg. 370-373: 3-5, 7, 14, 16, 18, 20-25, 27-28, 31-33, 35-36, 39 • Concept Review Chapter 10 Test/Concept Review Due: “A” Day: Thursday, 11-14 “B” Day: Friday, 11-15

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