Wednesday nov 6 th a day thursday nov 7 th b day 11 45 release agenda
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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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Wednesday nov 6 th a day thursday nov 7 th b day 11 45 release agenda

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


Lab calorimetry and hess s law

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!


Lab calorimetry and hess s law1

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)


Lab calorimetry and hess s law2

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)


Lab calorimetry and hess s law3

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.


Lab calorimetry and hess s law4

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 =


Lab calorimetry and hess s law5

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


Lab calorimetry and hess s law6

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 =


Lab calorimetry and hess s law7

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


Lab calorimetry and hess s law8

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)


Lab calorimetry and hess s law9

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


Lab calorimetry and hess s law10

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)


Lab calorimetry and hess s law11

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


Lab calorimetry and hess s law12

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


Lab calorimetry and hess s law13

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


Lab calorimetry and hess s law14

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.


Lab calorimetry and hess s law15

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)


Lab calorimetry and hess s law16

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.


Lab calorimetry and hess s law17

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.


Lab calorimetry and hess s law18

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.


Lab calorimetry and hess s law19

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.


Lab calorimetry and hess s law20

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.


Chapter review concept review work time

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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