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

  • Published on
    22-Feb-2016

  • View
    46

  • Download
    0

DESCRIPTION

Wednesday, Nov. 6 th : A Day Thursday, Nov. 7 th : B Day (11:45 release) Agenda. Lab: Calorimetry and Hesss 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 - PowerPoint PPT Presentation

Transcript

  • Wednesday, Nov. 6th: A DayThursday, Nov. 7th: B Day (11:45 release)AgendaLab: Calorimetry and Hesss LawComplete Calculations/Analysis/Hand InStart Ch. 10 ReviewConcept Review Work Time

    Chapter 10 Test/Concept Review Due:A day: Thursday, Nov. 14thB day: Friday, Nov. 15th

  • Lab: Calorimetry and Hesss LawWe 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!

    Dont forget the reflection statement!

  • Lab: Calorimetry and Hesss LawAnalysis1. Organizing DataWrite 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 Hesss Law2. Analyzing ResultsAdd 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 Hesss Law3. Explaining EventsWhy 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 Hesss Law4. Organizing DataCalculate the change in temperature (T) for each of the reactions.T = Tfinal TinitialExample:T1 = 26.5C 21.5C = 5.0CT2 = T3 =

  • Lab: Calorimetry and Hesss Law5. 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 Hesss Law6. Analyzing ResultsUse the calorimetry equation, q = mcpT, to calculate the heat released by each reaction. (cp water = 4.180 J/gC)Example: q = mcpTq1 = (100 g) (4.180 J/gC) (5.0C) = 2,090 J= 2.09 kJq2 =q3 =

  • Lab: Calorimetry and Hesss Law7. 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 Hesss Law8. Analyzing Results Calculate the H values in kJ/mol of NaOH for each of the 3 reactions.Since the reactions release heat (exothermic), H will be negative.The heat released by the reactions was transferred to the water, so H = -q

    Example reaction #1:H1 = - 2.09 kJ (from #6) = - 41.8 kJ/mol .05 mol NaOH (from #7)

  • Lab: Calorimetry and Hesss Law9. Analyzing ResultsBased on what you know about Hesss Law, how should the enthalpies for the 3 reactions be mathematically related?

    H1 + H2 = H3

  • Lab: Calorimetry and Hesss Law10. Analyzing ResultsWhich 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 Hesss LawConclusions11. Evaluating MethodsFind H for the reaction of solid NaOH with HCl solution by direct measurement and by indirect calculation.Direct measurement:H3 = -91.96 kJ/mol (from #8)

    Indirect Calculation:H3 = H1 + H2 - 41.8 kJ/mol + (- 51 kJ/mol) = -92.8 kJ/mol

  • Lab: Calorimetry and Hesss Law12. Drawing ConclusionsCould 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 Hesss LawNext, use the mole ratio from the balanced reaction between NaOH and HCl to convert moles HCl in the container moles NaOH:NaOH + HCl NaCl + H2O1.35 moles HCl = 1.35 moles NaOH

    Reaction #2: 1.35 mol NaOH X 51 kJ = 68.9 kJ 1 mol NaOHTotal heat of reaction: 58.5 kJ + 68.9 kJ = 127.4 kJ OR127,400 J

  • Lab: Calorimetry and Hesss LawFinally, use the calorimetry equation, q = mcpT to find T:

    127,400 J = (450 g) (4.180 J/gC) TT = 67.7CInitial temp = 25C + 67.7C = 92.7C

    Water hotter than 60C 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 Hesss Law13. Applying ConclusionsWhich 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 Hesss Law14. Evaluating ResultsWhen 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 Hesss Law15. Evaluating MethodsCould the same type of procedure be used to determine T 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 H would be positive.

  • Lab: Calorimetry and Hesss Law16. Drawing ConclusionsWhich is more stable, solid NaOH or NaOH solution?

    NaOH solution is more stable because solid NaOH absorbs water from the atmosphere.

  • Lab: Calorimetry and Hesss LawExtensions1. Applying ConclusionsExplain 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 Hesss Law2. Designing ExperimentsHow 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 TimeUse 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, 39Concept Review

    Chapter 10 Test/Concept Review Due:A Day: Thursday, 11-14B Day: Friday, 11-15

Recommended

View more >