Download ppt - 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 ReviewConcept 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”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 Law”Analysis1. 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 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 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 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 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 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 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 Law”8. Analyzing Results Calculate the ΔH values in kJ/mol of 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 mol NaOH (from #7)

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


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 Law”Conclusions11. 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 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 Law”Next, 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 NaOH

Total heat of reaction: 58.5 kJ + 68.9 kJ = 127.4 kJ OR

127,400 J


Finally, use the calorimetry equation, q = mcpΔT to find ΔT:

127,400 J = (450 g) (4.180 J/g·°C) ΔTΔT = 67.7°CInitial 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 Law”13. 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)


14. 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 Hess’s Law”15. Evaluating MethodsCould 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 Law”16. Drawing ConclusionsWhich is more stable, solid NaOH or NaOH

solution?

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


Extensions1. 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.


2. 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 Time

Use the rest of the time to work on the following:1.Ch. 10 review, pg. 370-373: 3-5, 7, 14, 16, 18, 20-

25, 27-28, 31-33, 35-36, 392.Concept Review

Chapter 10 Test/Concept Review Due:

“A” Day: Thursday, 11-14“B” Day: Friday, 11-15