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
“B” day: Friday, Nov. 15 th

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”
Analysis
1. Organizing Data
 Write a balanced chemical equation for each of the 3 reactions.
#1: NaOH
(s)
+ H
2
O
(l)
→ NaOH
(aq)
+ H
2
O
(l)
#2: HCl
(aq)
+ NaOH
(aq)
→ NaCl
(aq)
+ H
2
O
(l)
#3: HCl
(aq)
+ NaOH
(s)
→ NaCl
(aq)
+ H
2
O
(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)
+ #2: HCl
(aq)
+ H
2
O
(l)
→ NaOH
(aq)
+ H
2
O
(l)
+ NaOH
(aq)
→ NaCl
(aq)
+ H
2
O
(l)
#3 NaOH
(s)
+ HCl
(aq)
→ NaCl
(aq)
+ H
2
O
(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 = T final
– T initial
Example:
ΔT
1
= 26.5°C – 21.5°C = 5.0°C
ΔT
2
=
ΔT
3
=

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 H
2
O
(from data table) 1 mL

Lab: “Calorimetry and Hess’s Law”
6. Analyzing Results
 Use the calorimetry equation, q = mc
(c p water = 4.180 J/g·°C) p
ΔT, to calculate the heat released by each reaction.
Example: q
1 q = mc p
ΔT
= (100 g) (4.180 J/g·°C) (5.0°C)
= 2,090 J
= 2.09 kJ q
2
= q
3
=

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:
ΔH
1
= - 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?
ΔH
1
+ ΔH
2
= ΔH
3

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 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:
ΔH
3
= -91.96 kJ/mol (from #8)
Indirect Calculation:
ΔH
3
= ΔH
1
+ ΔH
2
- 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 + H
2
O
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 Law”
Finally, use the calorimetry equation, q = mc p to find ΔT:
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 3 rd 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 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 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 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 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 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 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 moistureabsorbing 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, 39
2. Concept Review
Chapter 10 Test/Concept Review Due:
“A” Day: Thursday, 11-14
“B” Day: Friday, 11-15