Transcript Lab Calc/Ch. Review Day - Morrison Community Unit District #6

Wednesday, Nov. 6th: “A” Day
Thursday, 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”
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) + 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
1,000 mL
1.0 mol NaOH = .05 mol NaOH
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
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:
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 + 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 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 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