5/4/2014
Calorimetry & Enthalpy
Measuring and Quantifying Heats of Reactions

Today:
◦ Summary of the First Law of
Thermodynamics

See Canvas Exam 3 Scores
Challenge Problem Set 13 DUE
this Thursday, May 8th @ 11 pm
Please Read Chp. 8: Sections 8.8◦ Enthalpy: Heats of Reactions & 8.11, pp. 290-299
Phase Changes

 Refrigerators, Heat Pumps & Heat

Engines
 Specific Heat Capacity
 Calorimetry
 Introduction to Hess’s Law
“The visible world is the
invisible organization of
energy.”
—Heinz Pagels
Thermodynamics of Refrigerators and Heat Pumps
A
D
B
C
ADIABATIC Expansion: A→B
•
•
•
•
Volume increases↑, Pressure decreases↓
q = 0 (Adiabatic = No heat flow)
w = Negative (−): gas expands against external pressure
ΔE = q + w = Negative (−) so TEMPERATURE DECREASES
collisions at the atomic scale transfer
net kinetic energy from HOT object to COLD object.
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5/4/2014
Thermodynamics of Refrigerators and Heat Pumps
A
D
B
C
Heat
absorbed
inside
refrigerator
ISOTHERMAL Expansion: B→C
•
•
•
•
q = Positive (+): HEAT IS ABSORBED: “Endothermic”
Constant Temperature means ΔE = q + w = 0
w = Negative (−): gas expands against Pexternal & breaks IMFs
Volume increases↑, Pressure decreases↓, ΔT = 0 (isothermal)
Thermodynamics of Refrigerators and Heat Pumps
A
D
B
C
Heat
absorbed
inside
refrigerator
ADIABATIC Compression: C→D
•
•
•
•
Volume decreases ↓, Pressure increases ↑
q = 0 (Adiabatic = No heat flow)
w = Positive (+): work done by compressor adds energy to gas
ΔE = q + w = Positive (+) so TEMPERATURE INCREASES
Thermodynamics of Refrigerators and Heat Pumps
Heat released
outside
refrigerator
A
D
B
C
ISOTHERMAL Compression: D→A
•
•
•
•
Heat
absorbed
inside
refrigerator
q = Negative (−): HEAT IS RELEASED: “Exothermic”
Constant Temperature (ΔT = 0 ) means ΔE = q + w = 0
w = Positive (+): gas condenses due to attractive IMFs
Volume decreases ↓, Pressure increases ↑
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5/4/2014
Calorimetry: Measuring the Heat of Reactions
• 1.594 g of methane (CH4) is burned (with excess oxygen) inside
a calorimeter containing 1037.4 g of water.
• If the water absorbs all of the heat from
the combustion, warming from 22.495
oC to 34.700 oC, calculate the heat
released in the reaction (qreaction).
qsystem + qsurroundings = 0
qreaction + qH2O = 0
• Calculate the Enthalpy of the reaction (ΔHreaction).
𝚫Hreaction =
𝐪𝐫𝐞𝐚𝐜𝐭𝐢𝐨𝐧
𝐦𝐨𝐥𝐞𝐬 𝐫𝐞𝐚𝐜𝐭𝐢𝐧𝐠
In 2012, the average energy consumption for a U.S. household
was 30 kilowatt hours (kWh) per day. What quantity of methane
(in grams) could be burned to release an equivalent amount of
energy as heat?
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
ΔHoreaction = -802.2 kJ
1 kWh = 3600 kJ
• How much heat would be released (in Joules), if 2.0 mL of
isopropyl alcohol (IPA) was burned according to the combustion
reaction below? Density of IPA = 0.786 g/mL.
2 CH3CH(OH)CH3(l) + 9 O2(g) → 6 CO2(g) + 8 H2O(g) ΔHo=-3659 kJ
• Imagine ALL of the heat from the combustion above was
absorbed by 1.00 L of H2O(l), what would be the expected
change in temperature (ΔT) for the water?
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5/4/2014
2 CH3CH(OH)CH3(l) + 9 O2(g) → 6 CO2(g) + 8 H2O(g)
(Heat of Combustion)
H2O(l) → H2O(g) ΔHo = + 44 kJ
(Heat of Vaporization)
Consider the fact that the isomerization reaction below is an
endothermic process.
Based on this information, which combustion will be MORE
EXOTHERMIC?
A. 2 CH3CH(OH)CH3(g) + 9 O2(g) → 6 CO2(g) + 8 H2O(g)
B. 2 CH3CH2CH2OH(g) + 9 O2(g) → 6 CO2(g) + 8 H2O(g)
C. They will both release the SAME AMOUNT OF HEAT
D. Cannot determine from information provided.
CO2(g) + H2O(g)
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5/4/2014
iClicker Participation Question:
Hess’s Law & Heats of Vaporization
Which reaction below is the MOST EXOTHERMIC?.
A. C5H12(l) + 8 O2(g) → 5 CO2(g) + 6 H2O(g)
B. C5H12(g) + 8 O2(g) → 5 CO2(g) + 6 H2O(g)
C. C5H12(l) + 8 O2(g) → 5 CO2(g) + 6 H2O(l)
D. C5H12(g) + 8 O2(g) → 5 CO2(g) + 6 H2O(l)
E. They all will release the SAME AMOUNT OF ENERGY
C5H12(g)
Enthalpy Diagram:
8 O2(g) + C5H12(l)
Fuel VAPORS are
more flammable
than liquid fuels
5 CO2(g) + 6 H2O(g)
6 H2O(l)
Hess’s Law:
The OVERALL enthalpy change for a reaction is equal to the sum of
enthalpy changes for the individual steps in the reaction.
C5H12(l) + 8 O2(g) → 5 CO2(g) + 6 H2O(g) ΔHo1 = -3509 kJ
C5H12(l) → C5H12(g) ΔHo2 = +26.2 kJ
H2O(l) → H2O(g) ΔHo3 = +44.0 kJ
C5H12(g) + 8 O2(g) → 5 CO2(g) + 6 H2O(l) ΔHoOVERALL = ?
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5/4/2014
Hess’s Law:
Adjust the reactions to eliminate the undesired substances.
THEN add the reactions up to achieve the OVERALL desired reaction.
C5H12(l) + 8 O2(g) → 5 CO2(g) + 6 H2O(g) ΔHo1 = -3509 kJ
C5H12(l) → C5H12(g) ΔHo2 = +26.2 kJ
H2O(l) → H2O(g) ΔHo3 = +44.0 kJ
C5H12(g) + 8 O2(g) → 5 CO2(g) + 6 H2O(l) ΔHoOVERALL = ?
Hess’s Law:
REVERSE the second reaction to eliminate C5H12(l) from the equation.
The reverse of an endothermic reaction is an exothermic one.
Change the sign of ΔH when reversing a reaction.
C5H12(l) + 8 O2(g) → 5 CO2(g) + 6 H2O(g) ΔHo1 = -3509 kJ
C5H12(g) → C5H12(l) ΔHo2’ = -26.2 kJ
H2O(l) → H2O(g) ΔHo3 = +44.0 kJ
C5H12(g) + 8 O2(g) → 5 CO2(g) + 6 H2O(l) ΔHoOVERALL = ?
Hess’s Law:
Reverse the 3rd reaction AND MULTIPLE BY 6 to eliminate 6 H2O(g).
Remember to change the sign of ΔH when reversing a reaction.
If a reaction is multiplied by a factor, so is the corresponding ΔH.
C5H12(l) + 8 O2(g) → 5 CO2(g) + 6 H2O(g) ΔHo1 = -3509 kJ
C5H12(g) → C5H12(l) ΔHo2’ = -26.2 kJ
6 H2O(g) → 6 H2O(l) ΔHo3’ = (-44.0 kJ)·6 = -264 kJ
C5H12(g) + 8 O2(g) → 5 CO2(g) + 6 H2O(l) ΔHoOVERALL = ?
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5/4/2014
Hess’s Law:
The OVERALL enthalpy change for a reaction is equal to the sum of
enthalpy changes for the individual steps in the reaction
C5H12(l) + 8 O2(g) → 5 CO2(g) + 6 H2O(g) ΔHo1 = -3509 kJ
C5H12(g) → C5H12(l) ΔHo2’ = -26.2 kJ
6 H2O(g) → 6 H2O(l) ΔHo3’ = (-44.0 kJ)·6 = -264 kJ
C5H12(g) + 8 O2(g) → 5 CO2(g) + 6 H2O(l)
ΔHoOVERALL = ΔHo1 + ΔHo2’ + ΔHo3’ = -3799 kJ
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