South Pasadena • Honors Chemistry
Name
6 • Thermochemistry
Period
STATION
1
–
Date
EXOTHERMIC/ENDOTHERMIC
PROCESSES
Determine if each statement describes an exothermic process (EXO) or endothermic process (ENDO).
_EXO_ H is negative
_EXO_
ENDO q > 0
_EXO_ surroundings get warmer
_EXO_ molecular attractions strengthen
ENDO PE diagram is uphill
ENDO surroundings get colder
_EXO_ combustion of propanol
_EXO_ H2(g) + ½O2 (g)  H2O (ℓ) + heat
ENDO H is positive
ENDO products have more energy than reactants
_EXO_ reactants have more energy than products
_EXO_ condensation of water
ENDO CaCO3(s) + heat  CaO(s) + CO2(g)
ENDO H2O (ℓ)  H2O (g)
_EXO_ q < 0
_EXO_ water, when placed in the freezer
South Pasadena • Honors Chemistry
6 • Thermochemistry
STATION
2
–
H
r x n
CALCULATIONS
Consider the following balanced equation for the combustion of propane, C3H8. Use the Chart of Thermodynamic
Values for ∆Hf and bond energies.
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (ℓ)
Products – Reactants Method
Fill in the values of ∆Hf for each substance.
Compound
Hf (kJ/mol)
C3H8 (g)
O2 (g)
CO2 (g)
H2O (ℓ)
–104
0
–393.5
–286
Bond Energy Method
Given the following structural formulas.
Substance
C3H8
Structures
Calculate the Hrxn using the ∆Hf values.
∆Hrxn = [3(–393.5) + 4(–286)] – [1(–104) + 5(0)]
= –1180.5 – 1144 + 104 = –2220.5 kJ/mol rxn
Calculate the Hrxn using bond energies.
Break 2 C–C bonds +2(347) = +694
8 C–H bonds +8(413) = +3304
5 O=O bonds +5(495) = +2475
Form 6 C=O bonds –6(799) = –4794
8 O–H bonds –8(467) = –3736
∆Hrxn = –2057 kJ/mol rxn
O2
CO2
H—O—H
H2O
The two answers above do not match. Which one is more accurate? Why? Products – Reactants is more accurate.
Bond Energy Method is less accurate because bond energies are averages and substances are in gas phase.
South Pasadena • Honors Chemistry
6 • Thermochemistry
STATION
3
– CALORIMETRY
A 100. g aluminum block (C = 0.900 J/g·ºC) in boiling water is added to an insulated cup containing 50.0 grams of water
(C = 4.184 J/g·ºC) at 5.00ºC. Calculate the final temperature of the mixture
Aluminum
Water
m = 100. g
m = 50.0 g
C = 0.900 J/g·°C
C = 4.184 J/g·°C
∆T = x – 100°C
∆T = x – 5.00°C
–qAl = +qH2O
–(mAl)(CAl)(∆TAl) = +(mH2O)(CH2O)(∆TH2O)
–(100 g)(0.900 J/g·°C)(x – 100°C) = +(50.0 g)(4.184 J/g·°C)(x – 5.00°C)
x = 33.6°C
South Pasadena • Honors Chemistry
6 • Thermochemistry
STATION
4
–
ENERGY FLOW
Heat always flows from an object of HIGHER temperature to an object of LOWER temperature.
Both blocks have a temperature of 22.5°C. Use arrows on the top picture to show the flow of heat and why the ice acts
the way it does. Provide a brief explanation for the observed differences.
Time
Block A
Block B
Explanation
0 minutes
Blocks A and B are different
materials. Block B transfers
energy to the ice at a faster rate
than Block A, so the block of ice
on Block B melts faster than that
on Block A.
2 minutes
South Pasadena • Honors Chemistry
6 • Thermochemistry
STATION
5
–
HEATS
OF
FUSION &
VAPORIZATION
kJ
kJ
J
J
J
For water, ∆Hfus = 6.01 mol , ∆Hvap = 40.68 mol , Cice = 2.10 g·°C , Cwater = 4.18 g·°C , Csteam = 2.08 g·°C
What is the value of q when 45.0 g of water freezes at 0°C?
1 mol 
n = 45.0 g
q = n·∆Hfus = (2.50 mol)(–6.01 kJ/mol) = –15.0 kJ
18.02 g = 2.50 mol
∆H = –∆Hfus = –6.01 kJ/mol
What mass of water can be vaporized with 75.0 kJ of energy?
q
+75.0 kJ
q = +75.0 kJ
n=
=
= 1.84 mol
∆Hvap +40.68 kJ/mol
18.02 g
∆H = +∆Hvap = +40.68 kJ/mol
m = 1.84 mol 
 1 mol  = 33.2 g
South Pasadena • Honors Chemistry
6 • Thermochemistry
STATION
6
–
H
r x n
FROM DATA
Write the balanced thermochemical equation for the combustion of pentane, C5H12.
C5H12 + 8 O2 → 5 CO2 + 6 H2O + heat
When 10.0 g of C5H12 is burned, 453 kJ of energy is released. What is the ∆Hrxn for the combustion of pentane?
Molar Mass = 5(12.01) + 12(1.008) = 72.1 g/mol
1 mol C5H12   1 mol rxn 
nrxn = 10.0 g C5H12 
= 0.139 mol rxn
72.1 g C5H12 1 mol C5H12
q
–453 kJ
∆Hrxn =
=
= –3270 kJ/mol rxn
nrxn 0.139 mol rxn
Write the balanced thermochemical equation for the melting of aluminum. (Include states.)
Heat + Al (s) → Al (ℓ)
When 10.0 grams of aluminum melts, 3.929 kJ of energy is required. What is the Hfus of Al?
Molar Mass = 26.98 g/mol
1 mol Al  1 mol rxn
nrxn = 10.0 g Al 
26.98 g Al  1 mol Al  = 0.371 mol
q
–3.929 kJ
∆Hrxn =
=
= –10.6 kJ/mol
nrxn 0.371 mol
South Pasadena • Honors Chemistry
6 • Thermochemistry
STATION
7
– COMPARING SUBSTANCES
Blocks X and Z have the same mass. When 500 J of heat is added to each, the temperature of Block X rises 10 ºC while
the temperature of Block Z rises 30 ºC. Which block has the larger specific heat, C? Justify your answer without math.
Block X. With the same amount of energy, the temperature of Block X rises less than Block Z, It is harder to
change the temperature of Block X (it is more resistant to change in temperature) and it has a larger specific
heat.
0.15 mole samples of Liquid A and Liquid B were kept at their boiling points. When 400 J of energy is added to each,
Liquid A completely vaporizes while some of Liquid B remained. Which liquid has a greater ∆Hvap? Justify your answer
without math.
Liquid B. With the same amount of energy, less of Liquid B vaporized than Liquid A. It is harder to
vaporize Liquid B, so it has a greater ∆Hvap.
South Pasadena • Honors Chemistry
6 • Thermochemistry
STATION
8
–
HEATING
CURVE
Temp (°C)
Sketch the graph when 40.0 g of ice at –30°C
is heated to steam at 140°C. Label the axes
(with units), and the Freezing and Boiling points.
Label the segments (a) through (e). Identify
the state(s) of matter for each segment.
(e) gas
(d) liquid/gas
100 ––
(c) liquid
0 –– (b) solid/liquid
(a) solid
In which segments: (circle)
 are molecular attractions weakening?
 is the kinetic energy increasing?
 are the processes endothermic?
 are the processes exothermic?
 do you use q = m C ∆T to calculate the heat?
 do you use q = n ∆H to calculate the heat?
–
a
a
–
a
–
b
–
b
–
–
b
–
c
c
–
c
–
d
–
d
–
–
d
–
e
e
–
e
–
South Pasadena • Honors Chemistry
6 • Thermochemistry
STATION
For water, ∆Hfus = 6.01
9
–
HEATING
CURVE CALCULATIONS
kJ
kJ
J
J
J
, ∆Hvap = 40.68
, C = 2.10
,C
= 4.18
,C
= 2.08
mol
mol ice
g·°C water
g·°C steam
g·°C
A 50.0 gram sample of ice at 0.0 ºC is heated until it is liquid at 80.0 ºC.
Indicate on the heating curve where the heating process begins and ends.
Calculate the heat absorbed by this process.
1 mol  6.01 kJ 1000 J
q2 = (50.0 g)
18.02 g  1 mol   1 kJ  = 16700 J
q3 = (50.0 g)(4.184 J/g·°C)(80°C – 0°C) = 16700 J
q = q2 + q3 = 16,700 J + 16,700 J = 33,400 J