Worked Answers:
g
kJ
13. Hvap (methanol) = 32.0 1.18 = 37.8 kJ/mol (e)
mol
g
120 g
16. 160 kJ = 246 kJ (c)
(2 mol 39.1 g/mol)
CHAPTER 5 REVIEW
(Page 356)
Understanding Concepts
1.
Physical
Chemical
Nuclear
(a) change in state or change in
arrangement of atoms in molecules
change in arrangement of atoms
in molecules
change in arrangement of nuclei
(b) about 10 kJ/mol
about 103 kJ/mol
about 1011 kJ/mol
(c) freezing water or melting butter
burning gasoline or cooking food
uranium decay or hydrogen fusion
in Sun
q
2. c = mT
16 000 J
= 938 g (35.0ºC – 19.5ºC)
c = 1.10 J/g•ºC
The specific heat capacity of the brick is 1.10 J/g•ºC.
3. We assume that no heat is lost to the environment, negligible heat is lost to the calorimeter materials unless specific
information is given about the container, and dilute aqueous solutions have density and specific heat capacity of water.
4. qwater = mc∆T
= 500 g 4.18 J/g•ºC (80ºC – 20ºC)
qwater = 1.25 105 J
qcopper = mc∆T
= 2000 g 0.385 J/g•ºC (80ºC – 20ºC)
qcopper = 4.6 104 J
qtotal = qwater + qcopper
qtotal = 1.7 105 J, or 170 kJ
170 kJ of heat is required.
5. mwater = d V
= 1.00 g/mL 200 000 g
mwater = 2.00 105 g
qwater = mc∆T
= 2.00 105 g 4.18 J/g•ºC (65°C – 20°C)
qwater = 3.76 104 kJ
186
Chapter 5
Copyright © 2003 Nelson
qwater
n = ∆Hcomb
3.76 104 kJ
= 2200 kJ/mol
n = 17.1 mol
MC H = 32.0 g/mol
3 8
mC H = 17.1 mol 32.0 g/mol
3 8
mC H = 547 g propane
3 8
You would have to burn 547 g of propane to heat the water.
6.
Potential Energy Diagram of
the Combustion of Nonane
C9H2O(l) + O2(g)
Ep
∆H
CO2(g) + H2O( l )
Reaction Progress
7. (a) The enthalpy changes for a reaction may be represented as an energy term in the equation, a ∆H value, the molar
enthalpy, or a potential energy diagram.
(b) NH4Cl(s) + 14 kJ → NH4Cl(aq)
NH4Cl(s) → NH4Cl(aq) ∆H = +14 kJ
∆Hsol (NH Cl) = +14 kJ/mol
4
Since the reaction is endothermic, the potential energy diagram will resemble Figure 6(b) on p. 318. Reactant is
ammonium chloride solid; product is aqueous ammonium chloride.
8. (a) H2O(l) + 44 kJ → H2O(g) or
H2O(l) → H2O(g)
∆H = + 44 kJ
(b) H2O(g) → H2O(l) + 44 kJ or
H2O(g) → H2O(l)
∆H = –44kJ
(c) 2 H2O(l) + 88 kJ → 2 H2O(g)
or 2 H2O(l) → 2 H2O(g)
∆H = +88 kJ
9. (a) 3 C(s) + 3 H2(g) + 1/2 O2(g) → C3H6O(l)
(b) (1) H2(g) + 1/2 O2(g) → H2O(l)
∆H = –285.8 kJ
(2) C(s) + O2(g) → CO2(g)
∆H = –393.5 kJ
(3) C3H6O(l) + 4 O2(g) → 3 CO2(g) + 3 H2O(l)
3
(c) 3 x (1): 3 H2(g) + O2(g) → 3 H2O(l)
2
3 x (2): 3 C(s) + 3 O2(g) → 3 CO2(g)
∆H = –1784 kJ
∆H = (3) (– 285.8) kJ
∆H = (3) (– 393.5) kJ
–1 x (3): 3 CO2(g) + 3 H2O(l) → C3H6O(l) + 4 O2(g)
∆H = (–1) (– 1784) kJ
______________________________________________________________________
1
3 C(s) + 3 H2(g) + O2(g) → C3H6O(l)
∆H = –253.9 kJ
2
Copyright © 2003 Nelson
Thermochemistry
187
10. The known equations are:
(1) C6H12O6(l) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)
∆H = –2813 kJ
(2) C2H6O(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)
∆H = –1369 kJ
The target equation is C6H12O6(l) → 2 C2H6O(l) + 2 CO2(g)
1 (1): C6H12O6(l) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)
∆H = (1) (– 2813) kJ
–2 (2): 4 CO2(g) + 6 H2O(l) → 2 C2H6O(l) + 6 O2(g) ∆H = (–2) (–1369) kJ
__________________________________________________________________
C6H12O6(l) → 2 C2H6O(l) + 2 CO2(g)
∆H –74 kJ
or ∆Hferm = –74 kJ/mol glucose
molar mass of glucose, M = 180 g/mol
1 mol
nglucose = 500 g 180 g
nglucose = 2.78 mol
∆H = n∆Hferm
= 2.78 mol 74 kJ/mol
∆H = –206 kJ
The enthalpy change would be –206 kJ.
11. qwater = mc∆T
= 3770 g 4.18 J/g•ºC (98.6 16.8)ºC
qwater = 1.29 103 kJ
n∆Hcomb = qwater
qwater
n = ∆Hcomb
1.29 103 kJ
= 802 kJ/mol
n = 1.61 mol
MCH = 16.0 g/mol
4
m = nM
= 1.61 mol 16.0 g/mol
m = 25.7 g
The minimum mass of natural gas that must be burned to heat 3.77 L of water from 16.8°C to 98.6°C is 25.7 g.
5
∆H = –1275 kJ
12. (1) CH3COCOOH + O2 → 3 CO2 + 2 H2O
2
(2) CH3COOH + 2 O2 → 2 CO2 + 2 H2O
∆H = –875.3 kJ
1
(3) CO + O2 → 1 CO2
∆H = –282.7 kJ
2
5
1 (1): CH3COCOOH + O2 → 3 CO2 + 2 H2O ∆H = (1) (–1275) kJ
2
–1 (2): 2 CO2 + 2 H2O → CH3COOH + 2 O2
∆H = (–1) (–875.3) kJ
1
–1 (3): 1 CO2 → CO + O2
∆H = (–1) ( –282.7) kJ
2
___________________________________________________________________
CH3COCOOH → CH3COOH + CO
188
Chapter 5
∆H = –117 kJ
Copyright © 2003 Nelson
13. (a) N2(g) + 3 H2O(l) → 2 NH3(g) + 3/2 O2(g)
3
∆H = 2 ∆H°f(NH ) + ∆H°f(O ) – 1 ∆H°f(N ) – 3 ∆H°f(H O )
3(g)
2(g)
2(g)
2 (l)
2
3
= 2 (–45.9) + (0) – 1 (0) – 3 (–285.8)
2
∆H = 765.6 kJ
∆Hr = 765.6 kJ/2 mol NH3
∆Hr = 382.8 kJ/mol NH3
m
(b) nammonia = M
1000 g
= 17.0 g/mol
nammonia = 58.8 mol
∆H = n∆Hr
= 58.8 mol NH3 382.8 kJ/mol NH3
∆H = 2.25 104 kJ
2.25 104 kJ of solar energy is needed to produce 1.00 kg of ammonia.
2.25 104 kJ
(c) area, a = (3.60 103 kJ/m2)
a = 6.25 m2
An area of 6.25 m2 would be needed to produce 1.00 kg of ammonia in one day.
(d) The assumption is made that all of the solar energy will go into the reaction and none will be lost to the
surroundings.
14. 2 NaHCO3(s) → Na2CO3(s) + H2O(g) + 2 CO2(g)
∆H = –74 kJ
∆H = 2 ∆H°f(CO
2(g))
+ 1 ∆H°f(H O
2 (g))
+ 1 ∆H°f(Na CO
2
3(s))
– 2 ∆H°f(NaHCO
3(s))
= 2 (–393.5) + 1 (–241.6) + 1 (–1131) – 2 (–947.7)
∆H = –264 kJ
The enthalpy change for the reaction is –264 kJ.
15
15. C6H5COOH(s) + O2(g) + → 7 CO2(g) + 3 H2O(l)
2
15
∆H = 7 ∆H°f(CO ) + 3 ∆H°f(H O ) – 1 ∆H°f(acid ) – ∆H°f(O )
2(g)
2 (l)
(s)
2(g)
2
15
–3223.6 = 7 (–393.5) + 3 (–285.8) – 1 (∆H°f(acid )) – (0)
(s)
2
1 (∆H°f(acid )) = 7 (–393.5) + 3 (–285.8) + 3223.6 kJ
(s)
∆H°f(acid
(s))
= –388.3 kJ/mol
The molar enthalpy of formation of benzoic acid is –388.3 kJ/mol.
Applying Inquiry Skills
16. (1) CaCO3(s) + 2 HCl(aq) → CO2(g) + H2O(l) + CaCl2(aq)
∆H1 = ? kJ
(2) CaO(s) + 2 HCl(aq) → H2O(l) + CaCl2(aq)
∆H2 = ? kJ
The target equation is
CaCO3(s) → CO2(g) + CaO(s)
∆H = ? kJ
Copyright © 2003 Nelson
Thermochemistry
189
Experimental Design
(a) +1 (1): CaCO3(s) + HCl(aq) → CO2(g) + H2O(l)
–1 (2): H2O(l) + CaCl2(aq) → CaO(s) + HCl(aq)
(1) ∆H1
(–1) ∆H2
_________________________________________________________
CaCO3(s) → CO2(g) + CaO(s)
∆Htarget = ∆H1 – ∆H2
Procedure
(b) • Find the mass of a Styrofoam cup, add about 170 mL of dilute acid, and find the mass of the cup again.
• Find the mass accurately of about 4 g of CaCO3 solid.
• Measure the initial temperature of the acid solution.
• Add the solid to the acid and stir until a maximum temperature is reached. Record this temperature.
• Repeat the previous steps for a new cup, sample of acid, and mass of calcium oxide solid.
Analysis
(c) In all three experiments, assume that the acid solution has the same density as water: 1 g/mL.
Experiment 1:
mass of acid solution, m = 173.2 g – 3.0 g
m = 170.2 g
qwater = mc∆T
= 170.2 g 4.18 J/g•ºC (31.0 – 29.0)°C
qwater = 1.4(2) kJ
n∆H1 = qwater
n∆H1 = 1.4(2) kJ
m
nCaCO = 3
M
1 mol
= 4.2 g CaCO3 100.1 g CaCO3
nCaCO = 0.042 mol
3
qwater
∆H1 = nCaCO
3
1.4(2) kJ
∆H1 = 0.042 mol
∆H1 = 34 kJ/mol CaCO3
Because the reaction is exothermic and is written for one mole of CaCO3, ∆H1 = –34 kJ.
Experiment 2:
mass of acid solution, m = 158.6 g – 3.1 g
m = 155.5 g
qwater = mc∆T
= 155.5 g 4.18 J/g•ºC (36.0 – 29.0)°C
qwater = 4.5(5) kJ
n∆H2 = qwater
n∆H2 = 4.5(5) kJ
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Chapter 5
Copyright © 2003 Nelson
m
nCaO = M
1 mol
= 4.6 g CaO 56.1 g CaO
nCaO = 0.0820 mol
qwater
∆H2 = nCaO
4.5(5) kJ
= 0.0820 mol
∆H2 = 55 kJ/mol CaO
Because the reaction is exothermic and is written for one mole of CaO, ∆H2 = – 55 kJ.
(d) ∆Htarget = ∆H1 – ∆H2
= – 34 kJ – ( – 55 kJ)
∆Htarget = + 19 kJ
(e) (i) Less solid would react, so the observed temperature change, calculated q, and ∆H would be smaller than expected.
(ii) Less heat would go into the water, so the observed temperature change, calculated q, and ∆H would be smaller
than expected.
17. (a) One possibility is to burn a mass of the alcohol in a bomb calorimeter and determine its heat of combustion, which
could be compared to a tabulated value. Another is to dissolve a known mass in water and determine its heat of
solution, which could be compared to a tabulated value.
(b) The boiling point of the liquid could be determined. Various diagnostic organic tests could be used to confirm its
identity as an alcohol. Derivatives could be made of the alcohol (for example, esters) whose melting and boiling
points could be compared to tabulated values.
Making Connections
18. (a) H2O(g) → H2O(l)
∆H = ∆H°f(H O
2 (l))
– ∆H°f(H O
2 (g))
= (–285.8) – (– 241.8)
∆H = –44 kJ
1
(b) H2(g) + O2(g) → H2O(l)
2
∆H°f(H O ) = –285.5 kJ/mol
2 (l)
(c) ∆H fusion = –1.7 109 kJ
(d) ∆H condensation would be about 44/100 cm, or 0.4 cm.
∆H°f(H O
2 (l))
would be about 285.5/100 cm, or 3 cm.
∆H fusion would be about –1.7 109/100 cm, or 108 cm, or 1000 km.
(e) (Answers will vary.) 0.4 g is about the mass of a fingernail; 3 g is about the mass of a teaspoon of sugar; 1000 kg
is about the mass of a small car.
19. Answers will vary, but could include the special demands of geography (Canada is a large country requiring transportation over great distances), temperature (Canada is a cool country requiring heating of homes and businesses), and
the economy (Canada is an economically strong country with heavy industry that demands power). Furthermore,
Canadians enjoy and expect a high standard of living, with many consumer goods.
Copyright © 2003 Nelson
Thermochemistry
191