(b)
The empirical molar mass of CH is 13.02 g/mol.
molar mass
76 g
5.8
6
empirical molar mass
13.02 g
=
=
≈
Therefore, the molecular formula is C
6
H
6
, and the hydrocarbon is benzene.
The combustion reaction is:
2C
6
H
6
(
l
)
+
15O
2
(
g
)
→
12CO
2
(
g
)
+
6H
2
O(
l
)
17.55 kJ of heat is released when 0.4196 g of the hydrocarbon undergoes combustion.
We can now
calculate the enthalpy of combustion (
rxn
°
Δ
H
) for the above reaction in units of kJ/mol.
Then, from the
enthalpy of combustion, we can calculate the enthalpy of formation of C
6
H
6
.
6
6
6
6
6
6
6
6
78.11 g C H
17.55 kJ
2 mol C H
6534 kJ/mol
0.4196 g C H
1 mol C H

×
×
= 
rxn
f
2
f
2
f
6
6
(12)
(CO )
(6)
(H O)
(2)
(C H )
°
°
°
°
Δ
=
Δ
+
Δ

Δ
H
H
H
H
f
6
6
6534 kJ/mol
(12)( 393.5 kJ/mol)
(6)( 285.8 kJ/mol)
(2)
(C H )
°

=

+


Δ
H
f
6
6
(C H )
°
Δ
=
49 kJ/mol
H
6.141
If the body absorbs all the heat released and is an isolated system, the temperature rise,
Δ
t
, is:
q
=
ms
Δ
t
7
1.0
10
J
(50,000 g)(4.184 J/g
C)
×
=
=
=
⋅°
48 C
q
ms
Δ
°
Δ
°
°
°
t
If the body temperature is to remain constant, the heat released by metabolic activity must be used for the
evaporation of water as perspiration, that is,
4
2
1 g H O
(1.0
10
kJ)
2.41 kJ
×
×
=
3
2
4.1
10 g H O
×
Assuming that the density of perspiration is 1 g/mL, this mass corresponds to a volume of 4.1 L.
The actual
amount of perspiration is less than this because part of the body heat is lost to the surroundings by convection
and radiation.
CHAPTER 6:
THERMOCHEMISTRY
223
6.142
(a)
Heating water at room temperature to its boiling point.
(b)
Heating water at its boiling point.
(c)
A chemical reaction taking place in a bomb calorimeter (an isolated system) where there is no heat
exchange with the surroundings.
6.143
Begin by using Equation (6.20) of the text,
Δ
H
soln
=
U
+
Δ
H
hydr
, where
U
is the lattice energy.
(1)
Na
+
(
g
)
+
Cl

(
g
)
→
Na
+
(
aq
)
+
Cl

(
aq
)
Δ
H
hydr
=
(4.0

788) kJ/mol
=

784.0 kJ/mol
(2)
Na
+
(
g
)
+
I

(
g
)
→
Na
+
(
aq
)
+
I

(
aq
)
Δ
H
hydr
=
(

5.1

686) kJ/mol
=

691.1 kJ/mol
(3)
K
+
(
g
)
+
Cl

(
g
)
→
K
+
(
aq
)
+
Cl

(
aq
)
Δ
H
hydr
=
(17.2

699) kJ/mol
=

681.8 kJ/mol
Adding together equation (2) and (3) and then subtracting equation (1) gives the equation for the hydration of KI.
(2)
Na
+
(
g
)
+
I

(
g
)
→
Na
+
(
aq
)
+
I

(
aq
)
Δ
H
=

691.1 kJ/mol
(3)
K
+
(
g
)
+
Cl

(
g
)
→
K
+
(
aq
)
+
Cl

(
aq
)
Δ
H
=

681.8 kJ/mol
(1)
Na
+
(
aq
)
+
Cl

(
aq
)
→
Na
+
(
g
)
+
Cl

(
g
)
Δ
H
=
+
784.0 kJ/mol
K
+
(
g
)
+
I

(
g
)
→
K
+
(
aq
)
+
I

(
aq
)
Δ
H
=

588.9 kJ/mol
We combine this last result with the given value of the lattice energy to arrive at the heat of solution of KI.
Δ
H
soln
=
U
+
Δ
H
hydr
=
(632 kJ/mol

588.9 kJ/mol)
=
43 kJ/mol
6.144
A
→
B
w
=
0, because
Δ
V
=
0
B
→
C
w
=

P
Δ
V
=

(2 atm)(2

1)L
=

2 L
⋅
atm
C
→
D
w
=
0, because
Δ
V
=
0
D
→
A
w
=

P
Δ
V
=

(1 atm)(1

2)L
=
+
1 L
⋅
atm
The total work done
=
(

2 L
⋅
atm)
+
(1 L
⋅
atm)
=

1 L
⋅
atm
Converting to units of joules,
101.3 J
1 L atm
1 L atm

⋅
×
=
⋅
101.3 J

In a cyclic process, the change in a state function must be zero.
We therefore conclude that work is not a
state function.
Note that the total work done equals the area of the enclosure.
6.145
C (graphite)
→
C (diamond)
H
=
U
+
PV
Δ
H
=
Δ
U
+
P
Δ
V
Δ
H

Δ
U
=
P
Δ
V
The pressure is 50,000 atm.
From the given densities, we can calculate the volume in liters occupied by one
mole of graphite and one mole of diamond.
Taking the difference will give
Δ
V
.
We carry additional
significant figures throughout the calculations to avoid rounding errors.
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 Spring '16
 Chemistry, Thermodynamics, Enthalpy, Thermochemistry, mol Ag, Standard enthalpy change of formation, kJ/mol, ∆H rxn