Chemistry 360
Dr. Jean M. Standard
Problem Set 5 Solutions
1.
Determine the amount of pressure-volume work performed by 1 mole of water freezing to ice at 0°C and
1 atm pressure. The density of liquid water at 0°C is 0.99984 g/mL and the density of ice is 0.91680
g/mL.
Going from liquid water to ice involves a volume change, so PV work must be determined. The definition of
work w is
w = −
∫ Pext dV .
At constant pressure (standard, 1 atm), we can pull the pressure out of the integral and integrate the volume
from the initial to the final value,
€
w = − Pext
V2
∫V
dV
1
w = − Pext (V2 − V1) .
To determine the volumes, we must use the densities given. For 1 mole of liquid water, the volume is
€
" mL %
V1 = (1 mol)(18.015 g/mol)$
'
# 0.99984 g &
V1 = 18.018 mL = 0.018018 L .
For 1 mole of ice, the volume is
€
" mL %
V2 = (1 mol)(18.015 g/mol)$
'
# 0.9168 g &
V2 = 19.650 mL = 0.019650 L .
The work is therefore
€
w = − Pext (V2 − V1)
= − (1 atm)( 0.019650 L − 0.018018 L)
# 101.325 J &
= − 0.001632 L atm %
(
$ 1 L atm '
w = − 0.165 J.
€
2
2.
One of the primary components of nuclear fuel rods is uranium(IV) oxide, UO2. The melting point of
UO2 is 3140 K and the enthalpy of fusion is 70.0 kJ/mol. Determine the enthalpy change per mole for the
process
UO 2 (s, 3100 K )
UO 2 ( ℓ, 3200 K ) .
→
The molar heat capacities are C p,m (s) = 211.0 J mol−1K −1 and C p,m ( ℓ) = 134.2 J mol−1K −1 . Assume 1 bar
pressure and that the molar heat capacities of solid and liquid UO2 are independent of temperature.
For a physical change involving heating/cooling and/or phase transitions, we can break the process up into steps
and then calculate the enthalpy changes for each of the steps. The overall enthalpy change is then just the sum
of the enthalpy changes for the steps since H is a state function. For this particular process, the steps are:
(1) heating solid UO2 from 3100 K to 3140 K,
(2) phase transition from solid to liquid at 3140 K,
(3) heating liquid UO2 from 3140 K to 3200 K,
UO 2 (s, 3100 K )
→ UO 2 (s, 3140 K ) .
UO 2 (s, 3140 K )
→ UO 2 ( ℓ, 3140 K ) .
UO 2 ( ℓ, 3140 K )
→ UO 2 ( ℓ, 3200 K ) .
The overall molar enthalpy change for the process is then just equal to the sum of the molar enthalpy changes
for each of the steps,
ΔH m = ΔH1,m + ΔH 2,m + ΔH 3,m .
In order to calculate the molar enthalpy changes for each of the steps, the molar heat capacities for the liquid
and solid phases are required, along with the molar enthalpy of fusion; these quantities are provided in the
problem.
The steps listed above can be divided into two types: heating steps and phase transition steps. For the heating
steps, we can begin with the exact differential for H,
# ∂H &
dH = C p dT + %
( dP .
$ ∂P 'T
Since the process occurs at constant pressure, dP = 0 , the expression becomes
€
dH = C p dT .
€
Integrating in order to calculate the enthalpy change for the heating process from T1 to T 2 ,
€
ΔH =
T2
∫T
C p dT .
1
€
€
If we assume that the heat capacity is independent of temperature, then we can integrate to get
€
ΔH = C p (T 2 − T1) .
Expressing this relation in terms of molar quantities, we divide both sides by the number of moles to give
€
ΔH m = C p,m (T2 − T1 ) .
3
2.) Continued
For the steps involving phase transitions, the enthalpy change is just the moles of substance multiplied by the
standard molar enthalpy change (in this case, fusion),
ΔH m = ΔH !fus .
Using these expressions, the enthalpy changes for each of the steps may be calculated.
Step 1
In this step, solid UO2 is heated from 3100 K to the transition temperature of 3140 K. Substituting the
appropriate heat capacity value and temperatures, we have
ΔH1,m = C p,m ( UO 2 , s) ⋅ (T2 − T1 )
(
)
= 211.0 J mol–1K –1 (3140 K − 3100 K )
ΔH1 = 8440 J/mol .
Step 2
In this step, UO2 undergoes a transition from solid to liquid at the transition temperature of 3140 K.
Substituting the appropriate standard molar enthalpy change, we have
ΔH 2,m = ΔH !fus
= ( 70.0 kJ/mol)
ΔH 2 = 70000 J/mol .
Step 3
In this step, liquid UO2 is heated from the transition temperature of 3140 K to 3200 K. Substituting the
appropriate heat capacity value and temperatures, we have
ΔH 3,m = C p,m ( UO 2 , ℓ) ⋅ (T2 − T1 )
(
)
= 134.2 J mol–1K –1 (3200 K − 3140 K )
ΔH 3 = 8052 J .
The overall enthalpy change for the process is then just equal to the sum of the enthalpy changes for each of the
steps,
ΔH m = ΔH1,m + ΔH 2,m + ΔH 3,m .
Substituting, we have
ΔH m = ΔH1,m + ΔH 2,m + ΔH 3,m
= 8440 + 70000 + 8052 J/mol
ΔH = 86492 J/mol or 86.49 kJ/mol .
4
3.
Determine the enthalpy change for one mole of water undergoing the following transformation,
H 2O s, – 30! C
(
)
H 2O g, 300! C .
(
→
)
The pressure is constant at 1 bar, and assume that the molar heat capacities are independent of
temperature.
€
For a physical change involving heating/cooling and/or phase transitions, we can break the process up into steps
and then calculate the enthalpy changes for each of the steps. The overall enthalpy change is then just the sum
of the enthalpy changes for the steps. For this particular process, the steps are:
(1)
heating solid water from – 30! C to 0! C,
H 2O s, – 30! C
(
phase transition from solid to liquid at 0! C, H 2O s, 0! C
(
)
→ H 2O ℓ, 0! C .
)
( 3) heating liquid water from 0 C to 100 C, H O (ℓ, 0 C) → H O (ℓ, 100 C) .
( 4) phase transition from liquid to gas at 100 C, H O (ℓ, 100 C) → H O ( g, 100 C) .
( 5) heating gaseous water from 100 C to 300 C, H O ( g, 100 C) → H O ( g, 300 C) .
( 2)
(
→ H 2O s, 0! C .
)
!
!
!
!
2
!
2
!
(
!
2
!
)
!
2
!
2
!
2
The overall enthalpy change for the process is then just equal to the sum of the enthalpy changes for each of the
steps,
€
ΔH = ΔH 1 + ΔH 2 + ΔH 3 + ΔH 4 + ΔH 5 .
In order to calculate the enthalpy changes for each of the steps, the molar heat capacities for all three phases are
required, along with the molar enthalpies of fusion and vaporization. These values are available from standard
€
reference sources such as your textbook, the CRC, or the NIST Webbook, and are tabulated below.
C p,m
Substance
–1
(Jmol K )
H2O (s)
H2O (l)
H2O (g)
36.2
75.3
33.6
€
ΔH R!
Phase change
for H2O
–1
(kJ/mol)
6.01
44.0
fusion
vaporization
The steps listed above can be divided into two types: heating steps and phase transition steps. For the heating
steps, we can begin with the exact differential for H,
# ∂H &
dH = C p dT + %
( dP .
$ ∂P 'T
Since the process occurs at constant pressure, dP = 0 , the expression becomes
€
dH = C p dT .
€
Integrating in order to calculate the enthalpy change for the heating process from T1 to T 2 ,
€
ΔH =
T2
∫T
1
€
C p dT .
€
€
5
3.) Continued
If we assume that the heat capacity is independent of temperature, then we can integrate to get
ΔH = C p (T 2 − T1) .
Expressing this relation in terms of the molar heat capacity, we have
€
ΔH = nC p,m (T 2 − T1) .
For the steps involving phase transitions, the enthalpy change is just the moles of substance multiplied by the
€ the phase transition,
standard molar enthalpy change for
ΔH = nΔH !phase change .
Using these expressions, the enthalpy changes for each of the steps may be calculated.
Step 1
In this step, solid water is heated from –30ºC to the transition temperature of 0ºC. Substituting the appropriate
heat capacity value and temperatures, we have
ΔH 1 = nC p,m ( H 2O, s) ⋅ (T 2 − T1)
(
)
= (1 mol) 36.2 J mol–1K –1 ( 273.15 K − 243.15 K)
ΔH 1 = 1090 J .
Step 2
In this step, water undergoes a transition from solid to liquid at the transition temperature of 0ºC. Substituting
€
the appropriate standard
molar enthalpy change, we have
ΔH 2 = nΔH !fus
= (1 mol)( 6.01 kJ/mol)
ΔH 2 = 6010 J .
Step 3
In this step, liquid water is heated from 0ºC to the transition temperature of 100ºC. Substituting the appropriate
€
heat capacity value and temperatures,
we have
ΔH 3 = nC p,m ( H 2O,ℓ) ⋅ (T 2 − T1)
(
)
= (1 mol) 75.3 J mol–1K –1 ( 373.15 K − 273.15 K)
ΔH 3 = 7530 J .
Step 4
In this step, water undergoes a transition from liquid to vapor at the transition temperature of 100ºC.
€
Substituting appropriate
standard molar enthalpy change, we have
!
ΔH 4 = nΔH vap
= (1 mol) ( 44.0 kJ/mol)
ΔH 4 = 44000 J .
6
3.) Continued
Step 5
Finally, in this step, gaseous water is heated from 100ºC to the final temperature of 300ºC. Substituting the
appropriate heat capacity value and temperatures, we have
ΔH 5 = nC p,m ( H 2O, g ) ⋅ (T 2 − T1)
(
)
= (1 mol) 33.6 J mol–1K –1 ( 573.15 K − 373.15 K)
ΔH 5 = 6720 J .
The overall enthalpy change for the process is then just equal to the sum of the enthalpy changes for each of the
€
steps,
ΔH = ΔH 1 + ΔH 2 + ΔH 3 + ΔH 4 + ΔH 5 .
Substituting, we have
€
ΔH = ΔH1 + ΔH 2 + ΔH 3 + ΔH 4 + ΔH 5
= 1090 + 6010 + 7530 + 44000 + 6720 J
ΔH = 65350 J
or
65.35 kJ .
7
4.
4
Metabolic activity in the human body releases about 1×10 kJ of heat per day. Since the body is made up
of mostly water, approximate a body as consisting of 50 kg of water.
(a) How fast would the body temperature rise (in degrees K/day) if the body is considered to be an
isolated system at constant pressure (1 bar)? The molar heat capacity of liquid water is
C p,m = 75.291J mol−1K−1 .
Since the process occurs at constant pressure,
€
q p = ΔH =
T2
∫T
C p dT .
1
If we assume that the heat capacity of the body (i. e., the 50 kg of water) is independent of temperature,
then we can integrate to get
€
q p = ΔH = C p (T 2 − T1) = nC p,m (T 2 − T1) ,
in terms of the molar heat capacity.
€ heat is produced per day, and we know the number of moles of water in 50 kg is
We are told how much
" 1000 g %" 1 mol %
'$
' = 2775 mol.
# 1 kg &# 18.015 g &
( 50 kg)$
We need to figure out the temperature change. Solving the equation above for the temperature difference,
€
q p = nC p,m ΔT
or ΔT =
qp
nC p,m
.
Substituting,
ΔT =
€q
p
nC p,m
=
1× 10 7 J
( 2775mol)( 75.291J mol−1K−1)
ΔT = 48 K .
Thus, the temperature rise per day is ΔT = 48 K per day.
€
(b) How much water in kg per day must the body eliminate as perspiration to maintain normal body
€
temperature of 98.6°F? Assume that the energy required to vaporize water is 2.41 kJ/g.
4
To maintain a constant body temperature of 98.6°F, the body would have to eliminate 1 × 10 kJ of heat per
day. If this were done by using the heat generated to vaporize water, each g of water would require 2.41 kJ
of heat. Thus, the amount of water vaporized per day would be
1× 10 4 kJ
= 4150 g.
2.41kJ/g
€
In other words, 4.15 kg of water would be vaporized as perspiration every day in order to maintain constant
body temperature!
€
8
5.
Consider the following compounds: NaHCO3 (s) , Na 2CO3 (s) , CO 2 (g) , and H 2O ( ℓ) .
(a) Write out the formation reactions for each of the compounds listed above.
The formation reaction of NaHCO3 (s) is
Na (s) +
1
2
H 2 ( g) + C (s) +
3O
2 2
( g)
( g)
→ Na 2CO 3 (s) .
→ NaHCO 3 (s) .
The formation reaction of Na2CO3 (s) is
€
2 Na (s) + C (s) +
3O
2 2
The formation reaction of CO2 (g) is
€
C (s) + O 2 ( g) → CO 2 ( g) .
The formation reaction of H2O (l) is
€
1O
2 2
H 2 ( g) +
( g)
→ H 2O ( ℓ) .
€
(b) Show that the formation reactions for the compounds listed above may be combined to produce an
overall reaction given by
2 NaHCO3 (s)
→
Na 2CO3 (s) + CO 2 (g) + H 2O ( ℓ) .
Now we have to take the formation reactions from part (a) in the appropriate combination to form the
overall reaction given here. Taking twice the reverse of the first reaction and adding all the others yields
2
[ NaHCO 3 (s)
→ Na (s) +
2 Na (s) + C (s) +
3O
2 2
1
2
( g) →
C (s) + O 2 ( g) → CO 2 ( g)
H 2 ( g) + 12 O 2 ( g) → H 2O ( ℓ)
2 NaHCO 3 (s)
€
→
H 2 ( g) + C (s) +
3O
2 2
( g) ]
Na 2CO 3 (s)
Na 2CO 3 (s) + CO 2 ( g) + H 2O ( ℓ)
9
5.) Continued
(c) Use values of standard enthalpies of formation from the CRC or NIST to calculate ΔH R! in kJ/mol at
25°C for the reaction given in part (b).
The standard enthalpies of formation at 25°C as obtained from the CRC are:
ΔH !f
Compound
(kJ/mol)
NaHCO3 (s)
Na2CO3 (s)
–950.81
€
–1130.77
CO2 (g)
–393.51
H2O (l)
–285.83
The overall enthalpy of reaction is therefore given by
ΔH R! = ΔH !f ( Na 2CO3 ) + ΔH !f (CO 2 ) + ΔH !f ( H 2O) − 2 ΔH !f ( NaHCO3 )
= (−1130.77kJ/mol) + (−393.51kJ/mol) + (−285.83kJ/mol) − 2 (−950.81kJ/mol)
ΔH R! = 91.51kJ/mol.
10
6.
The following may be considered as reactions used to power rockets,
( a)
( b)
( c)
H 2 ( g) +
1O
2 2
CH 3OH ( ℓ) +
( g)
→ H 2O ( g)
3O
2 2
( g)
→ CO 2 ( g) + 2 H 2O ( g)
H 2 ( g) + F2 ( g) → 2 HF ( g) .
(a) Using values from the tables in the Appendix of your textbook, calculate the molar enthalpies of
reaction at 25°C for each of these reactions. Convert your results to units of kJ per total mass of
€
reactants (in kg).
For reaction (a), H 2 ( g) + 12 O 2 ( g) → H 2O ( g) , the enthalpy of reaction is
ΔH R! = ΔH !f ( H 2O, g) − ΔH !f ( H 2 ) − 12 ΔH !f (O 2 )
= − 241.8kJ/mol − 0 − 0
€
ΔH R!
= − 241.8kJ/mol.
Expressing this per kg of reactants, the weight of 1 mole of H2 and 1/2 mole of O2 is 18.015 g, or 0.018015
kg. Thus,
# 1molreactants &
ΔH R! = (−241.8kJ/mol) %
(
$ 0.018015kg '
ΔH R! = −13420 kJ/kg reactants.
For reaction (b), CH 3OH ( ℓ) + 23 O 2 ( g) → CO 2 ( g) + 2 H 2O ( g) , the enthalpy of reaction is
ΔH R! = ΔH !f (CO 2 ) + 2 ΔH !f ( H 2O, g) − ΔH !f (CH 3OH ) − 23 ΔH !f (O 2 )
€
= − 393.5kJ/mol + 2 (−241.8kJ/mol) − (−239.2 kJ/mol) − 0
ΔH R! = − 637.9 kJ/mol.
Expressing this per kg of reactants, the weight of 1 mole of CH3OH and 3/2 mole of O2 is 80.041 g, or
0.080041 kg. Thus,
# 1molreactants &
ΔH R! = (−637.9 kJ/mol) %
(
$ 0.080041kg '
ΔH R! = − 7970 kJ/kg reactants.
For reaction (c), H 2 ( g) + F2 ( g) → 2 HF ( g) , the enthalpy of reaction is
ΔH R! = 2 ΔH !f ( HF ) − ΔH !f ( H 2 ) − ΔH !f ( F2 )
€
= 2 (−273.3kJ/mol) − 0 − 0
ΔH R! = − 546.6 kJ/mol.
11
6
a). Continued
Expressing this per kg of reactants, the weight of 1 mole of H2 and 1 mole of F2 is 40.012 g, or 0.040012
kg. Thus,
# 1molreactants &
ΔH R! = (−546.6 kJ/mol) %
(
$ 0.040012 kg '
ΔH R! = −13661kJ/kg reactants.
(b) Since the thrust in a rocket is greater when the molar mass of the exhaust gas is lower, divide the
absolute enthalpy per kilogram by the average molar mass of the products. Arrange the reactions in
order of effectiveness on the basis of thrust. What is the most effective reaction?
The thrust T is the absolute enthalpy per kilogram divided by the average molecular weight of the products.
For reaction (a), H 2 ( g) + 12 O 2 ( g) → H 2O ( g) , the thrust T is
T=
€
=
ΔH R! ( kJ/kg)
M products
−13420 kJ/kg
0.001801kg/mol
T = 7.451×10 6 kJ kg−2 mol.
For reaction (b), CH 3OH ( ℓ) + 23 O 2 ( g) → CO 2 ( g) + 2 H 2O ( g) , the thrust T is
€
T=
=
ΔH R! ( kJ/kg)
M products
1
3
−7970 kJ/kg
0.04401kg/mol
(
) + 23 (0.01801kg/mol)
T = 2.988 ×10 5 kJ kg−2 mol.
For reaction (c), H 2 ( g) + F2 ( g) → 2 HF ( g) , the thrust T is
€
T=
=
ΔH R! ( kJ/kg)
M products
−13661kJ/kg
0.02001kg/mol
T = 6.827 ×10 5 kJ kg−2 mol.
Therefore, the reactions in order of largest magnitude of thrust are: a > c > b. Of the three, reaction (a) is
the most effective for providing rocket thrust by this measure.
12
7.
Consider the reactions
(a )
( b)
( c)
C (graphite) + O 2 (g) → CO 2 (g)
H 2 (g ) +
1O
2 2
(g )
→ H 2 O ( ℓ)
ΔH R! = −393.5kJ/mol
ΔH R! = −285.5kJ/mol
2 C2 H 6 (g) + 7O 2 (g) → 4CO 2 (g) + 6 H 2O ( ℓ)
ΔH R! = −3119.6 kJ/mol.
Using these reactions in conjunction with Hess' Law to determine the enthalpy of formation of ethane.
The formation reaction of ethane is
2C (s, graphite) + 3H 2 ( g) → C 2 H 6 ( g) .
Note that reaction (a) above contains graphite as a reactant; the reaction must be doubled, though, since the
formation reaction involves 2 moles of graphite. Also, reaction (b) contains hydrogen gas as a reactant; this
€
reaction must be multiplied by a factor of 3 to match the molar ratio needed in the formation reaction. Finally,
reaction (c) involves ethane, but as a reactant; thus, the reverse of reaction (c) is required, multiplied by 1/2 to
get the correct stoichiometry:
2 [C ( graphite) + O 2 ( g) → CO 2 ( g)]
[
]
3 H 2 ( g) + 12 O 2 ( g) → H 2O ( ℓ)
1
2
[4 CO 2 (g)
+ 6 H 2O ( ℓ) → 2C 2 H 6 ( g) + 7O 2 ( g)]
2C ( graphite) + 3H 2 ( g) → C 2 H 6 ( g)
From the combination of reactions, the enthalpy of formation of ethane can be calculated from Hess' Law,
€
ΔH !f = 2 ΔH R! ( a ) + 3ΔH R! ( b) − 12 ΔH R! ( c )
= 2 (−393.5kJ/mol) + 3 (−285.5kJ/mol) − 12 (−3119.6 kJ/mol)
ΔH !f = − 83.7kJ/mol.
13
8.
The combustion of 0.4862 g naphthalene in a constant volume bomb calorimeter caused a temperature
rise in the water bath of 1.707°C. The final temperature was 298.00 K. The heat capacity of the
calorimeter was 10290 J/K.
(a) Determine the molar internal energy of combustion of naphthalene and its molar enthalpy of
combustion.
Using the bomb calorimeter equation developed in class,
ΔU comb = − C v,cal ΔT
= − (10290 J/K)(1.707 K)
ΔU comb = − 17565J .
To get the molar internal energy of combustion, we divide by the number of moles of naphthalene.
€
ΔU comb
n
−17565 J
=
$ 0.4862 g '
&
)
% 128.174 g/mol (
ΔU comb,m =
= − 4.6306 × 10 6 J/mol
or ΔU comb,m = −4.6306 × 10 3 kJ/mol.
Finally, to get the molar enthalpy of combustion, we use
€
ΔH comb,m = ΔU comb,m + Δn gas RT .
For this conversion, the balanced reaction for the combustion of naphthalene is required,
€
C 10 H 8 (s) + 12 O 2 ( g) → 10 CO 2 ( g) + 4 H 2O ( ℓ) .
From the balanced reaction, the change in number of moles of gas is
€
Δn gas = 10 − 12
Δn gas = − 2 .
Substituting,
ΔH comb,m = ΔU€
comb,m + Δn gas RT
(
)
= − 4.6306 × 10 6 J mol−1 + (−2) 8.314 J mol−1K−1 ( 298.00 K)
ΔH comb,m = − 4.6356 × 10 6 J mol−1
ΔH comb,m = − 4.6356 × 10 3 kJ mol−1 .
€
14
8.
Continued
(b) Using the results from part (a), calculate the enthalpy of formation of naphthalene.
The formation reaction of naphthalene is
10C ( graphite) + 4 H 2 ( g) → C 10 H 8 (s) .
Since naphthalene appears on the product side of the formation reaction, we will have to somehow use the
reverse of the combustion reaction. The reverse of the combustion reaction is
€
10 CO 2 ( g) + 4 H 2O ( ℓ) → C 10 H 8 (s) + 12 O 2 ( g) .
We can use this to construct the formation reaction, along with the formation reactions for water and
carbon dioxide,
€
10 CO 2 ( g) + 4 H 2O ( ℓ) → C 10 H 8 (s) + 12 O 2 ( g)
( g) → H 2O ( ℓ)]
10 [C ( graphite) + O 2 ( g) → CO 2 ( g)]
10 C ( graphite) + 4 H 2 ( g) → C 10 H 8 (s)
4
[H 2 (g)
+
1O
2 2
Thus, we see that the enthalpy of formation of naphthalene is equal to 4 times the enthalpy of formation of
water, plus 10 times the enthalpy of formation of carbon dioxide, minus the enthalpy of combustion of
naphthalene, €
ΔH !f (C 10 H 8 ) = 4 ΔH !f ( H 2O) + 10 ΔH !f (CO 2 ) − ΔH comb,m (C 10 H 8 )
= 4(−285.8 kJ/mol) + 10 (−393.5 kJ/mol)
− (−4635.6 kJ/mol)
ΔH !f
€
(C10 H 8 )
= − 442.6 kJ/mol.
15
9.
A sample of solid KNO3 (1.668 g) is dissolved in 100 mL of water initially at 24.85°C in a constant
pressure solution calorimeter. The final temperature was 23.76°C. Assume the constant pressure heat
capacity of the calorimeter is 101.3 cal/K. Determine the molar enthalpy of solution of KNO3.
Using the calorimeter equation for solution calorimetry,
ΔH soln = − C p,cal ΔT
= − (101.3cal/K)( 23.76 − 24.85K)
ΔH soln = 110.4 cal.
To get the molar enthalpy of solution, we divide by the number of moles of KNO3,
€
ΔH soln
n
110.4 cal
=
# 1.668 g &
%
(
$ 101.10 g/mol '
ΔH soln,m =
= 6692 cal/mol
or ΔH soln,m = 6.69 kcal/mol.
€
16
10. Using the values in the table below, reported at 25°C, determine ΔH R! at 500ºC for the reaction
CH 3OH ( ℓ) → CH 4 ( g) +
1
O
2 2
( g)
.
Assume that C p,m is independent of temperature for all species in the reaction.
€
€
CH 3OH ( ℓ)
CH4 (g)
O2(g)
ΔH !f
C p,m
(kJ/mol)
–238.66
(Jmol–1K–1)
€
81.6
€
74.81
0
35.31
29.36
€
The first step is to determine the enthalpy of reaction at 298 K:
ΔH R! = ΔH !f (CH 4 ) + 12 ΔH !f (O 2 ) − ΔH !f (CH 3OH )
= 74.81kJ/mol + 12 ( 0 ) − (−238.66 kJ/mol)
ΔH R! = 313.47kJ/mol.
The temperature dependence of the molar enthalpy of reaction is given by
ΔH R! (T2 ) = ΔH R! (T1 ) + ΔC p,m ⋅ (T2 − T1 ).
This equation assumes that the molar heat capacities are independent of temperature. In the equation, ΔC p,m is
the difference in molar heat capacities between products and reactants with the appropriate signed
stoichiometric coefficients included,
€
ΔC p,m =
∑ν C
i
p,m
(i )
.
i
For this reaction, the heat capacity difference is
€
ΔC p,m =
ν iC p,m (i )
∑
i
= C p,m (CH 4 ) +
= 35.31 +
1
2
1
C
2 p,m
( 29.36)
(O 2 )
− C p,m (CH 3OH)
− 81.6 Jmol−1K −1
ΔC p,m = − 31.61 Jmol−1K −1.
Substituting, the molar enthalpy of reaction at 500°C (773 K) is
€
ΔH R! ( 773 K ) = ΔH R! ( 298 K ) + ΔC p,m ⋅ ( 773 − 298 K )
$ 1 kJ '
= 313.47 kJ/mol + −31.61 Jmol−1K −1 ( 773 − 298 K ) &
)
% 1000 J (
= 313.47 - 15.01 kJ/mol
(
ΔH R! ( 773 K ) = 298.46 kJ/mol .
)
17
11. Calculate the standard molar enthalpy change at 600 K for the reaction
N 2 ( g) + 3H 2 ( g) → 2 NH 3 ( g) .
The standard molar enthalpies of formation and molar heat capacities at 25°C are listed below. Assume
that the molar heat capacities are independent of temperature.
€
N2 (g)
H2 (g)
NH3 (g)
€
ΔH !f
C p,m
(kJ/mol)
0
0
–45.9
(Jmol–1K–1)
29.1
28.8
35.1
€
First, the enthalpy of reaction at 298 K is calculated,
ΔH R! = 2ΔH !f ( NH 3 ) − ΔH !f ( N 2 ) − 3ΔH !f ( H 2 )
= 2 (−45.9 kJ/mol) − 0 − 3 ( 0 )
ΔH R!
= − 91.8kJ/mol.
The temperature dependence of the molar enthalpy of reaction is given by
ΔH R! (T2 ) = ΔH R! (T1 ) + ΔC p,m ⋅ (T2 − T1 ).
For this reaction, the heat capacity difference is
ΔC p,m = ∑ν i C p,m (i )
i
= 2C p,m ( NH 3 ) − C p,m ( N 2 ) − 3C p,m ( H 2 )
= 2 (35.1) − 29.1 − 3 ( 28.8) Jmol−1K −1
ΔC p,m = −45.3 Jmol−1K −1.
Substituting, the molar enthalpy of reaction at 600 K is
ΔH R! ( 600 K ) = ΔH R! ( 298 K ) + ΔC p,m ⋅ ( 600 − 298 K )
$ 1 kJ '
= −91.8 kJ/mol + −45.3 Jmol−1K −1 ( 600 − 298 K ) &
)
% 1000 J (
= −91.8 − 13.7 kJ/mol
(
ΔH R! ( 773 K ) = −105.5 kJ/mol .
)