Download Document

Chapter 4
Thermochemistry
Physical Chemistry 2nd Edition
Thomas Engel, Philip Reid
Objectives
• Discussion of Hess’s Law.
• Derive property that allows ∆H and ∆U to be
calculated without experiments.
Chapter 4: Thermochemistry
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Outline
1. Energy Stored in Chemical Bonds Is Released or
Taken Up in Chemical Reactions
2. Internal Energy and Enthalpy Changes Associated
with Chemical Reactions
3. Hess’s Law Is Based on Enthalpy Being a State
Function
4. The Temperature Dependence of Reaction
Enthalpies
5. The Experimental Determination of ∆U and ∆H for
Chemical Reactions
6. Differential Scanning Calorimetry
Chapter 4: Thermochemistry
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
4.1 Energy Stored in Chemical Bonds Is Released or Taken Up in Chemical
Chemical
Reactions
• The change in enthalpy or internal energy
due to temperature result in heat flow and/or
in the form of expansion or non-expansion
work.
• The focus is on using
measurements of heat
flow to determine
changes in U and H
due to chemical
reactions.
Chapter 4: Thermochemistry
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
4.2 Internal Energy and Enthalpy Changes Associated with Chemical
Chemical
Reactions
• The enthalpy of reaction, ∆H°reaction, is
the heat withdrawn from the surroundings as
the reactants are transformed into products
at constant T and P.
• The standard enthalpy of formation,
∆Hf°, is the enthalpy associated with the
reaction of 1 mol of the species under
standard state conditions.
• The enthalpy change associated with
reaction
Chapter 4: Thermochemistry
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
4.2 Internal Energy and Enthalpy Changes Associated with Chemical
Chemical
Reactions
• The enthalpy change associated with a
chemical reaction is
∆ H °reaction = ∑ vi ∆H ° f , i
i
Chapter 4: Thermochemistry
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
4.3 Hess’s Law Is Based on Enthalpy Being a State Function
• Hess’s law states that the enthalpy change
for any sequence of reactions that sum to
the same overall reaction is identical.
• It is useful to have tabulated values of ∆Hf°
for chemical compounds at one fixed
combination of P and T.
• ∆H°reaction can then be calculated for all
reactions among these compounds at the
tabulated values of P and T.
Chapter 4: Thermochemistry
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Example 4.1
The average bond enthalpy of the O-H bond
in water is defined as one-half of the enthalpy
change for the reaction H2O(g) → 2H(g) O(g).
The formation enthalpies, ∆H °f , for H(g) and
O(g) are 218.0 and 249.2 kJ mol-1, respectively,
°
∆
H
at 298.15 K, and
f for H2O(g) is 241.8 kJ
mol-1 at the same temperature.
Chapter 4: Thermochemistry
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Example 4.1
a. Use this information to determine the
average bond enthalpy of the O-H bond in
water at 298.15 K.
b. Determine the average bond energy, ∆U,
of the O-H bond in water at 298.15 K.
Assume ideal gas behavior.
Chapter 4: Thermochemistry
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Solution
a. We consider the sequence
H 2O( g ) → H(g) + 1 2 O 2 ( g )
∆H ° = 241.8kJmol −1
H 2 ( g ) → 2H(g)
∆H ° = 2 × 218.0kJmol−1
1 2 O 2 ( g ) → O(g)
∆H ° = 249.2 kJmol−1
________________________________________________
H 2 )( g ) → 2H(g) + O (g)
Chapter 4: Thermochemistry
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
∆H ° = 927.0kJmol−1
Solution
a. This is the enthalpy change associated with
breaking both O-H bonds under standard
conditions. We conclude that the average bond
enthalpy of the O-H bond in water is
1
× 927.0 = 463. 5kJmol−1
2
We emphasize that this is the average value
because the values of for the transformations
H2O(g) → H(g) OH(g ) and OH(g ) → O(g) H(g)
differ.
Chapter 4: Thermochemistry
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Solution
b.
∆U ° = ∆H ° − ∆(PV ) = ∆H ° − ∆nRT
= 927.0 − 2 × 8.314 × 298.15 = 922.0kJmol −1
The average value for ∆U ° for the O-H bond in
1
water is 2 × 922.0 = 461.0kJmol . The bond energy and
the bond enthalpy are nearly identical.
−1
Chapter 4: Thermochemistry
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
4.3 Hess’s Law Is Based on Enthalpy Being a State Function
• Values of bond energies
tabulated in the format
of the periodic table
together with the
electro- negativities
are shown.
Chapter 4: Thermochemistry
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
4.4 The Temperature Dependence of Reaction Enthalpies
• ∆H°T at elevated temperature is used to
determine a reaction to be endothermic or
exothermic.
• The enthalpy for each reactant and product
at temperature T relating to the value at
298.15 K is
T
∆H °T = ∆H ° 298 .15 K +
∫ ∆C (T ')dT '
p
298.15K
where
∆C p (T ') = ∑ vi C p ,i
Chapter 4: Thermochemistry
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
i
Example 4.2
Calculate the enthalpy of formation of HCl(g) at 1450 K
and 1 bar pressure given that ∆H ° f ( HCl, g ) = −92.3kJmol−1
at 298.15 K and that
over this temperature range. The ratios T/K and T2/K2
appear in these equations in order to have the right
units for the heat capacity.
Chapter 4: Thermochemistry
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Solution
By definition, the formation reaction is written
as
1 2 H 2(g) + 1 2 Cl 2(g) → HCl( g ) and
1450
∆H °1450 K = ∆H ° 298.15 K +
∫ ∆C° (T )dT
P
298 .15

28.165 + 1.809 × 10 −3 + 15.464 × 10− 7

1

∆C° P (T ) = − 29.064 − 0.8363 × 10 −3 + 20.111 ×10 − 7
 2
 1
− 31.695 + 10.143 ×10 −3 + 40.73 × 10− 7
 2
(
(
(
)
)
= − 2.215 − 2.844 × 10 −3 + 25.595 × 10− 7 JK −1mol −1
Chapter 4: Thermochemistry
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd








)
Solution
Solution:
∫ (− 2.215 − 2.844 ×10
1450
∆H °1450 K = −92.3 +
298. 15
= −92.3 − 2 .836 = −95 .1kJmol −1
Chapter 4: Thermochemistry
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
−3
+ 25.595 ×10
−7
)
T 
× d 
K
4.5 The Experimental Determination of ΔU and ΔH for
Chemical Reactions
• ∆U°reaction can be determined through
experiment in a bomb calorimeter, which
carried out in constant volume.
m
∆ U ° = s ∆ U ° reaction
Ms
,m
+
mH20
M
Chapter 4: Thermochemistry
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
H20
∆ T + C calorimete r ∆ T = 0
Example 4.3
When 0.972 g of cyclohexane undergoes complete
combustion in a bomb calorimeter, ∆T of the inner
water bath is 2.98°C. For cyclohexane,∆U ° reaction, m is
-3913 kJ mol-1. Given this result, what is the value for
∆U reavction ,m for the combustion of benzene if ∆T is
2.36°C when 0.857 g of benzene undergoes complete
combustion in the same calorimeter? The mass of the
water in the inner bath is 1.812 × 103 g, and the CP,m
of water is 75.291 J K-1 mol-1
Chapter 4: Thermochemistry
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Solution
To calculate the calorimeter constant through
the combustion of cyclohexane,
Ccalorimeter =
mH 2 O
ms
−
∆U °reaction −
C H 2 O ∆T
Ms
M H 2O
∆T
3
0.972
1
.
812
×
10
× 3919 ×103 −
× 75.291 × 2.98°C
18.02
= 84.16
2.98°C
= 7.59 ×10
3
Chapter 4: Thermochemistry
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
−1
(
)
J °C
Solution
In calculating
∆U ° reaction
for benzene,
 mH 2 O


∆U °reaction
C H 2O ∆T + C calorimeter ∆T 
 MH O

2



78.12  1.812 ×10 3

=−
×
× 75.291× 2.36 
0.857  18.02

M
=− s
ms
= − 3.26 × 106 Jmol −1
Chapter 4: Thermochemistry
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
4.5 The Experimental Determination of ΔU and ΔH for
Chemical Reactions
• When ∆U°reaction is obtained, we can get
∆H °reaction = ∆U °reaction + ∆nRT
∆n = number of moles of gas change in
reaction
• For constant pressure calorimetry involving
the solution of a salt in water:
∆H ° reaction
mH 2 0
ms
=
∆H ° solution ,m +
C H 2 O ,m ∆T + Ccalorimete r ∆T = 0
Ms
M H20
Chapter 4: Thermochemistry
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
4.5 The Experimental Determination of ΔU and ΔH for
Chemical Reactions
• For constant pressure calorimetry involving
the solution of a salt in water:
∆H ° reaction
mH 2 0
ms
=
∆H ° solution ,m +
C H 2 O ,m ∆T + Ccalorimete r ∆T = 0
Ms
M H20
Chapter 4: Thermochemistry
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Example 4.4
The enthalpy of solution for the reaction
is determined in a constant pressure calorimeter.
The calorimeter constant was determined to be 342.5
J K-1. When 1.423 g of Na2SO4 is dissolved in 100.34
g of H2O(l), ∆T = 0.037 K. Calculate ∆H m for Na2SO4
from these data. Compare your result with that
calculated using the standard enthalpies of formation
in Table 4.1 (Appendix B, Data Tables) and in Chapter
10 in Table 10.1.
Chapter 4: Thermochemistry
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Solution
 mH 2O


∆H ° solution, m
C H 2O ∆T + C calorimeter ∆T 
 MH O


2

142.04  100.34

=
×
× 75.3 × 0.037 + 342.5 × 0.037 
1.423  18.02

= −2.8 × 103 Jmol −1
M
=− s
ms
(
)
(
)
∆H ° solution,m = 2∆H ° f Na + , aq + ∆H ° f SO42− , aq − ∆H ° f (Na2 SO4 , s )
= 2 × (− 240 .1) − 909.3 + 1387 .1
= −2.4kJmol −1
The agreement between the calculated and
experimental results is satisfactory.
Chapter 4: Thermochemistry
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
4.6 Differential Scanning Calorimetry
• Enthalpy of fusion ∆Hfusion of a dozen
related solid materials is determined by
differential scanning calorimetry.
• The temperatures of each of the samples
and the reference are measured with a
thermocouple.
Chapter 4: Thermochemistry
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
4.6 Differential Scanning Calorimetry
• ∆Hfusion can be ascertained by determining
the area under the curve from the output
data.
Chapter 4: Thermochemistry
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd