Download Chapter 8 - Clayton State University

Chapter 8
Thermochemistry
Thermodynamics
Study of the changes in energy and transfers of
energy that accompany chemical and physical
processes.
 address 3 fundamental questions
Will two (or more) substances react when they are
mixed under specified conditions?
If they do react, what energy changes and transfers
are associated with their reaction?
If a reaction occurs, to what extent does it occur?
The First Law of Thermodynamics
Exothermic reactions
combustion of propane
C 3 H 8 g  + 5 O 2 g   3 CO 2 g  + 4 H 2 O  l  + 2.22  10 kJ
3
combustion of n-butane
2 C4H10g + 13 O2g  8CO2g +10 H2Ol  + 5.78  10 kJ
3
The First Law of Thermodynamics
Exothermic
reactions release
specific amounts of
heat as products
Potential energies
of products are
lower than potential
energies of
reactants.
The First Law of Thermodynamics
2 basic ideas of importance
systems tend toward a state of minimum potential energy
a. H 2 O flows downhill
b. objects fall when dropped
c. E potential  mgh
d. E potential  mg h
The First Law of Thermodynamics
2 basic ideas of importance
systems tend toward a state of maximum disorder
a. mirror shatters when dropped
b. easy to scramble an egg
c. food coloring disperses in water
The First Law of Thermodynamics
Also known as Law of Conservation of Energy
The total amount of energy in the universe is
constant.

Energy can be converted from one form to another but
cannot be created.
Some Thermodynamic Terms
System - substances involved in the chemical and
physical changes under investigation

for us this is what is happening inside the beaker
Surroundings - rest of the universe

outside the beaker
Universe - system plus surroundings
Some Thermodynamic Terms
Thermodynamic State of a System - set of conditions that
describe and define the system




number of moles of each substance
physical states of each substance
temperature
pressure
State Functions - properties of a system that depend only on
the state of the system

capital letters
Some Thermodynamic Terms
state functions are independent of pathway

climbing a mountain, taking two different paths
E1 = energy at bottom of mountain
E1 = mgh1
E2 = energy at top of mountain
E2 = mgh2
E  E2- E1 = mgh2 - mgh1 = mg(h)
Some Thermodynamic Terms
Properties that depend only on values of state
functions are also state functions

examples:
•
•
•
T
P
V
Enthalpy Change, H
Commonly, chemistry is done at constant pressure

open beakers on a desk top are at atmospheric pressure
H - enthalpy change
change in heat content at constant pressure
H = qp
Hrxn - heat of reaction
Hrxn = Hproducts - Hreactants
Hrxn = Hsubstances produced - Hsubstances consumed
Calorimetry
coffee-cup
calorimeter - used to
measure the amount
of heat produced (or
absorbed) in a
reaction at constant P

measures qP
Calorimetry
exothermic reaction - heat evolved by reaction is
determined from the temperature rise of the solution

2 part calculation
 Amount of heat
  Amount of heat
  Amount of heat 

 
 

 released by reaction  gained by calorimeter  gained by solution
Amount of heat gained by calorimeter is the heat capacity
of the calorimeter or calorimeter constant

value determined by adding a specific amount of heat to
calorimeter and measuring T rise
Calorimetry
When 3.425 kJ of heat is added to a calorimeter containing
50.00 g of water the temperature rises from 24.000C to
36.540C. Calculate the heat capacity of the calorimeter in
J/0C. The specific heat of water is 4.184 J/g 0C.
Four part calculation
Calorimetry
 Find the temperature change
T = 36.54 - 24.00 C = 12.54 C
0
0
 Find the heat absorbed by the water in going from 24.000C to
36.540C.
q P  mCT

= 50.00 g 4184
.
J
12.54 C
0
0
g C
 262337
. J  2623 J
Calorimetry
 Find the heat absorbed by the calorimeter.
total amount of heat added to calorimeter - heat absorbed by water
3.425 kJ = 3425 J
3425 J
- 2623 J  = 802 J
 Find the heat capacity of the calorimeter
(heat absorbed by the calorimeter)/(temperature change)
802 J
J 0  64.00 J 0

63
.
955
0
C
C
12.54 C
Calorimetry
A coffee-cup calorimeter is used to determine the
heat of reaction for the acid-base neutralization
CH3COOHaq + NaOHaq  NaCH3COOaq + H2Ol 
When we add 25.00 mL of 0.500 M NaOH at
23.0000C to 25.00 mL of 0.600 M CH3COOH
already in the calorimeter at the same temperature,
the resulting temperature is observed to be
25.9470C.
Calorimetry
The heat capacity of the calorimeter has previously
been determined to be 27.8 J/0C. Assume that the
specific heat of the mixture is the same as that of
water, 4.18 J/g0C and that the density of the mixture
is 1.02 g/mL.
Two part calculation:
a
Calculate the amount of heat given off in the reaction.
Calorimetry
temperature change
T = 25.947 - 23.000 C = 2.947 C
0
0
heat absorbed by calorimeter


q = 2.947 C 27.8
0
J0
C
  819. J
mass of solution in calorimeter
102
. g
 510
. g
25.00 mL + 25.00 mL
mL
Calorimetry
heat absorbed by solution
q = mCT

q = 51.0 g  418
.
J
2.947 C  628 J
0
g0 C
total amount of heat produced by reaction
q = 81.9 J + 628 J = 709.9 J
Calorimetry
b
Determine H for the reaction under the conditions of the
experiment.
•
•
must determine the number of moles of reactants consumed
limiting reactant calculation
Calorimetry
C H 3C O O H
aq 
+ N aO H
aq 
 N a C H 3C O O
aq 
+ H 2O
0 .5 0 0 m m o l N a O H 
 
 2 5 . 0 0 m L N a O H  

1 m L N aO H
 1 m m o l N a C H 3 C O O   1 2 .5 m m o l N a C H C O O


3


1 m m ol N aO H
 2 5 .0 0
 0 .6 0 0 m m o l C H 3 C O O H 
m L C H 3C O O H 
 


1 m L C H 3C O O H
 1 m m o l N a C H 3C O O 

  1 5 .0 m m o l N a C H 3 C O O
 1 m m o l C H 3C O O H 
l
Calorimetry

finally, calculate H based on the limiting reactant calculation
H rxn
12.5 mmol = 0.0125 mol
709.9 J

 56792 J / mol  56.8 kJ / mol
0.0125 mol
Thermochemical Equations
Thermochemical equations are a balanced chemical
reaction plus the H value for the reaction.

for example:
C 5 H12 l   8 O 2 g   5 CO 2 g  + 6 H 2 O  l   3523 kJ
1 mol
8 mol
5 mol
6 mol
coefficients in thermochemical equations must be
interpreted as numbers of moles
1 mol of C5H12 reacts with 8 mol of O2 to produce 5 mol
of CO2, 6 mol of H2O, and releasing 3523 kJ is referred to
as one mole of reactions
Thermochemical Equations
Equivalent method of writing thermochemical
equations
C5 H12l   8 O2g  5 CO2g + 6 H2Ol  H = -3523 kJ
H < 0 designates an exothermic reaction
H > 0 designates an endothermic reaction
Thermochemical Equations
Write the thermochemical equation for the previous
reaction.
CH 3COOH  aq  + NaOH  aq   NaCH 3COO  aq  + H 2 O  l  H = -56.8 kJ / mol
Standard States &
Standard Enthalpy Changes
Thermochemical standard state conditions
T = 298.15 K
P = 1.0000 atm
Thermochemical standard states
pure substances in their liquid or solid phase - standard state is the
pure liquid or solid
gases - standard state is the gas at 1.00 atm of pressure
•
gaseous mixtures - partial pressure must be 1.00 atm
aqueous solutions - 1.00 M concentration
Standard Molar Enthalpies of
Formation, Hfo
Standard molar enthalpy of formation
symbol is Hfo
defined as the enthalpy for the reaction in which
one mole of a substance is formed from its
constituent elements
for example:
Mg  s   Cl 2 g   MgCl 2s   6418
. kJ
o
H f MgCl 2  s 
 6418
. kJ / mol
Standard Molar Enthalpies of
Formation, Hfo
 Standard molar enthalpies of formation have been determined for
many substances and are tabulated in Table 8.3 and Appendix 1 in the
text.
 Standard molar enthalpies of elements in their most stable forms at
298.15 K and 1.000 atm are zero.
 Example: The standard molar enthalpy of formation for phosphoric
acid is -1281 kJ/mol. Write the equation for the reaction for which
Horxn = -1281 kJ.
P in standard state is P4
phosphoric acid in standard state is H3PO4(s)
Standard Molar Enthalpies of
Formation, Hfo
Hess’s Law
Hess’s Law of Heat Summation - enthalpy change for a
reaction is the same whether it occurs by one step or by any
(hypothetical) series of steps
true because H is a state function
we know the following Ho’s
1 4 FeO s  O2 g  2 Fe2 O3
2 2 Fe s  O2 g  2 FeO s
3 4 Fe s  3 O2 g  2 Fe2 O3 s
Ho   560 kJ
Ho   544 kJ
Ho  1648 kJ
Hess’s Law
We could calculate the Ho for [1] by properly using the
0
Ho’s for [2] and [3]
H
2 x [2] 2(2 FeO s   2 Fe s   O 2 g  )
2( 544 ) kJ
 3 4 Fes  3 O 2 g   2 Fe 2 O 3s
1 4 FeO s  O 2 g   2 Fe 2 O 3
 1648 kJ
 560 kJ
Hess’s Law
Example : Given the following equations and Ho values
H
o
 kJ 
[1] 2 N 2 g   O 2 g   2 N 2 O  g 
164.1
[2] N 2 g  + O 2 g   2 NO  g 
180.5
[3] N 2 g  + 2 O 2 g   2 NO 2  g 
66.4
calculate Ho for the reaction below.
N 2O
g
+ NO
2g 

3 NO
g
H
o
 ?
Hess’s Law
Use a little algebra and Hess’s Law to get the appropriate
Ho values
o
H (kJ)
1
2    1 N 2 O  g   N 2 g  +
3
2   2
1
2    3 NO 2 g  
3
2
3
N 2 g  +
1
2
2
1
2
O 2 g 
- 82.05
O 2 g   3 NO  g  270.75
N 2 g  + O 2 g 
N 2 O  g   NO 2 g   3 NO  g 
- 33.2
155.5
Hess’s Law
The + sign of the Ho value tells us that the reaction is
endothermic.
The reverse reaction is exothermic, i.e.,
3 NO  g   N 2 O  g  + NO 2  g 
H o = - 155.5 kJ
Hess’s Law
Hess’s Law in a more useful form
any chemical reaction at standard conditions, the standard enthalpy
change is the sum of the standard molar enthalpies of formation of
the products (each multiplied by its coefficient in the balanced
chemical equation) minus the corresponding sum for the reactants
o
H 298
 n
n
o
H f products
 n
n
o
H f reactants
Hess’s Law
Example: Calculate Ho298 for the following reaction from
data in Appendix 1.
C 3 H 8  g  + 5 O 2  g   3 CO 2  g  + 4 H 2 O  l 
Hess’s Law
Example: Calculate Ho298 for the following reaction from
data in Appendix 1.
C3H8g + 5 O2g  3 CO2g + 4 H2Ol 
o
H298


o
3Hf CO2 g
o
 4Hf H2Ol 
 
o
Hf C3H8 g
o
 5Hf O2 g
 3(3935
. )  4(2858
. )  (1038
. )  5(0) kJ
= -2211.9 kJ
Ho298   22119
. kJ, and so the reaction is exothermic.

Hess’s Law
Application of Hess’s Law and more algebra allows us to
calculate the Hfo for a substance participating in a reaction
for which we know Hrxno , if we also know Hfo for all
other substances in the reaction.
Hess’s Law
Example: Given the following information, calculate Hfo
for H2S(g)
2 H2Sg + 3 O2g 2 SO2g +2 H2Ol
Hof ?
(kJ / mol)
0
-296.8 -285.8
Ho298 = -1124 kJ
Hess’s Law

 
1124 kJ  2(2968
. )  2(2858
.   2H
o
H298

o
2Hf SO2g
o
 2Hf H2Ol 
o
2Hf H2Sg
o
f H2S g
now we solve for Hof H2Sg
2Hof H2Sg  412
. kJ
Hof H2Sg  206
. kJ
o
 3Hf O2g

 3(0) kJ

Bond Energies
Bond energy - amount of energy required to break the bond
and separate the atoms in the gas phase
A - B g   bond energy  A  g  + B g 
H - Cl  g   432 kJ mol  H  g  + Cl  g 
Bond Energies
Table of average bond energies
Molecule
Bond Energy (kJ/mol)
F2
159
Cl2
243
Br2
192
O2 (double bond)
498
N2 (triple bond)
946
Bond Energies
Bond energies can be calculated from other Ho298 values
Bond Energies
Example: Calculate the bond energy for hydrogen fluoride,
HF.
Bond Energies
Example: Calculate the bond energy for hydrogen fluoride,
HF.
H - F g   BE HF  H  g   F g  atoms NOT ions
or
H - F g   H  g   F g 

H o298  BE HF
 
H o298  H of H  g   H of F g   H of HF g 
o
H 298

 218.0 kJ + 78.99 kJ    271 kJ 
H o298  568.0 kJ
 BE for HF
Bond Energies
Example: Calculate the average N-H bond energy
in ammonia, NH3.
Bond Energies
NH 3g   N g  + 3 H g 

H
o
298
 3 BE N -H
 H

H
o
298
 H
H
o
298
 (472.7)  3(218)   46.11 kJ
H
o
298
 1173 kJ
o
f Ng 
average BE N -H
 3 H
o
f Hg 
o
f NH 3 g 
1173 kJ

 391 kJ mol N -H bonds
3
Bond Energies
In gas phase reactions Ho values may be related to bond
energies of all species in the reaction.
o
H 298   BE reactants   BE products
Bond Energies
Example: Use the bond energies listed in Table 8.4 to
estimate the heat of reaction at 25oC for the reaction below.
CH4g  2 O2g  CO2g  2 H2Og
o
H298
 4 BEC-H  2 BEO=O   2 BEC=O  4 BEO-H 
Bond Energies
Example: Use the bond energies listed in Table 8.4 to
estimate the heat of reaction at 25oC for the reaction below.
CH 4g  2 O2g  CO2g  2 H 2 Og
o
H 298
 4 BE C-H  2 BE O=O   2 BE C=O  4 BE O-H 
H o298  4(414)  2(498)  2(741)  4(464) kJ
o
H 298
 686 kJ