Chapter 6: Thermochemistry
Thermochemistry: energy considerations
associated with chemical and physical change
Energy: the capacity of a system to do work or
produce heat
potential energy – energy of position;
stored energy
kinetic energy – energy of motion
kinetic energy = ! mυ2
kinetic energy = 3⁄2 RT
In a system, energy can be transferred in the
form of heat or work:
heat, q – energy transfer results in
temperature change (ΔT);
ΔT = Tfinal – Tinitial
work, w – energy transfer during the act
of moving an object against an
opposing force
overall: ΔU = q + w
An Example of a System Configured for Energy
Transferred as Work:
The System, The Surroundings, The Universe
& The First Law of Thermodynamics
Units of Energy
SI unit of energy: Joule, J
1 Joule is the amount of energy required to raise
the temperature of 0.2390 g H2O by 1ºC.
1 kJ = 103 J
1 MJ = 106 J
thermochemical calorie, cal
1 calorie is the amount of energy required to
raise the temperature of 1 g of H2O by 1ºC.
1 cal = 4.184 J
1 kcal = 103 cal
1 nutritional calorie, 1 Cal = 1000 cal
Endothermic Change:
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energy is absorbed by the
system
energy flow is from the
surroundings into the system
Esys increases
Esurr decreases
for the system: Efinal > Einitial
if energy transferred in the form of heat,
Tsys increases and Tsurr decreases
E lost by surroundings = E gained by the system
The System:
◆
what we’re
interested in
The Surroundings:
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everything else
The Universe:
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System + Surroundings = the Universe
The 1st Law of
Thermodynamics:
The energy of the
Universe is constant.
Exothermic Change
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Changes in Thermchemical Quantities:
importance of sign and magnitude
energy is released by the
system
ΔX = Xfinal – Xinitial
energy flow is from the
system into the surroundings
if ΔX is + : Xfinal > Xinitial
if ΔX is ! : Xfinal < Xinitial
Esys decreases
Esurr increases
for the system: Efinal < Einitial
if energy transferred in the form of heat,
Tsys decreases and Tsurr increases
E lost by system = E gained by the surroundings
if q is +
qfinal > qinitial
if q is !
qfinal < qinitial
heat is absorbed by
the system
heat is released by
the system
endothermic
exothermic
if w is +
if w is !
surroundings do work on
the system
system does work on the
surroundings
endothermic
exothermic
if ∆E is +
Efinal > Einitial
if ∆E is !
Efinal < Einitial
energy is absorbed by
the system
energy is released by
the system
endothermic change
exothermic change
example:
Calculate the change in internal energy of a
system that releases 26.8 kJ of heat as it
does 68.7 kJ of work on the surroundings.
PV Work
expansion or compression of an ideal gas
w = !P∆V
example:
Calculate the quantity of work done (in kJ) by
an ideal gas as it expands from an initial volume
of 2.50 L to a final volume of 27.50 L against a
constant external pressure of 0.980 atm.
notes:
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w and ∆V are always
opposite in sign;
why?
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1 L•atm = 101.3 J
Enthalpy, H
H = E + PV
Enthalpy Changes for Chemical Reactions:
Thermochemical Equations
We will be focussing on changes in enthalpy that
accompany processes that occur at constant P.
∆Hrxn = Hfinal ! Hinitial
OR ∆Hrxn = Hproducts ! Hreactants
at constant P: ∆H = qP
∆H = Hfinal ! Hinitial
if ∆H is + : heat is absorbed by the system
endothermic process
if ∆H is ! : heat is released by the system
exothermic process
Thermochemical Equation:
balanced chemical equation + thermochemical data
example:
2 Na (s) + 2 H2O (l) ! 2 NaOH (aq) + H2 (g); ∆H = –367.5 kJ
notes:
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reaction is exothermic as written
specifically, 367.5 kJ of energy is released when:
2 mol Na & 2 mol H2O are consumed
2 mol NaOH & 1 mol H2 are formed
Enthalpy Changes for Chemical Reactions:
Thermochemical Equations
An Enthalpy Diagram for This Reaction:
interpretation of energy changes associated with a
reaction must take into consideration the following:
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phases of reactants and products
ex. 2 H2 (g) + O2 (g) ! 2 H2O (g); ∆H = –483.7 kJ
vs.
2 H2 (g) + O2 (g) ! 2 H2O (l); ∆H = –571.5 kJ
◆
direction of the reaction:
a reaction is endothermic in one direction and
exothermic in the reverse direction
ex. CH4 (g) + 4 Cl2 (g) ! CCl4 (l) + 4 HCl (g); ∆H = –433 kJ
vs.
CCl4 (l) + 4 HCl (g) ! CH4 (g) + 4 Cl2 (g); ∆H = +433 kJ
Stoichiometric Calculations:
Using ∆Hrxn as a Conversion Factor
Enthalpy Changes for Chemical Reactions:
Thermochemical Equations
◆
stoichiometry of balanced equation
energy released or absorbed is an extensive property
(depends on size of sample)
N2 (g) + 3 H2 (g) ! 2 NH3 (g);
∆H = –91.8 kJ
ex. combustion of propane
example:
C3H8 (g) + 5 O2 (g) ! 3 CO2 (g) + 4 H2O (g); ∆Hrxn = –2200 kJ
Determine the quantity of heat released (in kJ) in the
synthesis of 907 kg NH3.
combustion of 100 g (2.27 mol) C3H8 generates ~5000 kJ
vs.
combustion of 2.5 kg (56.7 mol) C3H8 generates ~125,000 kJ
example:
12.0 L N2 (g) at STP and 5.00 L H2 (g) at 30ºC and 3.02 atm
are combined and allowed to react. What is the maximum
amount of heat (in kJ) that can be generated?
Calorimetry
A laboratory technique in which a reaction
proceeds in an insulated container (i.e. calorimeter)
at constant P or constant V.
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adiabatic conditions for the system;
no heat is lost to or gained from the
surroundings
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record an observed ∆T
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calculate ∆H or ∆E
at constant P: ∆H = qP
at constant V: ∆E = qV
Heat Capacity (C) and Specific Heat (s)
How does a substance respond to heating?
Compare the observed temperature changes when
50.0 J energy are added to 10.0 g samples of
diamond and tungsten:
diamond
m = 10.0 g
q = 50.0 J
∆T = 9.8ºC
tungsten
m = 10.0 g
q = 50.0 J
∆T = 37.3ºC
Heat Capacity of a substance, C – heat required to
raise the temperature of a sample by 1º (C or K).
heat supplied
∆T produced
Heat Capacity = –––––––––––– ; units J∕ºC or K
OR
C = q∕∆T
Specific Heat, s – heat required to raise the
temperature of 1g of substance by 1º (C or K)
heat supplied
(∆T)(mass)
Specific Heat = –––––––––––– ; units J/g•ºC or K
OR
s = q/(m•∆T)
note:
use molar mass to convert between specific heat
and molar heat capacity of a substance
Constant Pressure Calorimetry to Determine ∆H
example:
adiabatic conditions:
◆ no heat exchanged between
system and surroundings
Calculate the amount of heat required (in J)
to raise the temperature of a 25.0 g block of
nickel from 22ºC to 104ºC. For Ni,
s = 0.44 J/g•ºC.
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∆Hsys = 0
if: ∆Hsys = ∆H1 + ∆H2
then: ∆H1 = –∆H2
example:
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30.0 g H2O at 280 K is mixed with 50.0 g
H2O at 330 K. What will be the final
temperature of the mixture? For H2O,
s = 4.184 J/g•ºC.
for an exothermic reaction:
∆H1 = heat released by chemical
reaction occurring in
calorimeter
∆H2 = heat absorbed by solution
& calorimeter resulting in
increase in T
example:
What do you measure in lab?
◆ amounts of reactants (mass, volume, mol)
◆ observed ∆T
order of determination (for the following examples):
∆H2
+ , units kJ
∆H1
! , units kJ
molar enthalpy of reaction
! , units kJ/mol
Consider the neutralization reaction of hydrochloric
acid and sodium hydroxide:
HCl (aq) + NaOH (aq) ! NaCl (aq) + H2O (l)
Determine the molar enthalpy change (in kJ/mol) for
this reaction if, in a constant pressure calorimetry
experiment 33. 0 mL of 1.20 M HCl (aq) is combined
with 42.0 mL of 1.20 M NaOH (aq) resulting in an
increase in temperature from an initial 25.00ºC to a
final 31.80ºC.
You can assume that, for the solutions, volume is
additive; d = 1.00 g/mL; s = 4.184 J/gºC.
Constant Volume Calorimetry to Determine ∆E
example:
When 23.6 g CaCl2 is dissolved in water in a
constant pressure calorimeter, the temperature
rose from 25.0ºC to 38.7ºC.
If the heat capacity of the calorimeter and
solution is 1258 J/ºC (i.e. Ccal = 1258 J/ºC),
determine the heat released by the dissolution
of 1.20 mol CaCl2.
adiabatic conditions:
◆ no heat exchanged between
system and surroundings
◆
∆Esys = 0
if: ∆Esys = ∆E1 + ∆E2
then: ∆E1 = –∆E2
◆
for an exothermic reaction:
∆E1 = heat released by chemical
reaction occurring in
calorimeter
CaCl2 (s) ! Ca2+ (aq) + 2 Cl– (aq)
∆E2 = heat absorbed by solution
& calorimeter resulting in
increase in T
Hess’s Law of Heat Summation
example:
The combustion of ethanol is studied in a
constant volume (bomb) calorimeter with a
calorimeter constant of 9.63 kJ/ºC (i.e Ccal =
9.63 kJ/ºC).
The combustion of 2.84 g C2H5OH results in an
increase in temperature from Ti = 25.00ºC to
Tf = 33.73ºC.
Determine the energy released in the
combustion of 1 mol of C2H5OH.
Determine the heat of combustion per gram of
ethanol (in kJ/g)
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We can determine of ∆H for one chemical
reaction based on the ∆H values for related
reactions using Hess’s Law:
For a chemical reaction that can be written
as the sum of 2 or more steps, ∆H for the
overall reaction is equal to the sum of ∆H’s
for the individual steps.
so: for a reaction that is a result of the sum of
3 individual steps:
∆Hrxn = ∆H1 + ∆H2 + ∆H3
Hess’s Law of Heat Summation
Why does Hess’s Law
of Heat Summation
work?
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Hess’s Law of Heat Summation
Determine ∆H for the following reaction using
equations 1 & 2 given below:
Enthalpy is a state
property:
dependent only on
initial and final states
independent of path
An Enthalpy Diagram for This Process:
2 C (s) + O2 (g) ! 2 CO (g)
∆H = ????
use:
(1) 2 C (s) + 2 O2 (g) ! 2 CO2 (g)
∆H1 = –787 kJ
(2) 2 CO2 (g) ! 2 CO (g) + O2 (g)
∆H2 = +566 kJ
overall:
2 C (s) + O2 (g) ! 2 CO (g)
∆H = –221 kJ
To solve a Hess’s Law of Heat Summation problem:
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consider the equations you have to work with
identify how to manipulate them so that when
you add them together, you end up with your
target, overall equation
your options are:
multiply all stoichiometric coefficients in the
equation by some factor, n " multiply ∆H by
the same factor, n
write an equation in reverse (i.e. change
direction) " change the sign on ∆H
solution:
example:
Determine ∆H for:
W (s) + C (s) ! WC (s); ∆H = ????
using:
(1) 2 W (s) + 3 O2 (g) ! 2 WO3 (s) ∆H = –1680.6 kJ
(2) C (s) + O2 (g) ! CO2 (g)
∆H = –393.5 kJ
(3) 2 WC (s) + 5 O2 (g) ! 2 WO3 (s) + 2 CO2 (g) ∆H = –2391.6 kJ
solution:
using:
(1) 2 W (s) + 3 O2 (g) ! 2 WO3 (s) ∆H = –1680.6 kJ
(2) C (s) + O2 (g) ! CO2 (g)
∆H = –393.5 kJ
(3) 2 WC (s) + 5 O2 (g) ! 2 WO3 (s) + 2 CO2 (g) ∆H = –2391.6 kJ
actions to take:
reverse equation (3)
multiply equation (1) by factor of "
multiply equation (3) by factor of "
do nothing to equation (2)
Standard Reaction Enthalpies
Determine ∆H for:
W (s) + C (s) ! WC (s); ∆H = ????
using:
(1) W (s) + 3/2 O2 (g) ! WO3 (s)
∆H = –840.3 kJ
(2) C (s) + O2 (g) ! CO2 (g)
∆H = –393.5 kJ
(3) WO3 (s) + CO2 (g) ! WC (s) + 5/2 O2 (g)
∆H = +1195.8 kJ
–––––––––––––––––––––––––––––––––––––––––––––––––––––––––
overall: W (s) + C (s) ! WC (s)
Determine ∆H for:
W (s) + C (s) ! WC (s); ∆H = ????
∆H = –38.0 kJ
Standard Reaction Enthalpies, ∆Hº:
reaction enthalpy corresponding to reactants in
their standard states forming products in their
standard states
standard state: the physical state of a pure
substance at P = 1 atm and a defined
temperature
Standard Enthalpies of Formation, ∆Hºf
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∆Hºf corresponds to the formation of 1 mol of a
substance in its standard state from its elements
in their standard states.
ex:
Na (s) + " Cl2 (g) ! NaCl (s)
6 C (s) + 6 H2 (g) + 3 O2 (g) ! C6H12O6 (s)
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∆Hºf of an element in its most stable form = 0
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units of ∆Hºf are kJ/mol
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tabulated in Table 6.2 and Appendix C of text
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use ∆Hºf’s to calculate ∆Hºrxn
Using Standard Enthalpies of Formation, ∆Hºf
∆Hºrxn = ∑n•∆Hºf products – ∑n•∆Hºf reactants
example:
Calcualte ∆Hº (in kJ) for the following reaction
using the given ∆Hºf values:
3 NO2 (g) + H2O (l) ! 2 HNO3 (aq) + NO (g)
∆Hºf: 33.2
–285.8
–206.6
90.3
(in kJ/mol)
example:
Determine ∆Hºf (in kJ/mol) for HCl (g) using the
following data:
CH4 (g) + 4 Cl2 (g) ! CCl4 (l) + 4 HCl (g); ∆Hº = –433.3 kJ
∆Hºf : –74.9
0
–139
???
(in kJ/mol)
example:
Determine ∆Hºf (in kJ/mol) for HCl (g) using the
following data:
CH4 (g) + 4 Cl2 (g) ! CCl4 (l) + 4 HCl (g); ∆Hº = –433.3 kJ
∆Hºf : –74.9
0
–139
???
(in kJ/mol)
solution:
let x = ∆Hºf of HCl (g)
–433.3 kJ = [(1 mol CCl4)(–139 kJ/mol) + (4 mol HCl)(x)] !
[(1 mol CH4)(–74.9 kJ/mol)]
–508.2 kJ = –139 kJ + (4 mol)(x)
–369.2 kJ = 4 mol(x)
so x = ∆Hºf for HCl (g) = –92.3 kJ/mol