The energy aspect of chemistry
Thermochemistry
Thermochemistry
1
Thermochemical Data and Services
Energy is a quantity that can be stored in a chemical form, and
chemical reaction involve energy. There is an extensive
thermochemical data for various scientific and engineering
applications.
NIST site has thermochemical data for over 6000 organic and small
inorganic compounds. It also has reaction thermochemistry data for
over 9000 reactions and ion energetics data for over 16,000
compounds.
Thermochemical services you may need or provide
Calculation services
Consulting
Experimental services
Technical literature search service
Thermochemistry
2
Temperature Energy & Heat
Temperature is a measure of hot and cold – an intensive property.
Thermometers in various scales are used to compare temperature.
Recall: urms = ( 3 k T / m)1/2 = (3 RT / M)1/2
Energy is the driving force for changes. A change often is associated
with a certain amount of energy and amounts of energy can be
measured according to the quantities changed. Like other quantities,
energy is an extensive property, unlike temperature.
Heat is energy in transfer or energy transferred and heat is often used
as another term for energy. In this aspect, it is the quantity of energy
transferred or involved.
Describe hotness and heat
Thermochemistry
3
A system for Thermochemistry
For the study of energy, there must be
a reference for its amounts transferred
into or out. Such a reference is called
a system.
The boundary isolates the system
from its surrounding, so that the
amount of energy or heat transferred
into or out can be identified.
Surrounding
System
Environment
Picture a system with boundary, surrounding,
and heat transfer in your mind.
Thermochemistry
4
Measuring Heat (Energy)
Heat = Energy = Work
Heat is energy – the ability to do work. Units for work, heat or energy is J
1J=1Nm
(in honor of James Joule 1818 – 1889)
1 cal = 4.184 J (defined. The amount of heat to raise 1 g water for 1 K)
Gaining heat increase T, as the kinetic energy of molecules increase.
Remember these from discussion of gases?
Kinetic energy = ½ m u 2
Meaning of T?
3 R T = 1/3 NA m u 2
Heat capacity: the amount of heat required to raise T by 1 K
Specific heat: heat capacity (no unit) compare to water (1.000 cal K-1 g-1)
Thermochemistry
5
Conservation of Energy
Energy can be converted from one form to another at a fixed rate,
and it cannot be destroyed or created. In other words, energy is
conserved. Quantity of heat q follows
no creation & no destruction
qsystem + qsurrounding = 0
gain of q is +ve
loss of q is -ve
What can be created and destroyed?
1 cal = 4.184 J (heat-work conversion)
1 J = 1 kPa L
Has this principle been proved?
What is a perpetual motion?
How can this principle be disproved?
What’s the principle of conservation of matter?
Thermochemistry
6
Calorimeter Experiment
q = C * T
Heat gained
Temperature difference
Heat capacity
Heat capacity of water = 1 cal deg-1 g-1
molar heat capacity of water
= 18 * 4.184 J deg-1 mol-1
= 75.3 J deg-1 mol-1
Specific Heat (no unit)
water 1
copper 0.385
aluminum 0.903
Thermochemistry
What to use for a miracle thaw?
7
Bomb and Coffee-cup Calorimeter
q = C * T
Use a known amount of q and
measure T to determine C.
By measuring T , calculate
C = q / T
(corection)
Once C is known, and a known amount
of reactants, heat of reaction, Hreaction,
can be evaluated
Thermochemistry
8
How do you solve this?
Two 100.0-mL solutions of NaOH and HCl are both 1.00
mol L-1. When they are mixed in a coffee cup, the
temperature raised from 20.0 to 26.7o C. What is the heat
released.
Solution:
What’s the reaction?
What assumptions to make?
Steps?
Thermochemistry
9
Determine Heat of Reaction
Two 100.0-mL solutions of NaOH and HCl are both 1.00 mol L-1. When
they are mixed in a coffee cup, the temperature raised from 20.0 to 26.7o
C. What is the heat released.
Solution:
Assumptions: no heat loss,
consider solution as water, C = 200.0 g * 4.18 J g–1 K –1 = 836 J K–1.
Heat of reaction = 836 J K–1.* (26.7 – 20.0) K = 5601 J = 5.6 kJ.
Mole of H+ and OH–1 reacted = 1.0 mol L–1 * 0.100 L = 0.10 mol
Molar heat of neutralization, Hreaction = 5.6 kJ / 0.1 mol = 65 kJ mol –1
Reaction:
H+ + OH–1 = H2O
Hreaction = 65 kJ mol –1
Thermochemistry
10
Pressure-Volume Work
P
Pressure = P (N m-2)
V = AArea * h distance (m3 = 1000 L)
A
h
Fforce = P * A
WWork = F * h distance
= P * A * h distance
=PV
1J
1000 L
=1N
(
)
1 m3
= 1 1000 Pa L
m-2
= 1 kPa L
m3
Thermochemistry
Show
P V is work and 1 J = 1 kPa L
11
The Gas Constant R
The ABC laws of gases can be combined into one and the result is an
ideal gas equation.
PV=nRT
1 atm * 22.4140 L
R = ---------------------------- = 0.082057 L atm mol–1 K–1
1 mol 273.15 K
101.325 kPa * 22.414 L
R = ----------------------------------- = 8.3145 L kPa mol–1 K–1 = 8.3145 J mol–1 K–1
1 mol 273.15 K
1 kPa L = __?__ J
1 L = __?__ m3
1 L atm = __?__ J
Thermochemistry
1 m3 = __?__
L
12
The Gas Constant R
You blow a balloon for a volume of 1.00 L per breath. If the pressure
you are blowing against is 1.10 atm, how much work, in J, have you
done?
Solution:
P * V = 1.10 atm * 1.00 L = 1.10 L atm
–1 K–1
8.3145
J
mol
1.10 L atm
0.08205 L atm mol–1 K–1
R is used for unit conversion
= 111.4 J
8.3145 J mol–1 K–1
= 101.33
0.08205 L atm mol–1 K–1
Thermochemistry
13
First Law of Thermodynamics
To accommodate the principle of conservation of energy, we define an
Internal Energy U, as the total kinetic and potential energy.
The U cannot be measured or determined, thus we usually are only
interested in U, the difference in internal energy before and after a
change.
The w : work done onto the system
U = q + w
The q: heat absorbed by the system
The heat absorbed by the system and the work done onto the system
are all used to increase the internal energy of the system.
Thermochemistry
14
Evaluating Internal Energy
A balloon expanded against the atmosphere pressure performed 25 J
of work, and absorbed 30 J of heat, what is the change in internal
energy?
Solution:
U = q + w
= (30 – 25) J
=5J
Discussion:
A change in internal energy results in change in temperature T.
How does the average kinetic energy of the molecule change?
Thermochemistry
15
State and State Function
The state of a system is the description of T, P, amounts and kinds of
substances in the system, etc. The state defines all these quantities, and
these quantities are specific for the state.
A property that has a specific value for a state is called a state function
(or function of state).
The internal energy, U, is a state function, and the difference U is
evaluated from the final and initial state. It is independent of the path.
U = U f – U i
I will further explain the meaning of path independence using examples.
Heat q and work w depend on path, for example.
Explain state functions!
Thermochemistry
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Heats of Reaction, U & H
For a reaction
Reactants  Products
The change in internal energy is
U = UReactants – Uproducts
= qrxn + w
The change in internal energy U is the heat of reaction under constant
volume, qv, U = qv, because w = 0, no P-V work.
Enthalpy H is the heat of reaction under constant pressure qp, H = qp
Under this condition, work done on or by the system, is w = – P V.
Thus,
U = qp - P V = H – P V
or
H = U + P V
Explain U and H .
reactants
H
Thermochemistry
U + P V
products
17
An U, H Example
For the reaction,
2CO (g) + O2 (g)  2 CO2 (g), at 298 K
qp = -566.0 kJ, calculate qv and w.
Solution:
The work is due to change of volume,
w = PV = (nf – ni) R T
= ( 2 – 3) * 8.3145 J mol –1 K–1 * 298 K
= – 2500 J = – 2 .5 kJ mol –1
2CO (g) + O2 (g)
U
H
H = – 566.0 kJ mol –1 (of equation)
U = – 566.0 – (– 2.5) kJ = – 563.5 kJ
Note: 3 moles of gas is reduced to 2 moles,
work must be done to the system to compress
it back to the same P, w is positive.
Thermochemistry
w
2CO2 (g)
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Application of Enthalpy Given per
Given,
C12H22O11 (s) + 12 O2 (g)  12 CO2 (g) + 11 H2O (l),
rxn eqn
H = –5650 kJ
How much heat is released when 10.0 g of sucrose, (molar mass
342.3), after you have ingested, digested, and completely oxidized.?
Solution:
The condition given here applies to the human utilization of sucrose (S),
– 5650 kJ
10.0 g S 1 mol S
=–
kJ
342.3 g S
1 mol S
–ve, energy released
Evaluate U and w
Thermochemistry
19
Enthalpy Diagram for States
vapor
Explain these changes of state:
H2O (s)  H2O (l)
H = 6.01 kJ at 273 K
H2O (l)  H2O (g)
H = 44.0 kJ at 298 K
How much energy is required to convert 72 kg of
ice at 273 K to vapor at 298 K?
Solution:
1 mol
6.01 kJ
72e3 g ice
= __x_ ?___
18 g
1 mol
72e3 g water * 4.184 J (298 – 273) K = __y_ ?___
4e3 mol * 44.0e3 J / mol
= __z__ ?___
44.0 kJ
Water at 298 K
Water at 273 6.01
K kJ
18 g Ice at 273 K
Total = _x_ + _y_ + _z_ = ________________
Thermochemistry
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Endothermic and Exothermic Reactions
Use an energy level diagram to help visualize
endothermic and exothermic reactions.
Explain the sign changes in the following:
H2O (s)  H2O (l)
H = 6.01 kJ at 298 K
H2O (l)  H2O (s) H = – 6.01 kJ at 298 K
H2O (l)  H2O (g)
H = 44.0 kJ at 273 K
H2O (g)  H2O (l) H = – 44.0 kJ at 273 K
2 NO (g)  N2 (g) + O2 (g)
H = 181 kJ
N2 (g) + O2 (g)  2 NO (g)
H = – 181 kJ
Explain why the sign changesThermochemistry
when the process is reversed
21
Hess’s Law
Germain Hess
(1802 - 1850)
Like internal energy, enthalpy is also a state function.
Hess’s Law is based on the principle of conservation of energy.
It implies that if a process occurs in stages (even if only hypothetical), the
enthalpy change for the overall process is the sum of the enthalpy
changes for the individual steps.
Another statement: the heat evolved or absorbed in a chemical process
is the same whether the process takes place in one or in several steps.
Explain Hess’s law
Thermochemistry
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Hess’s Law Illustrated
A + B  AB
A+2B
H1
AB + B = AB2
H2
__add___________________
A + 2 B = AB2
H1 + H2
= H1 2
H1
AB + B
H1 2
H 2
AB2
Thermochemistry
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Chemical Energy and Hess’s Law
C(graphite) + 0.5 O2  CO
H° = – 110 kJ/mol.
2 C(graphite) + O2  2 CO
H° = – 220 kJ/mol (multiplied by 2)
6 C(graphite) + 3 O2  6 CO
H° = – 660 kJ/mol (multiplied by 6)
2 CO  C(graphite) + O2
H° = 220 kJ/mol (+ve)
CO (g) + 0.5 O2 (g)  CO2 (g)
H° = – 283 kJ/mol.
1 mol C and O2
A quantity difficult to measure: – 110 kJ
1 mol CO & 0.5 mol O2
– 393 kJ
– 283 kJ
1 mol CO2
Thermochemistry
24
Standard State
Standard temperature and pressure (STP) for gases are 273.15 K and
1 atm.
Standard state refers to states at pressure of 1.0 bar, and temperature
must be specified for data at standard state.
Standard enthalpy of reaction, standard enthalpy of formation,
standard enthalpy of combustion, and in general standard energy of
changes are referring to standard states.
Reactants at
standard states
Reaction at
other states
Thermochemistry
Products at
standard states
25
Standard Enthalpy of Formation
The standard enthalpy of formation, Hf°,
is the energy of reaction for the formation
of the substance at its standard state from
its elements at their standard states.
Enthalpies of elements are 0 at their
standard state, these are reference
points.
Discuss the significance and value in
everyday life of the data in the Table
on the right!
Thermochemistry
Hf° kJ mol–1 of compounds
H2O (l)
H2O (g)
CO (g)
CO2 (g)
CH4 (g)
C2H2 (g)
C2H4 (g)
C2H6 (g)
CH3OH (l)
C2H5OH (l)
– 285.8
– 241.8
– 110.5
– 393.5
– 74.8
+ 226.7
+ 52.4
– 84.7
– 238.7
– 277.7
26
Meaning of Some Special
Enthalpies of Formation
Cpd Hf .kJ Reaction (Explain)
Cdiamond
1.9 C (graphite)  C (diamond) (phase transition)
Br2(g) 30.9 Br2 (l)  Br2 (g) (evaporation at 298 K)
P4 (red) -17.6 P4 (s, white)  P4 (s, red)
(White phosphorus is considered the standard state)
H (g) 217.8 ½ H2 (g)  H (g) (½ H–H bond energy)
Thermochemistry
27
Find Standard Enthalpy of Formation
The enthalpy of combustion for H2, C(graphite) and CH4 are - 285.8,
- 393.5, and - 890.4 kJ/mol respectively. Calculate the standard
enthalpy of formation dHf for CH4. (correction made)
Solution: C (graphite) + 2 H2  CH4
Explain these from the data
2 H2(g) + O2(g)  2 H2O(l)
C(graphite) + O2(g)  CO2(g)
CO2(g) + 2H2O(l)  CH4(g) + 2O2(g)
Hf° = _____
2 *(-285.8)
H = - 571.6 kJ
H = - 393.5 kJ
H= 890.4 kJ (+ve)
(1) + (2) + (3)
C (graphite) + 2 H2  CH4
Hf° = - 74.7 kJ
Thermochemistry
Make an energy level diagram from
these to see the relationship.
28
Energy Level Diagram
Hf° = - 74.7 kJ
C(graphite) + 2 H2(g) + 2 O2(g)
CH4(g) + 2 O2(g)
H = - 965.3 kJ
H = - 965.3 kJ
CO2 + 2 H2O
Thermochemistry
29
Application of Energy Level Diagram
and Enthalpies of Formation
Hf° = - 74.7 kJ
If you know any
two of the three
thermodynamic
data, you can
evaluate the third.
C(graphite) + 2 H2(g) + 2 O2(g)
CH4(g) + 2 O2(g)
H = - 965.3 kJ
H = - 965.3 kJ
CO2 + 2 H2O
Thermochemistry
30
Applications of Hf°
Hf° can be very useful for the
calculation of enthalpy of reaction,
by applying Hess’s Law.
2 C (s) + 3 H2 (g) + 7/2O2(g)
Hf C2H6 = - 84.7
C2 H6 +
7/
2
O2
Calculate the H for the reaction:
C2H6 (g) + 7/2 O2 (g)  2 CO2 (g)
+ 3 H2O (g) H = ____?
Solution: look up Hf°s for C2H6 etc
Hrxn = products Hf° – reacts Hf°
H = 2*(– 393.5) + 3*(– 285.8)
– {1 * (– 84.7) + (7/2)*(0)}
= _______ kJ
Explain the generalized formula
Thermochemistry
2*Hf CO2
2*(-393.5)
= - 787.0
2 CO2 + 3 H2 + 3/2 O2
H = ______
3*Hf H2O
3*(-285.8)
= - 857.4
31
2 CO2 (g) + 3 H2O (g)
Enthalpy of Reaction and Fuel
6 CO2 + 6 H2O
Chlorophyll
Sun light h v
C6H12O6 + 6 O2, H = 2.8e3 kJ
Animal digestion
Oxidation
Fuels:coal, petroleum, natural gas, methanol, ethanol, hydrogen,
gasified coal, waste, organic matter, etc.
Please read about fuel and environmental impacts.
Thermochemistry
32
Chemistry: relationship of material and energy
Chemistry is the study of behavior of materials under the influence of
energy, which drives all changes. Without energy, there is no
behavior to study, and energy not only causes all changes, it is also
the source of life. Thus, life must constantly utilizes energy, almost
exclusively chemical energy.
Furthermore, energy is also required to keep and maintain a
comfortable and pleasant living. Chemical energy, usually known as
food and fuel, is one of the most important energy sources. Thus, the
skill to use thermodynamic data is important.
Thermochemistry
33
Review questions – 0
For an endothermic reaction, how does the internal energy, U,
change, positive, zero, or negative?
What is the units for heat capacity?
Look for the data and name H for the processes H2O (l)  H2O(g),
H2O (g)  H2O(l), H2O (s)  H2O(g), H2O (s)  H2O(l) at STP? For
which of the above has the smallest difference of U - H?
The reaction in a bomb calorimeter is carried out at what condition?
constant P, constant V, constant T, at STP or at 1 atm and 25 K
When 2.00 g of CH4 are burned in a bomb calorimeter (Cv = 2.677e3 J
/ K), the temperature rises from 24.00 to 27.06o C.
What is the q absorbed by the calorimeter?
What is the molar heat of combustion of CH4 under this condition?
Thermochemistry
34
Review questions - 1
When 0.0500 mol HCl reacts with 0.100 mol NaOH contained in 25.0
mL of solution, the temperature of the water increases by 13.7o C.
What is the heat of the reaction? (limiting reagent, heat capacity)
What is the molar heat of neutralization reaction? (stoichiometry and
thermal energy)
Which of the following are state functions?
U, H, qv, P, T, w (work), V
The standard enthalpy of formation for C and CO are –393 and 287 kJ
respectively. What is the standard enthalpy of formation for CO?
(Hess’s law, standard enthalpy of formation, energy level diagram slide 24, given any
two of these three, you should be able to find the third one.)
How much energy is released when 100.0 g CO is formed from
reaction of carbon and oxygen? (stoichiometry)
Thermochemistry
35
Review questions - 2
The standard enthalpy of formation of water (liquid H2O) is –286 kJ mol–1.
What is standard enthalpy of combustion (or oxidation) of H2?
What is the energy for the reaction, H2O (l)  H2(g) + ½O2(g)?
How much energy is release when 0.15 g H2 at 273K and 1 bar is ignited
in excess O2 in a bomb calorimeter in which the water vapor so formed is
condensed into liquid at at the same temperature? (calculation gives
–21.5 kJ, and the negative sign indicated energy released)
What are U, W and H for the reaction
0.1 H2O (l) = 0.1 H2O (g, 300 K and 1 bar) ?
(Note, the internal energy increases, W = P V = n R T; n = 0.1 mol; if hydrogen is
burned in air, this amount of energy and this energy is not released immediately from the
burning of H2 gas.)
Thermochemistry
36
Review questions - 3
When 4.765 g of MgCl2 is formed by burning Mg(s) in Cl2(g), 32.1 kJ of
energy is released. What is the enthalpy of formation of MgCl2?
The standard enthalpy of formation is zero for which ones of the
following, O2(g), O3(g), O(g), C(graphite), C(diamond), C(g), P4(white),
P4(red), Br2(s), Br2(l), Br2(g), I2(s), I2(l), I2(g), Hg(s), Hg(l), S(s), S(g)?
(standard states, which ones have positive enthalpies of formation?)
Given these data at STP,
S(s) + O2(g)  SO2 (g) H = –3 95 kJ
S(g) + O2(g)  SO2 (g) H = –618 kJ
What is the standard enthalpy of formation of S(g)? (+223 kJ mol-1)
Thermochemistry
37
Review questions - 4
For applications of standard enthalpy of formation and enthalpy of
reaction, try these:
The standard enthalpy of formation for CH3CHO(g), H2O(l) and CO2(g)
are –166, –286, and –394 kJ mol-1 respectively. Calculate the heat of
combustion of CH3CHO(g). (-1194 kJ mol-1, Hess’s law)
The standard enthalpy of formation of C2H2(g) is 227 kJ mol-1. What is
the enthalpy of combustion of C2H2(g)? (use data from above, Hess’s law)
Calculate the energy evolved in the complete oxidation of 5.55 g of Al at
298 K and 1 atm pressure. (Hof of Al2O3 is –1767 kJ mol-1)
What is the energy required to reduce 5.55 g of Al2O3? (use data from the
above question)
Thermochemistry
38
Review questions - 5
When concentrated sulfuric acid is diluted with water, the solution
becomes hot. Is the reaction of H2SO4 with H2O exothermic or
endothermic?
A synthetic gas contains 55.0% CO, 33.0% H2, and 12% CO2. If all the
energy from the combustion of 1.0 L of this gas is absorbed by 1.0 L of
H2O initially at 300 K, what is the final temperature? (find the required data
and carried out the calculation)
A and B are two different compounds with the same molecular formula,
C5H10. Both compounds burn in oxygen to give CO2(g) and H2O(g).
The enthalpies of combustion for A and B are –3050 and 3025 kJ mol1, respectively, at 298 K. What is the Ho for the reaction
A  B?
Thermochemistry
39
Review – 6
(see slide 27 and 28)
(see slide 7)
(review Ho and uo difference, see slides 16 - 18)
(see slide 25-29)
The standard enthalpy of formation of diamond is 1.9 kJ mol-1, and the standard enthalpy of combustion at 298 K
is 293.5 kJ mol-1 for graphite. What is the theoretical enthalpy of combustion of diamond at 298 K?
The heat capacity of platinum and liquid water are 0.133 J g-1, K-1 and 4.185 respectively. What mass of liquid
water at 283 K, is needed to cool 50.0 g of platinum from 323 K to 300 K when the platinum is placed in the
water?
The thermochemical equation of the reaction at 298 K is,
Al(s) + 3/2O2 (g)  Al2O3(s)
Ho = –1767 kJ
What is the standard internal energy change, Uo, at 298 K for this reaction?
With the following information, what is the standard enthalpy of ormation of C2H4(g)
C2H4(g) + 3 O2  2CO2 + 2 H2O(l), Ho = – x kJ
Data:
Hof = – y kJ mol for CO2
Hof = – z kJ mol for H2O
Solution: Hof = _______ kJ mol for C2H4
Thermochemistry
40
Weakness shown in Test 2 problems
3. A sample of gas absorbs exactly 200 J from its surroundings. The
gas expands from 2.75 L to 13.5 L against a constant 0.860 atm
pressure. What is the internal energy change for the gas?
5. Radiation with a wavelength of 217 nm causes electrons to be
ejected from the surface of lithium metal. If the maximum kinetic
energy of ejected electrons is 4.420e-19 J, what is the longest
wavelength radiation that can be used to disloge electrons from the
surface of lithium?
8. Number of humps p, d orbitals in 4r22 versus r2 plot.
(4r2R2 vs r2)
11. What is the electronic configuration of the transition ions, Fe3+?
Thermochemistry
41
Final Exam info
Chem120 Dec. 11, 14 – 17 pm
PAC all areas and MC4020, 4021, 4045, 4058, 4059, 4060, 4061
Thermochemistry
42