1. No category

Theromochemistry (The first law of
thermodynamics)
What is energy?
The ability to do work or produce heat: The sum of
all potential and kinetic energy in a system is known
as the internal energy of the system.
• Potential energy
In chemistry refers to the energy stored in
bonds.
• Kinetic energy
o Energy of motion, usually of particles,
proportional to Kelvin temperature.
KE =n 1/2mv2
Units of energy
Calorie 1 Calorie = 1000 calories
Joule (J) is the SI unit for energy
1 cal = 4.184 J
The First Law of Thermodynamics also known as
the Law of Conservation of energy states that energy
can by converted from one form to another but can be
neither be created nor destroyed.
The first law of thermodynamic holds for any
physical or chemical change. Ex. Chemical reaction,
change of state, solution formation
1
Explain exothermic and endothermic in terms of The
Law of conservation of Energy.
Explain as endothermic or as exothermic
H2O(g) à H2O(l)
H2O(s) à H2O(l)
Chemical energy
Energy absorbed or released by a chemical reaction.
Energy is released or absorbed from or to chemical
bonds
Formula Name Chemical Formula ΔfH (kcal/mol) CAS registry number
H2
Hydrogen [H][H]
0.0
CH4
Methane C
-17.9
000074-82-8
C2H6
Ethane
CC
-20.0
000074-84-0
C2H2
Acetylene CC
+54.2
C3H8
n-Propane CCC
-25.0
000074-98-6
C4H10
n-Butane CCCC
-30.0
000106-97-8
C5H12
n-Pentane CCCCC
-35.1
000109-66-0
C6H14
n-Hexane CCCCCC
-40.0
000110-54-3
C7H16
n-Heptane CCCCCCC
-44.9
000142-82-5
C8H18
n-Octane CCCCCCCC
-49.8
000111-65-9
C9H20
n-Nonane CCCCCCCCC
-54.8
000111-84-2
C10H22 n-Decane CCCCCCCCCC
-59.6
000124-18-5
2
Write out the combustion reaction for octane (C8H18)
How much energy is produced when a gallon of
octane is burned? Density = .703 g/ml, 1 gal = 3785
ml
Reactants + energy à Products
Reactants à products + energy
endothermic
exothermic
Where did the energy come from in gasoline?
Heat, q is the transfer of energy between two objects
due to a temperature difference. Direction is from a
hot object to a cold one.
Work is defined as a force acting over a distance.
Work is required to move an object from point A to
point B
Energy and Work
ΔE = q = w
3
Calculate ΔE for a system undergoing an
endothermic process in which 15.l6 kJ of heat flows
and where 1.4 kj of work is done on the system.
In a gas
W = -PΔV
Calculate the work associated with the expansion of a
gas from 46 L to 64 L at a constant external pressure
of 15 atm.
Energy is transferred by heat and by work
The internal energy (E) of a system is the sum of the
kinetic and potential energies of all the particles in
the system. This can be changed by the flow of work
or heat
Enthalpy (H)
The internal energy of a system
A state function
H = E + PV
4
E = energy, P = pressure, V = volume
ΔH = q at constant pressure
Enthalpy change (ΔH) (first law of
thermodynamics
Energy and changes
ΔH = Hend – H begining
ΔH is positive for an endothermic process
ΔH is negative for an exothermic process
Enthalpy changes
Heat of reactions ΔHrxn
Heat of combustion ΔHcomb
Heat of fusion ΔHfus
ΔH = Hice – H liquid
Heat of vaporization ΔHvap
ΔH = Hgas – H liquid
Bond dissociation energy ΔHBDE
Heat of formation ΔHf
Heat of solution ΔHsoln
ΔH = Hsolute + solvent – H pure solvent
5
The change in enthalpy for any process is directly
proportional to the amount of reactants or products
involved in the change.
Example
Write out the combustion reaction for butane (C4H10)
The molar heat of combustion for butane is 2859
kJ/mol. What is the enthalpy change if 1 kg of
butane burns.
Calorimetry
A device used to measure the heat involved in a
process. It measures heat flow.
6
7
Heat Capacity (C) is the heat required to raise the
temperature of an object by 1 Kelvin. The units of C
are J/K.
The molar heat capacity, Cmolar is the heat required
to raise the temperature of one mole of a substance
by 1 K.
Specific heat capacity is the heat required to raise
the temperature of one gram of a substance by 1 K.
The specific heat of water is 4.184 J/g K
The heat change in a process is
Energy released in a process is = to specific heat
capacity x mass of solution x increase in temperature
E = Cs x m x ΔT
Specific Heat capacities of some common substances
Substance Specific heat capacity
(J/oCg)
H2O(l)
4.18
H2O(l)
2.03
Al(s)
0.89
Fe(s)
0.45
Hg(l)
0.14
8
In calorimetry the heat lost by a substance is equal to
the heat gained by the water. We can measure the
change in heat of the process by measuring the heat
gained by the water.
1. In a coffee cup calorimeter, 100.0 mL of 1.0 M NaOH and
100.0 mL of 1.0 M HCl are mixed. Both solutions were originally
at 24.6°C. After the reaction, the final temperature is 31.3°C.
Assuming that all solutions have a density of 1.0 g/cm3 and a
specific heat capacity of 4.184 J/g°C, calculate the enthalpy change
for the neutralization of HCl by NaOH. Assume that no heat is lost
to the surroundings or the calorimeter.
2. When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0°C is mixed
with 1.00 L of 1.00 M Na2SO4 solution at 25°C in a calorimeter,
the white solid BaSO4 forms and the temperature of the mixture
increases to 28.1°C. Assuming that the calorimeter absorbs only a
negligible quantity of heat, that the specific heat capacity of the
solution is 4.18 J/°C ⋅ g, and that the density of the final solution
is 1.0 g/mL, calculate the enthalpy change per mole of BaSO4
formed.
9
3. It has been suggested that hydrogen gas obtained by the
decomposition of water might be a substitute for natural gas
(principally methane). To compare the energies of combustion of
these fuels, the following experiment was carried out using a bomb
calorimeter with a heat capacity of 11.3 kJ/°C. When a 1.50-g
sample of methane gas was burned with excess oxygen in the
calorimeter, the temperature increased by 7.3°C. When a 1.15-g
sample of hydrogen gas was burned with excess oxygen, the
temperature increase was 14.3°C. Calculate the energy of
combustion (per gram) for hydrogen and methane.
How much heat will 56 grams of water absorb if
its temperature is raised from 45 oC to 65 oC?
4.
5. 50 grams of a metal were heated and then placed
in 100 grams of water and the temperature of the
water changed from 20 oC to 30 oC. What is the
specific heat of this metal?
10
6. A 50 grams sample of water at 56 oC is added to a
75 grams sample of water at 60 oC. Calculate the
final temperature of the water if no energy is lost to
the surroundings.
11
Hess's Law:
Hess's Law says that the net change in enthalpy of a
reaction will be the sum of the enthalpy changes that
occur along the way. Many reactions contain several
steps. If so, then each step will have its own enthalpy
change.
If all of the changes are added together, then the sum of
that addition will be the net change for the system.
Solving Hess’s law problems
♦ First decide how to rearrange equations so reactants and products are on appropriate
sides of the arrows.
♦ If equations had to be reversed, reverse the sign of ΔH
♦ If equations had be multiplied to get a correct coefficient, multiply the ΔH by this
coefficient since ΔH’s are in kJ/MOLE (division applies similarly)
♦ Check to ensure that everything cancels out to give you the exact equation you
want.
For example the reaction: C2H4 (g) + H2 (g) → C2H6 (g), occurs with the following steps each with different enthalpy changes.
C2H4 (g) + 3 O2 (g) −−> 2 CO2 (g) + 2 H2O (l)
ΔH = -­‐1411. kJ
C2H6 (g) + 3½ O2 (g)-­‐-­‐> 2 CO2 (g) + 3 H2O (l)
ΔH = -­‐1560. kJ
H2 (g) + ½ O2 (g) −−> H2O (l)
ΔH = -­‐285.8 kJ
If all of the changes are added together, then the sum of
that addition will be the net change for the system.
12
1. From the following heats of reaction
2 SO2(g) + O2(g) → 2 SO3(g) ΔH = – 196 kJ
2 S(s) + 3 O2(g) → 2 SO3 (g) ΔH = – 790 kJ
calculate the heat of reaction for
S(s) + O2(g) → SO2(g)
ΔH = ? kJ
2. Calculate ΔH for the reaction: C2H4 (g) + H2 (g) → C2H6 (g), from the following data.
C2H4 (g) + 3 O2 (g) → 2 CO2 (g) + 2 H2O (l)
ΔH = -­‐1411. kJ
C2H6 (g) + 3½ O2 (g) → 2 CO2 (g) + 3 H2O (l)
ΔH = -­‐1560. kJ
H2 (g) + ½ O2 (g) → H2O (l)
ΔH = -­‐285.8 kJ 3. Calculate ΔH for the reaction 4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g), from the following data.
N2 (g) + O2 (g) → 2 NO (g)
ΔH = -­‐180.5 kJ
N2 (g) + 3 H2 (g) → 2 NH3 (g)
ΔH = -­‐91.8 kJ
2 H2 (g) + O2 (g) → 2 H2O (g) ΔH = -­‐483.6 kJ
4. Find ΔH° for the reaction 2H2(g) + 2C(s) + O2(g) → C2H5OH(l), using the following thermochemical data.
C2H5OH (l) + 2 O2 (g) → 2 CO2 (g) + 2 H2O (l)
ΔH = -­‐875. kJ
C (s) + O2 (g) → CO2 (g)
ΔH = -­‐394.51 kJ
H2 (g) + ½ O2 (g) → H2O (l)
ΔH = -­‐285.8 kj 13
5. Calculate ΔH for the reaction CH4 (g) + NH3 (g) → HCN (g) + 3 H2 (g), given:
N2 (g) + 3 H2 (g) → 2 NH3 (g)
ΔH = -­‐91.8 kJ
C (s) + 2 H2 (g) → CH4 (g)
ΔH = -­‐74.9 kJ
H2 (g) + 2 C (s) + N2 (g) → 2 HCN (g)
ΔH = +270.3 kJ 6. Calculate ΔΗ for the reaction 2 Al (s) + 3 Cl2 (g) → 2 AlCl3 (s) from the data. 2 Al (s) + 6 HCl (aq) → 2 AlCl3 (aq) + 3 H2 (g)
ΔH = -­‐1049. kJ
HCl (g) → HCl (aq)
ΔH = -­‐74.8 kJ
H2 (g) + Cl2 (g) → 2 HCl (g)
ΔH = -­‐1845. kJ
AlCl3 (s) → AlCl3 (aq)
ΔH = -­‐323. kJ
14
Enthalpies of Formation
A formation reaction is a synthesis reaction in which
a compound is formed from its elements when all
reactants and products are at their standard state.
A degree symbol indicates that a process has been
carried out at its standard state.
ΔHo
Definitions of standard states
A compound
The standard state of a gaseous substance is at 1 atm.
For a pure substance in a liquid or solid state the
standard states is the pure liquid or solid
For a substance present in a solution, the standard
state is a concentration of exactly 1 M.
An element
The standard state of an element is the form in which
the element exists under conditions of 1 atm and 25
o
C.
The change in enthalpy for a given reaction can be
calculated from the enthalpies of formation of
reactants and products.
15
Standard enthalpy of formation is the enthalpy
change for the formation of one mole of that
substance from its elements in their most stable
thermodynamic forms at 25 oC and one atm.
16
Calculate ΔH for the following reaction:
8 Al(s) + 3 Fe3O4(s) --> 4 Al2O3(s) + 9 Fe(s)
Solution
ΔH for a reaction is equal to the sum of the heats of
formation of the product compounds minus the sum
of the heats of formation of the reactant compounds:
ΔH = ΔHf products - ΔHf reactants
Omitting terms for the elements, the equation
becomes:
ΔH = 4 ΔHf Al2O3(s) - 3 ΔHf Fe3O4(s)
The values for ΔHf may be found in the Heats of
Formation of Compounds table in the back of the
textbook. Plugging in these numbers:
ΔH = 4(-1669.8 kJ) - 3(-1120.9 kJ)
ΔH = -3316.5 kJ
1)
Calcium carbonate decomposes at high temperature to form carbon dioxide and
calcium oxide:
CaCO3 à CO2 + CaO
Given that the heat of formation of calcium carbonate is –1207 kJ/mol, the heat of
formation of carbon dioxide is –394 kJ/mol, and the heat of formation of calcium
oxide is –635 kJ/mol, determine the heat of reaction.
17
2)
Carbon tetrachloride can be formed by reacting chlorine with methane:
CH4 + 2 Cl2 à CCl4 + 2 H2
Given that the heat of formation of methane is –75 kJ/mol and the heat of
formation of carbon tetrachloride is –135 kJ/mol, determine the heat of reaction.
3)
When potassium chloride reacts with oxygen under the right conditions,
potassium chlorate is formed:
2 KCl + 3 O2 à 2KClO3
Given that the heat of formation of potassium chloride is –436 kJ/mol and the heat
of formation of potassium chlorate is –391 kJ/mol, determine the heat of reaction.
18
Second Law Of Thermodynamics
System-The part of the universe that is under study.
(open or closed)
State Function- ΔH, ΔS, ΔG
Standard State-1 atm, 25 oC, 1 mole, ΔHO, ΔSO, ΔGO
Spontaneous Processes
• A spontaneous process occurs without any
outside intervention.
• Does not have to be fast.
Examples
• A rock rolling down a hill
• Combustion of gasoline
• Melting of ice above 0 oC
• Graphite turning to diamond (spontaneous even
though it is very slow.
In all spontaneous processes the reverse will not be
spontaneous
19
Entropy (S)
• Measurement of the disorder of a system
• The number of possible arrangements of the
particles in a system
Change of entropy (Δ S)
Δ S positive, more entropy in products
Δ S negative, less entropy in products
A process is more likely if the ΔS is positive
Predict the sign of the entropy change for the
following processes
a. Solid sugar is added to water to form a solution.
b. Iodine vapor condenses on a cold surface to form
crystals.
c. solid CO2 evaporates
d. The pressure of a gas changes from 1 atm to 1.0
x 10-2 atm
20
In all of these properties the driving force is an
increase in entropy
Changes that show increasing entropy
• Solidà liquid àgas
• Increasing the number of moles of a gas
Ex. C6H12O6(s) + O2(g) à 6CO2(g) + 6H2O(g)
• Increasing temperature
• Decrease in pressure
Entropy measures the spontaneous dispersal of energy: how
much energy is spread out in a process, or how widely spread
out it becomes — at a specific temperature
Entropy and the Second Law of Thermodynamics
The Second Law of Thermodynamics
•
•
The entropy of the universe increases in a spontaneous process.
Nature heads to increasing disorder.
Energy spontaneously disperses from being localized to
becoming spread out if it is not hindered from doing so.
21
Entropy changes in chemical reactions
Entropy changes in chemical reactions are calculated
by using the following:
ΔSo = Soproducts - Soreactants
Predict whether the entropy of each of the following systems should increase, decrease,
or remain approximately constant during the following reactions. If the entropy does
change, give the sign of S.
(a) MnO2(s) Mn(s) + O2(g)
(b) H2(g) + Br2(l)
2HBr(g)
(c) Cu(s) + S(g) CuS(s)
(d) 2LiOH(s) + CO2(g)
Li2CO3(s) + H2O(g)
(e) CH4(g) + O2(g)
C(s) + 2H2O(g)
(f) CS2(g) + 3Cl2(g)
CCl4(g) + S2Cl2(g)
Calculating entropy changes in chemical reactions
C6H12O6(s) + O2(g) à 6CO2(g) + 6H2O(g)
ΔSo = 212 205
214
70 (kj mol-1)
Calculate the entropy change in the reaction above
Predict ΔS for the following reactions and then
calculate using standard values
a. CO2(s) à CO2(g)
b. H2(g) + Cl2(g) à 2HCl(g)
c. KNO3(s) à KNO3(l)
d. C(diamond) à C(graphite)
22
Calculate the entropy change in J/K for each of the following reactions
using thermo. tables.
(a) CaO(s) + 2 HCl(g) ---> CaCl2(s) + H2O(l)
(b) C2H4(g) + H2(g) ---> C2H6(g)
.
23
Free energy (ΔG)
Free energy relates temperature and heat (ΔH
enthalpy) with entropy (ΔS).
ΔG = ΔH - TΔS
if Δ G is negative the process will be
spontaneous
if Δ G is positive the process will not be
spontaneous
Use standard free energy data to determine the free energy change for each of the
following reactions, which are run under standard state conditions. Appendix 4 in your
book
(a) MnO2(s) Mn(s) + O2(g)
(b) H2(g) + Br2(l)
2HBr(g)
(c) 2LiOH(s) + CO2(g) Li2CO3(s) + H2O(g)
(d) SnCl4(l)
SnCl4(g)
ii. For which of the reactions in this question does the free energy change indicate a
spontaneous reaction under standard state conditions?
The standard molar enthalpies of formation of NO(g), NO2(g), and N2O3(g) are 90.25 kJ
mol -1, 33.2 kJ mol -1, and 83.72 kJ mol -1, respectively. Their standard molar entropies
are 210.65 J mol -1 K-1, 239.9 J mol -1 K-1, and 312.2 J mol -1 K-1, respectively.
(a) Use these data to calculate the free energy change for the following reaction at
25.0oC.
N2O3(g)
NO(g) + NO2(g)
(b) Repeat the above calculation for 0.00o C and 100.0oC, assuming that the enthalpy and
entropy changes do not vary with a change in temperature. Is the reaction spontaneous at
0.00oC? At 100.0oC?
24
Free energy and spontaneity
ΔH
+
Fill out the table using ΔG = ΔH - TΔS
Spontaneous/
ΔS
ΔG
nonspontaneous
+
+
Free Energy changes in chemical reactions can be calculated the same as entropy and
enthalpy
Using the table, calculate ΔGo for the following reactions:
1. CH4 (g) + 2 O2 (g) --------->
CO2 (g)
+ 2 H2O (l)
ΔGoproducts
ΔGoreactants
(1 mol CH4)(-50.81 kJ/mol) = -50.81
(1 mol CO2)(-394.4 kJ/mol) = -394.4 kJ
kJ
(2 mol H2O)(-237.192 J/Kmol) = -474.384 kJ
(2 mol O2)(0 kJ/mol) = 0 kJ
o
ΣΔG products = -868.784 kJ
ΣΔGoreactants = -50.81 kJ
ΔGorxn= -868.784 kJ - (-50.81 kJ) = -817.974
spontaneous reaction
kJ
2. 2 MgO (s) -------->
ΔGoproducts
2 Mg (s) + O2 (g)
ΔGoreactants
(2 mol Mg)( 0 kJ/mol) = 0 kJ (2 mol MgO)(-569.0 kJ/mol) = -1138.0 kJ
(1 mol O2)( 0 kJ/mol) = 0 kJ
€ ΣΔGoproducts = 0 kJ
ΣΔGoreactants = -1138.0 kJ
25
ΔGorxn= 0 kJ - (-1138.0 kJ) = +1138.0 kJ
nonspontaneous reaction
Free energy determines whether the reaction will
occur not how fast it is. A spontaneous process can
be slow.
Negative ΔG AND large Keq do not affect the rate
Keq Is the equilibrium Constant
26
Free Energy and Equilibrium
The free energy change for any reaction under
nonstandard conditions, ΔG can be calculated from
the standard free energy change, ΔGo, by the
following equation:
ΔG = ΔGo + RT ln Q
ΔG = The free energy change under nonstandard
conditions
ΔGo =The free energy change under standard
conditions: 25 OC and one atm partial pressure of all
gases and 1 M concentration for all solutes
R = 8.314 J/mol K, the ideal gas constant
T = The temperature in Kelvin
Q = the reaction quotient
Q = [product 1][product 2]….
[reactant1][reactant 2]….
27
Calculate ΔG for the reaction:
NO(g) + O3 (g) à NO2(g) + O2 (g)
GokJ/mo
l87
163
52
0
PNO =1.00 x 10-6 atm, PO3 = 2.00 x 10-6 atm
PNO2 = 1.00 x 10-7 atm, PO2 = 1.00 x 10-3 atm
ΔG positive reaction occurs in the forward direction
ΔG negative reaction occurs in the backward
direction
ΔG = 0, reaction at equilibrium
At equilibrium ΔG is 0
Because
Gproducts = G reactans or ΔG = G producs – G reactants = 0
therefore:
ΔG = ΔGo + RT ln Q
becomes:
0 = ΔGo + RT ln K
or
ΔGo = - RT ln K
Substance
H2(g)
ΔHfo
kJ/mol
0
ΔSo
J/(mol K)
130.680
28
I2(s)
0
O2(g)
0
C(graphite) 0
C(diamond)
1.985
H2O(l)
-285.83
HI(g)
26.50
116.14
205.152
5.74
2.377
69.95
206.590
Find the enthalpy and entropy changes for the reaction:
H2(g) + I2(s) -> 2 HI(g)
Calculate the standard Gibb's energy of formation of HI,
½ H2(g) + ½ I2(g) -> HI(g),
and the equilibrium constant K for the above reaction.
Calculate the standard entropy of formation of H2O(l), its standard Gibb's
energy of formation, and the equilibrium constant K for the reaction.
H2(g) + ½ O2(g) -> H2O(l)
Estimate the standard Gibb's free energy of formation for ammonia.
.
Evaluate ΔGoreaction for the reaction:
4 NH3(g) + 3 O2(g) -> 2 N2 + 6 H2O(g)
29
Flash Card List
Thermodynamics
Energy
First Law of thermodynamics (law of conservation of mass)
Heat
calorimetry
Work
Hess’s Law
Enthalpy
Of reaction
Of combustion
Of formation
Fusion
Vaporization
System
Surroundings
Endothermic
Exothermic
State function
Entropy
Second Law of Thermodynamics
Free energy
Spontaneous reactions
Standard state
Nonspontaneous reactions
ΔH and ΔS and spontaneity
ΔG and equilibrium
ΔG and Q
Standard conditions
30
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