H/Chem 1213
Thermodynamics and Reaction Rates
Thermodynamics
Temperature
oC,
Heat
K
kJ, kcal (Cal) [1 kcal = 4.184 kJ]
measure of average KE
measure of total energy
of motion of particles
transferred from an object of high E  low E
Note: A change in T is accompanied by a transfer of heat energy.
Specific heat (c or cp):
(specific for each substance)
amount of energy required to raise the temperature of 1 g of that substance by 1 oC.
Thinking about specific heat:
List the following substances in terms of specific heat, from lowest to highest: water, gold,
ethanol, granite
Relationship between Heat and Temperature:
change in heat = specific heat x mass x change in T
q
= cp x m x t
Units: heat absorbed or released, in J;
cp in J/g.oC
m in g
t in oC or K
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H/Chem 1213
Thermodynamics and Reaction Rates
Examples
1.
How much heat energy is released to your body when a cup of hot tea containing
200. g of water is cooled from 65.0oC to body temperature, 37.0oC?
(cp of H2O =
4.18 J/goC)
A: 23.4 kJ
2.
A 4.50-g nugget of pure gold absorbed 276 J of heat. What was the final
temperature of the gold if the initial temperature was 25.0oC? (cp of gold = 0.129
J/goC)
A: 500. oC
3.
A 155-g sample of an unknown substance was heated from 25.0 oC to 40.0oC. In
the process, the substance absorbed 5696 J of energy. What is the specific heat
of the substance?
A: 2.45 J/goC
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H/Chem 1213
Thermodynamics and Reaction Rates
Calorimetry
Instrument: bomb calorimeter
Put dried food sample in “bomb”.
Burn it.
Use t of water to calculate H of the food.
The amount of energy in food is still measured in Calories, rather than kJ.
1 Calorie = 1 kcal = 4.184 kJ
Basis:
q = cpH2O x mH2O x tH2O
Step 1. Calculate q from calorimeter data.
Step 2. Relate q to the quantity of substance in the calorimeter.
Example 1
What is the heat of combustion, Hc , of sucrose, C12H22O11, if 1.500 g of sugar raises the
temperature of water (3.00 kg) in a bomb calorimeter from 18.00 oC to 19.97oC? (cp of H2O =
4.18 J/goC) (A: 16.5 kJ/g sucrose, 5650 kJ/mol sucrose
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H/Chem 1213
Thermodynamics and Reaction Rates
Example 2
You burned 1.30 g of peanuts below an aluminum can which contained 200.0 mL of ice water
(DH2O = 1.00 g/mL). The temperature of the water increased from 6.7 oC to 28.5oC. Assume
all of the heat energy produced by the peanuts went into the water in the can.
(cpH2O = 4.18 J/goC)
What was the energy content of peanuts, in kJ/g (i.e. Hcomb of peanuts?
A: 14.0 kJ/g
Heats of Hx
The change in heat energy which occurs during a variety of processes can be described as
Hx, where x = process
H = change in enthalpy (i.e. change in the amount of heat)
Hcomb = heat of combustion (kJ/mol)
energy released when 1 mol of a substance undergoes complete combustion
Hfus = heat of fusion (kJ/mol)
energy needed to melt 1 mol of a substance
Hvap = heat of vaporization (kJ/mol)
energy needed to boil 1 mol of a substance
Hrxn = heat of reaction (kJ)
heat energy absorbed/released during a reaction

o
Hf = heat of formation (kJ/mol)
energy absorbed/released during synthesis of 1 mol of a compound from its elements at 298
K and 1 atm pressure
Hsol = heat of solution (kJ)
energy absorbed/released when substance is dissolved in a solvent
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H/Chem 1213
Thermodynamics and Reaction Rates
Thermodynamics Word Problems Using Hx Data
Two of the following three factors will be given:
amount of substance of interest (usually in grams),
Hx for the substance
total kJ
e.g. Calculate the heat required to melt 25.7 g of solid methanol at its melting point.
Hfus of methanol = 3.22 kJ/mol (from Table 16-6 on p. 502)
1.
Convert mass of methanol to moles.
2.
Multiply moles of methanol by Hfus.
A: 2.58 kJ
Practice problem:
How much heat is needed to vaporize 343 g of liquid ethanol (CH3COOH)?
Hvap = 38.6 kJ/mol
(A = 287 kJ)
p. 525, #82
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H/Chem 1213
Thermodynamics and Reaction Rates
Calculating Energy Changes using Hx Data – Lab example
Determination of heat of Hx:
1.
Measure total # kJ (often by calorimetry), using
q = cp x m x t
2.
Measure mass of substance, e.g. paraffin lost when candle is burned
3.
Divide #kJ by mass of substance of interest  kJ/g
4.
Convert to kJ per mole by performing a mass  mole conversion:
___
(
)
kJ molar mass (g)
g
1 mol
 ____kJ/mol
e.g. Combustion of a Candle Lab
1.
Total # kJ was determined by calorimetry, using qH2O = cpH2O x mH2O x ΔTH2O
2.
Mass of wax burned was determined by difference in mass of the candle + card before
and after the experiment.
3.
#1  #2  kJ/g
4.
___
(note units include both kJ and g)
(
kJ molar mass of paraffin (g)
g
1 mol
(convert grams to moles)
Paraffin = C26H54
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)
 ____kJ/mol paraffin
H/Chem 1213
Thermodynamics and Reaction Rates
Heating Curves
1.
Use Table 16-6 on p. 502 for Hvapo and Hfuso for different substances.
Calculate the energy required to melt 8.5 g of ice at 0oC.
The molar heat of fusion for ice is 6.02 kJ/mol.
A: 2.8 kJ
2.
Calculate the total energy change (in kJ) required to
a) heat 25 g of liquid water from 25oC to 100.oC,
then
b) change it to steam at 100.oC.
cpliquid water = 4.18 J/goC, HvapH2O = 40.6 kJ/mol.
A: a) 7.8 kJ; b) 56 kJ; Total = 64 kJ
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H/Chem 1213
Thermodynamics and Reaction Rates
Thermochemical Stoichiometry
The amount of energy (in kJ) can be incorporated into mole ratios.
e.g. C6H12O6(aq) + 6 O2(g)  6 CO2(g) + 6 H2O(l) + 2870 kJ
H = -2870 kJ/mol glucose
Note that because H is a negative number, the kJ go on the products side of the equation.
Note that the number of moles has an unlimited number of sig figs while you must take the
number of sig figs in the energy term into consideration.
Examples of mole ratios derived from the above equation:
1 mol C6H12O6
6 mol CO2
2870 kJ
6 mol H2O
6 mol O2
6 mol H2O
1 mol C6H12O6
2870 kJ
Example Problems:
1.
How much energy (in kJ) will be released when 675 g of glucose
is burned? (A: 10,800 kJ)
2.
If 398 kJ is released when a certain amount of glucose is burned,
how many grams of oxygen are consumed? (A: 26.6 g)
3.
If 5782 kJ is released when a certain amount of glucose is
burned, how many liters of carbon dioxide are released,
assuming the reaction takes place at STP? (A: 271 L)
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H/Chem 1213
Thermodynamics and Reaction Rates
Enthalpy
enthalpein = to warm (Greek)
Enthalpy (H) = total amount of energy a substance contains
Enthalpy change (H)= comparison between enthalpy of products and reactants in a
chemical equation
H = Hproducts – Hreactants
Hrxn = +
if endothermic:
Hrxn = -
if exothermic: energy of reactants > energy of products
energy of products > energy of reactants
2 H2 + O2  2 H2O + 483.6 kJ
H = -483.6 kJ
Hess’s Law
(Version 1)
The total enthalpy change for a chemical or physical change is the same whether it takes
place in one or several steps.
Example:
What is the change in enthalpy for converting graphite to diamond?
Given:
C(graphite) + O2(g)  CO2(g)
H = -393.5 kJ/mol C
C(diamond) + O2(g)  CO2(g)
H = -395.4 kJ/mol C
Step 1: Write rxn.
C(graphite)  C(diamond)
Step 2: Add up rxns to get total rxn, canceling similar terms.
H = -393.5 kJ/mol C
C(graphite) + O2(g)  CO2(g)
CO2(g)
 C(diamond) + O2(g)
H = +395.4 kJ/mol C

C(graphite)
 C(diamond)
H = +1.9 kJ/mol C
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H/Chem 1213
Thermodynamics and Reaction Rates
Hess’s Law Version 1 Practice
Calculate the heat of formation of pentane (C5H12).
Given:
C(s) + O2(g)  CO2(g)
H = -393.51 kJ/mol
H2(g) + ½ O2(g)  H2O(l)
H = -285.83 kJ/mol
C5H12 + 8 O2(g)  5 CO2(g) + 6 H2O(l)
H = -3536.1 kJ/mol
_______________________________________________________
Final equation: 5 C(s) + 6 H2(g)  C5H12
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Hfo = ?
H/Chem 1213
Thermodynamics and Reaction Rates
Hess’s Law
(Version 2)
The enthalpy change of a rxn is equal to the heats of formation of the products minus the
heats of formation of the reactants (and taking stoichiometric relationships into account).
H = Hfo (products) – Hfo (reactants)
Example
Calculate the change in enthalpy for this reaction:
CH4 + 2 O2  CO2 + 2 H2O
Given:
Hfo CO2 = -393.5 kJ/mol
Hfo H2O(l) = -285.8 kJ/mol
Hfo CH4 = -74.86 kJ/mol
Hfo O2
=0
(Note: Hfo of free elements = 0)
Step 1: Calculate the total Hfo for the products and the reactants.
Hproducts = (-393.5 kJ)(1 mol CO2) + (-285.8 kJ)(2 mol H2O) = -965.1 kJ
1 mol
1 mol
Hreactancts = (-74.86 kJ)(1 mol CH4) = -74.86 kJ
1 mol
Step 2. Plug these values into the following equation.
Hrxn = Hproducts – Hreactants
= -965.1 kJ – (-74.86 kJ) = -890.2 kJ
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H/Chem 1213
Thermodynamics and Reaction Rates
Practice problems – Hess’s Law Version 2
Use standard enthalpies of formation from Appendix C, Table C-13 on p. 921 to calculate
Horxn for the following reaction:
4 NH3(g) + 7 O2(g)  4 NO2(g) + 6 H2O(l)
A: -1397.9 kJ
Use standard enthalpies of formation from Appendix C, Table C-13 to calculate Horxn for the
thermite reaction:
2 Al(s) + Fe2O3(s)  2 Fe(s) + Al2O3(s)
A: - 851.5 kJ
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H/Chem 1213
Thermodynamics and Reaction Rates
Entropy
entropy = en- (Greek, = in;
trope (Greek, = a turning)
entropy (S) = degree of disorder (“chaos”)

In nature, things tend toward disorder.
It is usually much easier to go from low S  high S

All chemical/physical changes involve changes in entropy.

This natural tendency to disorder can be overcome by enthalpy.
Entropy change (S) = comparison in entropy of the products and reactants in a chemical
equation
S = Sproducts – Sreactants
If S = +,
 in entropy,
entropy of products > entropy of reactants
If S = -,
 in entropy,
entropy of reactants > entropy of products
Units of S = J/mol.K
Always use K in S calculations (K = 273 + oC)
Factors That Increase Entropy
1.
change state: solid  liquid  gas
2.
when # product particles > # reactant particles (decomposition, dissolving)
3.
 T (more collisions, more chaos)
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H/Chem 1213
Thermodynamics and Reaction Rates
Calculating S from So
ΔS = S°Products - S°Reactants
Use the information in the table below to calculate ∆S for the following reaction:
Given: 2 SO2(g) + O2(g)  2 SO3(g)
Hfo
So
(kJ/mol) (J/K.mol)
SO2(g)
-297
248
SO3(g)
-396
257
0
205
O2(g)
Would this be favored?
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H/Chem 1213
Thermodynamics and Reaction Rates
Spontaneous Reactions
Nonspontaneous = does not happen
Spontaneous = happens! (not a function of how fast it happens)
If there is a change in conditions, nonspontaneous  spontaneous
e.g. T, P, concentration, catalyst
Spontaneity depends on:
1.
entropy
2.
H of reaction (endothermic or exothermic)
3.
Temperature (T in K)
e.g. Burn octane: 2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g) + kJ
entropy  due to (l)  (g) favored and more product particles
exothermic - favored
Gibbs Free Energy Calculation
G = H - TS
G = - kJ: yes! spontaneous
G = + kJ; no! not spontaneous
Relating Entropy and Enthalpy to Spontaneity
H
S
Spontaneity
(-) value
(+) value
Yes
exothermic
disorder
(-) value
(-) value
exothermic
order
(+) value
(+) value
endothermic
disorder
(+) value
(-) value
endothermic
order
Example
2 K(s) + 2 H2O(l)  2 KOH + H2(g)
Yes at low T
H2O(g)  H2O(l)
Yes at high T
H2O(s)  H2O(l)
16 CO2(g) + 18 H2O(l)  2 C8H18(l) + 25 O2(g)
No
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H/Chem 1213
Thermodynamics and Reaction Rates
Practice Problem – Gibbs Free Energy
Will the thermite reaction occur spontaneously at room temperature (298 K)?
2 Al(s) + Fe2O3(s)  2 Fe(s) + Al2O3(s)
H = -851.5 kJ
S = -0.038 kJ/K
G = ?
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H/Chem 1213
Thermodynamics and Reaction Rates
Reaction Pathways
Why don’t exothermic reactions occur spontaneously (e.g. with no spark)?
Energy is needed to overcome the mutual repelling of the atoms involved in the reaction.
Activation energy (Ea or Ae) = initial output of energy, results in an activated complex
Note: Different reactions have different activation energies.
Endothermic reaction
Exothermic reaction
(a) Activation energy
(b) Energy released by activated complex
(c) ∆Hrxn = Hproducts – Hreactants
Effect of Adding a Catalyst on Reaction Rate
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