Thermodynamics
Temperature Vs. Heat
• Temperature
– K, °C
– Measure of average KE
of motion of particles
• Heat
– kJ, kcal (Cal)
– 1kcal=4.184 kJ
– Measure of total
energy transferred
from an object of high
E to low E
Note: A change in T is accompanied
by a transfer of energy
Specific Heat
• c or cp
• specific for each substance
= amount of energy required to raise the
temperature of 1 g of a substance by 1°C
Thinking about specific heat
List the following substances in terms of
specific heat, from lowest to highest:
water, gold, ethanol, granite
Thinking about specific heat
List the following substances in terms of
specific heat, from lowest to highest:
Answer:
gold, granite, ethanol, water
0.129
0.803
2.44
4.184
J/goC
Relationship among
temperature, heat and mass
Change in heat = specific heat x mass x change in T
q
=
cp
x
m x
Units: heat absorbed or released, J
cp in J/g°C
m in g
ΔT in °C or K
Δt
Example Problems
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?
(cpH2O = 4.18 J/goC)
A: 23.4 kJ
Example Problems
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/goC)
Hint: begin by solving for DT
A: 500. oC
Example Problems
3. A 155 g sample of an unknown
substance was heated from 25.0oC 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
Using a Calorimeter (WS)
1. Put dried food sample in
“bomb”
2. Burn it
3. Use ∆T of water to
calculate ∆Hcomb of the
food.
4. The amount of energy in
food is measured in Cal,
rather than kJ.
5. 1 Calorie = 1 kcal =
4.184 kJ
Calorimetry
Step 1
Calculate q from calorimeter data.
q = cpH2O x mH2O x DtH2O
Step 2
Relate q to the quantity of substance in the
calorimeter.
Calorimetry Problem
What is the heat of combustion, DHc , of
sucrose, C12H22O11, if 1.500 g of sugar
raises the temperature of water (3.00 kg)
in a bomb calorimeter from 18.00oC to
19.97oC? (cpH2O = 4.18 J/goC)
A:
16.5 kJ/g sucrose, 5650 kJ/mol sucrose
Calorimetry Practice Problem
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
6.7oC 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.
DHcomb of peanuts?
A: 14.0 kJ/g
Calorimetry Problem, p. 881, #6
(Honors)
• A 75.0 g sample of a metal is placed in boiling water until its
temperature is 100.0oC.
• A calorimeter contains 100.0 g of water at a temperature of 24.4oC.
• The metal sample is removed from the boiling water and
immediately placed in water in the calorimeter.
• The final temperature of the metal and water in the calorimeter is
34.9oC.
• Assuming that the calorimeter provides perfect insulation, what is
the specific heat of the metal?
• A: 0.900 J/goC
Heats of …
• ΔH = change in
• ΔHcomb= … combustion
enthalpy (i.e. change
• ΔHfus= … fusion
in the amount of heat) • ΔH = … vaporization
vap
• The heat change
• ΔHrxn= … reaction
which occurs during
• ΔHf°= … formation
processes can be
• ΔHsol= … solution
described as ΔHx,
x=process name
Units: All in kJ/mol except
heats of reaction and
solution, which are just in kJ
Thermodynamics Word Problems
Using Heats of _____ Data
Two of the following three factors will be given:
• amount of substance of interest
(usually in grams),
• heat of _____ for the substance (kJ/g or kJ/mol)
• total kJ
Thermodynamics Word Problems
Using DHx Data
e.g. Calculate the heat required to melt 25.7 g of
solid methanol at its melting point.
DHfus 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 DHfus.
A: 2.58 kJ
Thermodynamics Word Problems
Using DHx Data
•
How much heat is needed to vaporize
343 g of acetic acid (CH3COOH)?
DHvap = 38.6 kJ/mol
(A = 220. kJ)
Calculating E change using ΔHx
Lab Data
Determining ΔHx
1. Measure total # kJ (calorimetry) using
q = cp x m x ΔT
2. Measure mass of substance
3. Divide # kJ by mass of substance  kJ/g
4. Convert to kJ/mol using massmole
conversion
Combustion of a Candle Lab
What is the DHcomb of paraffin wax?
1. Measure total # kJ using
qH2O = cpH2O x mH2O x ΔTH2O
(i.e. q of water in can)
2. Measure mass of substance:
mwax burned = mcandle+card before and after expt.
Combustion of a Candle Lab
What is the DHcomb of paraffin wax?
3. Divide # kJ by mass of substance:
#1 #2  kJ/g
(note units include both kJ and g)
4. Convert to kJ/mol using massmole conversion
__ kJ (MM of paraffin in g)  ____ kJ/mol paraffin
g
1 mol paraffin
paraffin = C26H54
Temperature Changes of Water
(WS)
Using Heat of Fusion
Use Table 16-6 on p. 502 for DHvapo and DHfuso
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
Using Heat of Vaporization
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, DHvapH2O = 40.6 kJ/mol.
A: a) 7.8 kJ
b) 56 kJ
Total = 64 kJ
Thermochemical Stoichiometry
• The amount of energy (in kJ) can be
incorporated into mole ratios.
• C6H12O6(aq) + 6 O2(g)  6 CO2(g) + 6 H2O(l) + 2870 kJ
• ΔH = -2870 kJ/mol glucose
• Mole Ratios Examples
1 mol C6H12O6
6 mol O2
6 mol CO2
6 mol O2
2870 kJ
1 mol C6H12O6
6 mol H2O
2870 kJ
Thermochemical Stoichiometry
Problems
C6H12O6(aq) + 6 O2(g)  6 CO2(g) + 6 H2O(l) + 2870 kJ
1. How much energy (in kJ) will be released
when 675 g of glucose is burned?
A: 10,800 kJ
Thermochemical Stoichiometry
Problems
C6H12O6(aq) + 6 O2(g)  6 CO2(g) + 6 H2O(l) + 2870 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
Thermochemical Stoichiometry
Problems
C6H12O6(aq) + 6 O2(g)  6 CO2(g) + 6 H2O(l) + 2870 kJ
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
Enthalpy
• enthalpien= to warm (Greek)
• Enthalpy (H) = total E of a substance
• Enthalpy change (ΔH) = comparison
between H of products and H of reactants
in a reaction (ΔH=Hproducts - Hreactants)
Enthalpy
• ΔHrxn is + if endothermic
(energy of products > energy of reactants)
• ΔHrxn is - if exothermic
(energy of reactants > energy of products)
2 H2 + O2  2 H2O + 483.6 kJ
DH = - 483.6 kJ
Hess’s Law
• The total enthalpy change for a chemical
or physical change is the same whether it
takes place in one or several steps.
• “state function”
Hess’s Law
• Version 1: Add equations together
• Version 2: Use heats of formation (DHfo)
of products and reactants
Hess’s Law Version 1
Example:
What is the change in enthalpy for converting graphite to diamond?
Given:
C(graphite) + O2(g)  CO2(g)
DH = -393.5 kJ/mol C
C(diamond) + O2(g)  CO2(g)
DH = -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.
C(graphite) + O2(g)  CO2(g)
DH = -393.5 kJ/mol C
CO2(g)
 C(diamond) + O2(g)
DH = +395.4 kJ/mol C
______________________________________________
C(graphite)
 C(diamond)
DH = +1.9 kJ/mol C
Note:
Focus is on the DH.
Adding equations is the way you get there.
Hess’s Law Version 1 Practice
Calculate the heat of formation of pentane (C5H12).
Given:
C(s) + O2(g)
 CO2(g)
DH = -393.51 kJ/mol
H2(g) + ½ O2(g)  H2O(l)
DH = -285.83 kJ/mol
C5H12(l) + 8 O2(g)  5 CO2(g) + 6 H2O(l)
DH = -3536.1 kJ/mol
Final equation:
5 C(s) + 6 H2(g)  C5H12(l)
(A: -146.4 kJ)
DHfo = ?
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).
DH = SDHfo (products) – SDHfo (reactants)
Hess’s Law (Version 2)
Example: Calculate ΔH for this reaction
CH4(g) + 2O29g)  CO2(g) + 2H2O(l)
Given:
DHfo CO2(g)= -393.5 kJ/mol
DHfo H2O(l) = -285.8 kJ/mol
DHfo CH4(g) = -74.86 kJ/mol
DHfo O2(g) = 0 (Note: DHfo of free elements = 0)
Hess’s Law (Version 2)
Step 1: Calculate the total DHfo for the products and the
reactants.
DHproducts = (-393.5 kJ)(1 mol CO2) + (-285.8 kJ)(2 mol H2O)
1 mol
1 mol
= -965.1 kJ
DHreactants = (-74.86 kJ)(1 mol CH4)
1 mol
= -74.86 kJ
Hess’s Law (Version 2)
Step 2: Plug these values into the following
equation.
DHrxn = DHproducts – DHreactants
= -965.1 kJ – (-74.86 kJ)
= -890.2 kJ
Hess’s Law (Version 2) Practice
Use standard enthalpies of formation from
Appendix C, Table C-13 on p. 921 to
calculate DHorxn for the following reaction:
4 NH3(g) + 7 O2(g)  4 NO2(g) + 6 H2O(l)
A: -1397.9 kJ
Hess’s Law (Version 2) Practice
Use standard enthalpies of formation from
Appendix C, Table C-13 to calculate DHorxn
for the thermite reaction:
2 Al(s) + Fe2O3(s)  2 Fe(s) + Al2O3(s)
A: - 851.5 kJ
Entropy: S
en- (Greek, = in) trope (Greek, = a turning)
Entropy: degree of disorder
or chaos
In nature, things tend toward
disorder
It is usually much easier to go
from low S to high S
All chemical/physical changes
involve changes in entropy.
This natural tendency to
disorder can be overcome
by enthalpy.
Change in Entropy: ∆S
Entropy change (∆S) = comparison in entropy of the
products and reactants in a chemical equation
∆S = Sproducts – Sreactants
+∆S increased entropy: S products > S reactants
-∆S decreased entropy: S products < S reactants
Units of S = J/mol•K
Always use K in S calculations
Factors that Increase Entropy
• state change from solid to liquid to gas
• increased temperature (more collisions, more
disorder)
• more particles of product than reactant
(decomposition, dissolving)
Calculating ∆S from So
ΔS = S°Products - S°Reactants
Use the information in the table below to calculate
∆S for the following reaction:
2 SO2(g) + O2(g)  2 SO3(g)
∆Hfo (kJ/mol)
So (J/K.mol)
SO2(g)
-297
248
SO3(g)
-396
257
O2(g)
0
205
[2 mol SO3(257 J/mol K)] –
[2 mol SO2 (248 J/mol K) + 1 mol O2(205 J/mol K)] =
514 J/K – 701 J/K = -187 J/K Would this be favored?
Spontaneity
Spontaneous = happens
Spontaneity depends on: ∆S and ∆H of
reaction
Example: Combustion of octane is spontaneous
2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g) + kJ
Exothermic Reaction (-∆H)
Entropy increased (+∆S) due to
– More particle energy (l)  (g)
– More product particles
Spontaneous Reactions
Spontaneity depends on
1. Entropy of reaction (DS)
2. Enthalpy of reaction (DH)
3. Temperature (T in K)
 Gibbs Free Energy calculation
Example
• Burn octane:
2 C8H18(l) + 25 O2(g)
 16 CO2(g) + 18 H2O(g) + kJ
• increased entropy due to (l)  (g) favored
and more product particles
• exothermic - favored
Gibbs Free Energy
ΔG = ΔH - (TΔS)
ΔG = - kJ; yes! spontaneous
ΔG = + kJ; no! nonspontaneous
Gibbs Free Energy: ∆G = ∆H - T∆S
Reaction is spontaneous if the sign of ∆G is negative.
Is this reaction spontaneous @ 25oC?
2 SO2(g) + O2(g)  2 SO3(g)
T
298 K
Favored?
∆H
- 198 kJ
Y
∆S
-187 J/K
N
To take both the change in enthalpy AND entropy into
account, calculate ∆G:
∆G = ∆H - T∆S
= -198 kJ – 298 K (-0.187 kJ/K) watch the units…
= -198 kJ – (-55.7 kJ) = -142 kJ
So, is this reaction spontaneous at 298 K? YES.
Gibbs Free Energy: ∆G = ∆H - T∆S
Is this reaction spontaneous @ 2000.oK?
2 SO2(g) + O2(g)  2 SO3(g)
T
2000 K
Favored?
∆H
- 198.2 kJ
Y
∆S
-187 J/K
N
∆G = ∆H - T∆S
= -198.2 kJ – 2000. K (-0.187 kJ/K)
= -198.2 kJ – (-374 kJ) = 176 kJ
So, is this reaction spontaneous at 2000. K? NO.
The reaction is spontaneous at low temps (enthalpy
wins) but not spontaneous at very high temps
(entropy wins) (decomposition occurs….)
Gibbs Free Energy
• Will the thermite reaction occur
spontaneously at room temperature (298 K)?
• 2 Al(s) + Fe2O3(s)  2 Fe(s) + Al2O3(s)
DH = -851.5 kJ
DS = -0.038 kJ/K
DG = ?
Gibbs Free Energy: ∆G = ∆H - T∆S
Will the thermite reaction occur spontaneously at
room temp (298 K)?
2 Al(s) + Fe2O3(s) ➝ 2 Fe(s) + Al2O3(s)
∆H = -851.5 kJ
∆S = -0.038 kJ/K
∆G = ?
-851.5 kJ – [298 K (-0.038 kJ/K)] = - 840 kJ
Yes, ∆G is negative
Free Energy Practice Problems
(extra)
1. For the reaction
NH4Cl(s)  NH3(g) + HCl
at 25oC, DH = 176 kJ/mol,
DS = 0.285 kJ/mol.K
Calculate DG and decide if the reaction
will be spontaneous in the forward
direction at 25oC.
Free Energy Practice Problems
extra
2. When graphite (C) reacts with hydrogen
at 27oC,
DH = -74.8 kJ/mol and
DS = -0.0809 kJ/mol.K.
Will this reaction occur spontaneously?
Free Energy Practice Problems
extra
3. Calculate the standard entropy change,
DS, that occurs when 1 mol H2O(g) is
condensed to 1 mol H2O(l) at 25oC.
So H2O(g) = 188.7 J/mol.K
So H2O(l) = 69.94 J/mol.K
Free Energy Practice Problems
extra
4. For the reaction
H2O(g) + C(s)  CO(g) + H2O(g),
DH = 131.3 kJ/mol, DS = 0.134 kJ/mol.K
a) Is the reaction spontaneous at 1350 K?
b) At what temperature is the reaction
spontaneous?
Relating Entropy and Enthalpy to
Spontaneity: DG = DH – (TDS)
DH
DS
Spontaneity
Example
(Is DG negative?)
(-)
(+)
exothermic disorder
(-) value
(-)
exothermic
order
(+) value
(+)
endothermic disorder
(+) value
endothermic
(-)
order
yes
2 K(s) + 2 H2O(l) 
2 KOH(aq) + H2(g)
Yes at low T
H2O(g)  H2O(l)
Yes at high T
(NH4)2CO3(s)
 2 NH3(g) + CO2(g)
+ H2O(g)
no
16 CO2(g) + 18 H2O(l)
 2 C8H18(l)+25 O2(g)
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.
Reaction Pathways
Activation energy (Ea) = initial output of
energy
Note: Different reactions have different
activation energies.
Reaction Profiles (WS)
(a) Activation energy
(b) Energy released by activated complex
(c) ∆Hrxn = Hproducts – Hreactants
Effect of Adding a Catalyst on
Reaction Rates