PP 16: Thermochemistry
Drill: Calculate the volume of gas released at 227oC under 83.1 kPa pressure when 320 kg of NH4NO3 is
exploded forming N2, O2, & H2O.
Drill: Draw LDDs for:
SCl2
CH2Cl2
HSO4-1
C4H6
Thermochemistry: The study of heat transfer in chemical reactions
Thermochemical Terms
Heat (H): A form of energy that can flow between samples of matter
energy
Heat cannot be measured directly; thus, we measure heat change (H)
Enthalpy Change: Heat that can flow in/out of a system (H)
System: That part of nature upon which attention is focused
Surroundings: That part of nature which we are not focused
Reaction Coordinate: A graph of energy change versus time in a
chemical reaction
Time
Exothermic Reaction: Chemical reactions that release, give off heat, or lose heat
 Products will contain less heat than the reactants or H < 0 or negative
Endothermic Reaction: Chemical reactions that absorb, take in heat, or gain heat
 Products will contain more heat than the reactants or H > 0 or positive
Thermochemistry Topics:
•
•
•
Heat Change
Calorimetry
Hess’s Law & the Thermochemical Equation (TCE)
Heat Change: Calculating the change of heat within a system
Specific Heat [C]: The heat required to raise one gram of a substance 1oC
Specific Heat Formula: H = mCT
Heat of Fusion (Hf):
• The heat required to melt one gram of a substance at its normal MP
• Hf: (J/g or J/kg)
Heat of Fusion Formula:
H = mHf
Heat of Vaporization (Hv):
• The heat required to boil one gram of a substance at its normal BP
• Hv: (J/g or J/kg)
Heat of Fusion Formula:
H = mHv
C: (J/goC, J/kgK)
Problem: Calculate the heat change when 10.0 g H2O goes from –10.0oC to 120.0oC.
MP = 0oC
cice = 2.06 J/gK
BP = 100oC
cwater = 4.18 J/gK
Hf = 334 J/g
HV = 2260 J/g
csteam = 2.02 J/gK
Problem: Calculate the heat change when 5.00 g H2O goes from –100.0oC to 200oC.
MP = 0oC
cice = 2.06 J/gK
BP = 100oC
cwater = 4.18 J/gK
Hf = 334 J/g
HV = 2260 J/g
csteam = 2.02 J/gK
Drill: Calculate the heat change when 25 g H2O goes from 140.0oC to 60.0oC.
MP = 0oC
cice = 2.06 J/gK
BP = 100oC
cwater = 4.18 J/gK
Hf = 334 J/g
HV = 2260 J/g
csteam = 2.02 J/gK
Calorimetry: The study of how to measure experimental heat transfer in a system
Calorimeter:
• The device used to measure heat transfer
• A calorimeter is an adiabatic system
• Adiabatic Systen
• A system that exchanges zero heat with its surroundings
• Hsystem = 0
• A calorimeter is used to measure experimental yields
Heat or enthalpy change (H ~ q): Hsystem = Hall parts
Calorimetry calculations:
 Hsystem = 0
 Hsys = Hcal + Hrxn
 Hrxn = -Hcal (all parts)
 Hrxn = -mcDTcal (all)
Problem: Calculate Hrxn when Q reacts in a 1.5 kg calorimeter containing 2.5 kg water. The temperature
changes from 22.50oC to 26.50oC. (Cwater = 4.18 J/gK & Ccal = 2.00 J/gK)
Problem: When X reacts in a 2.0 kg calorimeter containing 1.5 kg water, the temperature changes from
22.50oC to 26.50oC. Calculate Hrxn (Cwater = 4.18 J/gK & Ccal = 2.00 J/gK)
Problem: When Z reacts in a 2.0 kg calorimeter containing 2.0 kg water, T = 20.0oC to 0.0oC & 50.0 %
of the water freezes. Calculate Hrxn (Cwater = 4.18 J/gK, Ccal = 2.00 J/gK, & Cice = 334 J/g))
Drill: When Z reacts in a 1.0 kg calorimeter containing 2.0 kg water, T = 30.5oC to 20.5oC.
(Cwater = 4.18 J/gK & Ccal = 2.00 J/gK)
Problem: When 40.0 g NH4NO3 dissolves in 460.0 g water at 25.0oC, the temp. falls to 22.5oC.
Calculate the Ho per mole of NH4NO3.
Csoln = 4.00 J/gK
Thermochemical Equation (TCE)
Thermochemical Equation Terms:
Heat of Reaction (Hrxn): The heat or enthalpy change of a chemical reaction
Heat of Solution (Hsoln): The heat or enthalpy change when a substance is dissolved
Heat of Combustion (Hcombustion): The heat or enthalpy change when a substance is burned
Heat of Formation (Hof): The heat required to form one mole of a compound from pure elements
 Heat of formation is measured in kJ/mole.
 The Degree Symbol: Indicates standard conditions & molar quantities by itself or from
a balanced equation.
Gibb’s Free Energy (G):
• Energy of a system that can be converted to work
• Determines spontaneity
Exergonic Reaction: A reaction in which free energy is given off & G < 0 or G is negative.
Engergonic Reaction: A reaction in which free energy is absorbed & G > 0 or G is positive.
Reaction at Equilibrium: G = 0
Gibb’s Free Energy of Formation or Energy of Formation (Gof):
• The Gibb’s Free Energy required to form one mole of a compound from pure elements
• Gibb’s Free Energy of formation is measured in kJ/mole.
Entropy (S): A measure of disorder generally measured in J/K.
Entropy of Formation (Sof) measured in J/moleK: The entropy of one mole of a substance
Interrelation Term (G): G is the term that interrelates thermochemistry,
chemical equilibria, & electrochemistry
Problem: When 10.0 g NH4NO3 dissolves in 400.0 g water at 25.0oC, the temp. falls to 20.0oC. Calculate
the Ho per mole of NH4NO3.
CHOH = 4.18 J/gK
CAN = 0.50 J/gK
Drill: Define: Heat of reaction, Heat of solution, Heat of formation, Free Energy of formation, & Entropy
Thermochemical Equation (TCE): An equation to determine changes in heat, energy, etc
Thermochemical Equation (TCE):
 Horxn = Hfoproducts - Hforeactants
 Gorxn = Gfoproducts - Gforeactants
 Sorxn = Sfoproducts - Sforeactants
Thermochemical Equation (TCE):
 Is called the stoichiometry of heat change because this equation is used to determine the
theoretical change of a chemical reaction.
 The 3rd step in stoichiometry involves converting what’s given to what’s asked for.
 The TCE is used the determine the heat change per mole
 Use the TCE for the 3rd step of stoichiometry to convert moles given to heat or free energy.
Interrelating Equation:
G = H - TS
Drill: Identify type of rxn when: G < 0
G > 0
G = 0
Thermochemical Equation (TCE) & Interrelating Equation:
 Xorxn = Xfoproducts - Xforeactants
 G = H - TS
Problem: Use the table below to Calculate H, G, & S for the reaction when 13.6 g of CaSO4 is
converted to CaO + SO2 + O2 at 25oC
Cmpd
CaSO4
SO2
CaO
Hfo (kJ/mole)
-1434.1
-296.8
-635.1
Gfo (kJ/mole)
-1321.8
-300.2
-604.0
Problem: Use the table below to Calculate H, G, & S for the reaction when 19.7 kg of BaCO3 is
converted to BaO + CO2 at 25oC
Cmpd
BaCO3
CO2
BaO
Hfo (kJ/mole)
-1216.3
-393.5
-553.5
Gfo (kJ/mole)
-1137.6
-394.4
-525.1
Drill: Calculate Horxn & Gorxn for the following reaction: __A + __B2C3  __AC2 + __B
Cmpd
B2O3
AC2
Hfo (kJ/mole)
-150
-250
Gfo (kJ/mole)
-175
-225
Problem: Calculate Ho, Go, & S for the following reaction at -23oC: _AD2 + _BC  _AC2 + _BD
Also calculate SfoBD.
Cmpd
BC
AD2
AC2
BD
Hfo (kJ/mole)
-150
-250
-300
-175
Gfo (kJ/mole)
-125
-225
-250
-150
Sfo (J/moleK)
75
50
80
?
Drill: Calculate Horxn & DGorxn for the following reaction:
Cmpd
QR
PR2
Hfo (kJ/mole)
-250
-450
Gfo (kJ/mole)
-225
-425
_P + _QR  _PR2 + _Q
Practice for Lab calculations:
In lab, a calorimeter was assembled with some water in a cup. The initial temperature of the system was
measured. A small amount of solid KOH was added to the system, & the final temperature of the water was
measured. Use the results from the experimental data table to determine Hexperimental. Use the results from the
heat of formation data table to determine Htheoretical.
Hfo Tables
Experimental Data Table
App.
Cup
H2 O
KOH
Thermo
Cmpd
Hfo(kJ/mole)
Mass
5.00 g
50.00 g
2.00 g
20.00 g
KOH
-424.7
K+
-252.4
OH-
-230.0
Initial Temp. =- 22.0oC
Final Temp. = 30.0oC
Problem: Use the information from the table below to calculate the theoretical Ho, Go, & So for the
reaction involving the burning of C3H8 :
Cpd
Hfo
Gfo
C3H8
CO2
-103.8
-393.5
- 23.5
-394.4
All units (kJ/mole)
H2O
-241.8
-228.6
Drill: Calculate Ho, Go, S, & at -23oC when:
Compd
Hfo (kJ/mole)
Gfo (kJ/mole)
BC
-150
-175
_A + _BC 
_AC + _B
AC
-250
-225
Problem: When 35.0 g NaOH dissolves in 965 mL water in a 2.5 kg calorimeter, the temp. went from 22.5oC
to 26.5oC. Calculate: Hosoln
(Csoln = 4.00 J/gK) (Ccal = 2.00 J/gK)
Problem: Use the data in the table below & the following reaction to calculate Ho, Go, & So, Teq, &
Ho of 48 g PbO2:
PbO2 + CO  CO2 + Pb
Cpd
Hfo (kJ/mole)
Gfo (kJ/mole)
PbO2
CO
CO2
-277.4
-217.4
-110.5
-137.2
-393.5
-394.4
Problem: Use the data in the table below & the following reaction to calculate the potential H, G, & S
for the reaction & Sfo for O2 when burning 8.8 kg of C3H8:
Cpd
C 3 H8
-103.8
- 23.5
269.9
Hfo (kJ/mole)
Gfo (kJ/mole)
Sfo (J/mole K)
CO2
-393.5
-394.4
213.6
H2O
-241.8
-228.6
188.7
Drill: Use the data in the table to the right to calculate
H, G, & S at 298 K for the production of
3.4 g of NH3 from N2 & H2.
Compd
Hfo (kJ/mole)
Gfo (kJ/mole)
NH3
-46.1 kJ/mole
-16.5 kJ/mole
Test Review:
Problem: Use the data in the table below & the following reaction to calculate Ho, Go, & So:
__ N2O5 + __ H2O  __ HNO3
Hfo
Gfo
Cpd
(kJ/mole)
(kJ/mole)
N2O5
-11.3
-10.4
H2O
-285.8
-237.2
HNO3
-174.1
-151.5
Problem: When 10.8 g N2O5 is added to a 1.50 kg calorimeter containing 1.00 kg of H2O, the N2O5 reacts with
the water forming HNO3. The temperature of the system changes from 23.000oC to 23.750oC.
Calculate the Hrxn/mole N2O5.
(Ccal = 1.00 J/gK)
(C = 4.18 J/gK)
Problem: The solution for the 1st problem is the theoretical yield for the production of HNO3 in the balanced
reaction using molar quantities. The solution for the 2nd problem is the experimental yield for the
production of HNO3 from 10.8 grams of N2O5. Calculate the % yield of the reaction.
Problem: When 25.0 g CaCO3 is made from CaO & CO2 in a 500. g calorimeter containing 250.0 g of water,
the system changes from 25.0oC to 45.0oC. Calculate Hrxn/mole of CaCO3:
(Ccal = 1.50 J/gK)
(CH2O = 4.18 J/gK)
Problem: Use the information from the data table & the following reactions to solve the Hrxn for
the entire reaction mechanism:
Substance Hof (kJ/mole)
Reaction Mechanism
C + A 
2H+P
Sub.
Hof
Sub.
Hof
C + A 
2H+P
A
-100
H
-150
D + B 
2K+P
B
-150
K
-200
H+K 
P+Q
C
-50
P
-250
D
-125
Q
-300
Heat Change Problem: Calculate the heat change when the temperature of 1.0 kg H2O is changed
from –100.0oC to 200.0oC.
Problem: Use the information from the data table to calculate
Ho, Go, & S at 227oC for the
production of ammonia from N2 & H2.
Compd
Hfo (kJ/mole)
Gfo (kJ/mole)
NH3
-46.1
-16.5
Extra:
1st Law Thermodynamics: Total energy change = heat + work E = q + W)
Derivation from the work formulas:
•
•
•
•
W = Fd
P = F/A
V = Ad
W = PV = nRT
2nd Law Thermodynamics: Total entropy in a system always increases assuming no energy
is added to the system
Thermodynamic Rxns are State Rxns:
State Reaction: Reactions that are independent of the path; thus not dependent on intermediates