VI. Thermochemistry
Thermochemistry: the study of heat changes involved in physical/chemical processes.
Temperature: the average speed of particles. Measured in degrees Celsius (ºC) or Kelvin (K).
Heat: the transfer of energy between two object of different temperature. Measured in Joules (J).
I. Heating and Cooling Curves
Phase Changes
- at a phase change, the temperature of a substance will remain constant
- when a substance is melted or boiled, the heat energy added is used to break bonds between molecules
- when a substance is frozen or condensed, heat energy is released as bonds are formed between molecules
Melting and Freezing
∆H = H fus m
where:
∆H = Heat (J)
Hfus = Heat of Fusion (J/g)
m = mass (g)
Boiling and Condensing
∆H = H vap m
where: ∆H = Heat (J)
Hvap = Heat of Vapourization (J/g)
m = mass (g)
ex. Calculate the amount of heat required to change 60 g of ice to water (Hfus = 334 J/g) at 0 ˚C
ex. Calculate the amount of heat released to change 60 g of steam to water (Hvap = 2256 J/g) at 100 ˚C
Heating and Cooling
- when a substance is heated, the energy added is used to increase the movement of the molecules (∆H is positive)
- when a substance is cooled, heat energy is released as the movement of the molecules decreases (∆H is negative)
∆H = m c ∆ T
where: ∆H = Heat (J)
m = mass (g)
c = specific heat capacity (J/g˚C)
∆T = Temperature Change (Final Temperature – Initial Temperature) (˚C)
ex. Calculate the amount of energy required to heat 25 g of water (c = 4.18 J/g ºC) from 20 ˚C to 80 ˚C
ex. Calculate the mass of ice (c = 2.09 J/g ºC) that requires 130 J to heat from –40 ˚C to –10 ˚C
ex. Calculate the final temperature of 2.0 kg of steam (c = 2.00 J/g ºC) at 160 ˚C if 2.4x104 J of heat is released.
II. Calorimetry
A calorimeter is a device containing water that is used to measure the amount of heat given off in a chemical process.
ex. A reaction produces enough energy to heat 500 g of water in a calorimeter from 10 ˚C to 50 ˚C. How much heat was
released?
ex. What will be the final temperature if a reaction produces 335 kJ of heat in a calorimeter containing 4.0 kg of water at 20 ºC?
ex. A 35.0 g piece of iron (c = 0.450 J/g ºC) at 95.0 ºC is dropped into a calorimeter containing 200 g of water (c = 4.18 J/gºC) at
20.0 ºC. Determine the final temperature of the system.
III. Heat of Reaction
Heat of Reaction: the amount of energy released or absorbed by a chemical reaction, ∆H.
At a constant pressure, the heat of a reaction is also called Enthalpy
∆H = Hproducts – Hreactants
- energy can be released (an exothermic reaction) or absorbed (an endothermic reaction)
In an endothermic reaction, ∆H is POSITIVE
In an exothermic reaction, ∆H is NEGATIVE
Calculating the Heat of a Reaction
The heat of a reaction at 1.00 atm and 298 K is written as ∆Hº298.
(1) Heat of Formation
Heat of Formation, ∆Hºf: the heat released or absorbed when one mole of a compound is formed by a combination of its
elements
The heat of a reaction is the total of the heats of formation of products, minus the total of the heats of formation of the reactants.
∆Ho = total Hfo products – total Hfo reactants
The heat of formation of elements in standard state is equal to zero. The standard state if an element is the form in which it
exists at 25 ºC and 1.00 atm.
Standard State
Heat of Formation ∆Hºf (kJ/mol)
Solid metals
most metals
Solid non metals
C (s) or I2 (s)
Gases
H2 (g), F2 (g), N2 (g), O2 (g), Cl2 (g),
He (g), Ne (g), or Ar (g)
Liquids
Br2 (l), Hg (l)
The heat of formation of compounds can be found on tables. Heats of formation are given for compounds under standard
conditions (gases at 1.00 atm, solutions with concentration of 1.0 M, and pure liquids or solids).
Use the Heats of Formation to calculate ∆Hº for the following reactions:
ex. Cu (s) + H2O (l) → CuO (s) + H2 (g)
Substance
Cu (s)
H2O (l)
CuO (s)
H2 (g)
∆Hfº
(kJ/mol)
0
-286
-156
0
ex.4NH3 (g) + 7O2 (g) → 4NO2 (g) + 6H2O (g)
Substance
NH3 (g)
O2 (g)
NO2 (g)
H2O (g)
∆Hfº
(kJ/mol)
-80
0
34
-242
(2) Bond Energies
In a chemical reaction, the bonds of compounds on the reactant side are broken and the bonds of compounds on the product side
are formed.
Energy is required to break bonds and released to form bonds.
The heat of a reaction is the total energy of the bonds broken, minus the total energy of the bonds formed
∆Hº = total energy of bonds broken – total energy of bonds formed
ex. Use the bond energies to calculate ∆Hº for the following reactions:
2H2 (g) + O2 (g) → 2H2O (g)
Bond
H–H
O=O
H–O
Bond Energy
(kJ/mol)
432
495
467
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)
Bond
C–H
O=O
C=O
H–O
Bond Energy
(kJ/mol)
413
495
799
467
(3) Hess’ Law: The overall enthalpy change for a reaction is equal to the sum of enthalpy changes for individual steps in the
process.
ex. 2NO (g) + O2 (g) → 2NO2 (g)
Steps:
2NO (g) → N2 (g) + O2 (g)
∆Hº = -180.6 kJ/mol
N2 (g) + 2O2 (g) → 2NO2 (g)
∆Hº = 66.4 kJ/mol
∆Hº =
Overall:
Sometimes the steps need to be manipulated so that the overall reaction can be obtained.
If the reaction is reversed, the sign of ∆H is also reversed (ie. negative)
If the reaction is multiplied by a number, ∆H is also multiplied by the number
ex. 2Mg (s) + SiCl4 (l) → Si (s) + 2MgCl2 (s)
Steps:
Si (s) + 2Cl2 (g) → SiCl4 (l)
∆Hº = -687 kJ/mol
Mg (s) + Cl2 (g) → MgCl2 (s)
∆Hº = -642 kJ/mol
∆Hº =
Overall:
ex. 2C2H5OH (l) + 6O2 (g) → 4CO2 (g) + 6H2O (g)
Steps:
Overall:
4C (s) + O2 (g) + 6H2 (g) → 2C2H5OH (l)
∆Hº = - 555.2 kJ/mol
C (s) + O2 (g) → CO2 (g)
∆Hº = - 393.5 kJ/mol
2H2 (g) + O2 (g) → 2H2O (g)
∆Hº = - 571.6 kJ/mol
∆Hº =
IV. Entropy
Entropy: a measure of the disorder in a system, ∆S.
The phase of a substance provides a good indication of the entropy:
gases >> solutions > liquids >> solids
In general, more complex substances have a greater entropy.
Standard Entropy
The entropy of elements/compounds at standard conditions can be found on tables.
(Note: elements do NOT have an entropy equal to zero)
The entropy change in a reaction can be calculated according to the following equation:
∆So = total So products - total So reactants
Determine the standard entropy change of the reaction.
ex. NH4Cl (s) → HCl (g) + NH3 (aq)
Substance
NH4Cl (s)
HCl (g)
NH3 (aq)
∆Sº
(J/mol K)
96
187
111
ex. N2H4 (l) + 3O2 (g) → 2NO2 (g) + 2H2O (l)
Substance
N2H4 (l)
O2 (g)
NO2 (g)
2H2O (l)
∆Sº
(J/mol K)
121
205
240
70
V. Free Energy
Gibb’s Free energy: the total energy change in a chemical process, ∆G.
Standard Free Energy
The free energy of a reaction can be calculated according to the following equation:
∆Go = total Gof products - total Gof reactants
The free energy of elements in standard state is equal to zero. The free energy of compounds at standard conditions can be found
on tables.
Determine the standard free energy change of the reaction.
ex. C2H5OH (l) → C2H4 (g) + H2O (l)
Substance
C2H5OH (l)
C2H4 (g)
H2O (l)
∆Gºf
(kJ/mol)
-175
68
-237
ex. B2H6 (g) + 3O2 (g) → B2O3 (s) + 2H2O (l)
Substance
B2H6 (g)
O2 (g)
B2O3 (g)
H2O (l)
∆Gºf
(kJ/mol)
87
0
-1194
-237
Free Energy can also be calculated using a method similar to Hess’ Law
ex. 2CO (g) + O2 (g) → 2CO2 (g)
Steps:
2CH4 (g) + 3O2 (g) → 2CO (g) + 4H2O (l)
∆Gº = -1088 kJ/mol
CH4 (g) + 2O2 (g) →CO2 (g) + 2H2O
∆Gº = -801 kJ/mol
∆Gº =
Overall:
VI. Free Energy, Enthalpy, and Entropy
The free energy of a reaction can be calculated from the enthalpy, entropy, and temperature.
∆Gº = ∆Hº - T∆Sº
Determine ∆Hº, ∆Sº, and ∆Gº for each reaction at 298 K.
ex. 2SO2 (g) + O2 (g) → 2SO3 (g)
Substance
SO2 (g)
O2 (g)
SO3 (g)
∆Hºf
(kJ/mol)
-297
0
-396
∆Sº
(J/molK)
248
205
257
ex. 3NO2 (g) + H2O (l) → 2HNO3 (l) + NO (g)
Substance
NO2 (g)
H2O (l)
HNO3 (l)
NO (g)
∆Hºf
(kJ/mol)
34
-286
-174
90
∆Sº
(J/molK)
240
70
156
211
When a system is not a standard state, the free energy is represented as ∆G.
∆G = ∆H - T∆S
(note: ∆H and ∆S do not change significantly with temperature, ∆G does change significantly with temperature)
ex. Determine ∆G for the reaction: 3NO2 (g) + H2O (l) → 2HNO3 (l) + NO (g) at 200 K.
Free Energy and Spontaneity
Spontaneous: A reaction that will occur without an outside supply of energy.
A reaction is spontaneous when ∆G < 0
A reaction is non-spontaneous when ∆G > 0
A reaction is in equilibrium if ∆G = 0
Case
∆H< 0 and ∆S> 0
∆H> 0 and ∆S< 0
∆H< 0 and ∆S< 0
∆H> 0 and ∆S> 0
Result
Spontaneous at all
temperatures
Non-spontaneous at all
temperatures
Can be spontaneous at low
temperatures
Can be spontaneous at
high temperatures
VII. Free Energy and Equilibrium
The Free Energy of an equilibrium system can be related to the equilibrium constant according to the following equation:
∆Gº = -RT ln(K)
Case
∆Gº = 0
∆Gº > 0
∆Gº < 0
K
K=1
K<1
K>1
ex. Calculate ∆Gº and K for the equilibrium at 298 K.
2COF2 (g) U CO2 (g) + CF4 (g)
Substance
COF2 (g)
CO2 (g)
CF4 (g)
∆Hºf
(kJ/mol)
-640
-393.5
-925
∆Sº
(J/molK)
235
214
262
For an equilibrium, the relationship between ∆G and ∆Gº can be described according to the following equation:
∆G = ∆Gº + RT ln(Q)
Where Q is the reaction quotient.
For and equilibrium at 298 K, this equation can also be written as follows:
∆G = ∆Gº + 2.303 RT log (Q)
ex. Determine ∆G for the following equilibrium at 298 K if [NO] = 2.00 M, [O2] = 0.500 M, and [NO2] = 1.00 M
2NO (g) + O2 (g) U 2NO2 (g)
Substance
NO (g)
O2 (g)
NO2 (g)
∆Gºf
(kJ/mol)
87
0
52
VIII. Free Energy and Electrochemical Cells
The free energy produced by an electrochemical cell can be calculated according to the following equation:
∆Gº = -nFEºcell
Where: ∆Gº = standard free energy (J)
n = number of moles of electrons (mol e-)
F = Faraday’s constant (96500 (96500 C/mol e-)
Eºcell = standard cell potential (V)
ex. Calculate the standard free energy for the following reaction:
Cu2+ (aq) + Fe (s) → Cu (s) + Fe2+ (aq)
ex. Calculate the standard free energy of the following electrochemical cell.
The cell potential for a cell under non-standard conditions (ie. where the concentration of solutions is not equal to 1.0 M) can be
calculated according to the Nernst Equation:
Ecell = E o cell −
RT
ln(Q)
nF
Where Ecell = cell potential (V)
Eºcell = standard cell potential (V)
R = 8.314 J/mol K
T = temperature (K)
n = number of moles of electrons (mol e-)
F = Faraday’s constant (96500 C/mol e-)
Q = the reaction quotient
For a cell at 298 K, this equation can also be written as follows:
Ecell = E o cell −
0.0592
log(Q)
n
ex. Consider the following electrochemical cell: Fe (s) | Fe2+ (aq) || Ag+ (aq) | Ag (s)
Write the overall reaction and Determine the standard cell voltage and standard free energy
Determine the cell voltage (at 298 K) if [Fe2+] = 1.20 M and [Ag+] = 0.60 M
ex. Calculate the standard cell potential and standard free energy of the following electrochemical cell.
The cell operates until the [Al3+] = 1.24 M. Determine [Sn2+]. Determine Ecell (at 298 K).
IX. Free Energy and Electrolytic Cells
The free energy of an electrolytic cell can be calculated according to the following equation:
∆Gº = -nFEºcell
ex. Calculate the free energy of the following electrolytic cell.
Note: Electrochemical cells involve spontaneous reactions (∆G<0) and Electrolytic cells involve non spontaneous reactions
(∆G>0).