Thermodymanics
 Thermodynamics
is a branch of science that
focuses on energy changes that accompany
chemical and physical changes.
 Objective:
To calculate heat capacity.
 Heat
(q): The energy transferred between
objects that are at different temperatures.
 Unit:
joules (J)
 Molar
Heat Capacity: Energy (heat) needed
to increase the temperature of 1 mol of
substance by 1 K.
q = nC∆T
q = heat
n = # of moles
C = molar heat capacity
∆T = change in temperature
 The
molar heat capacity of water is larger
than the molar heat capacity of land. This
means that water does not heat up as easily
as land does. As a result, oceans can help
keep coastal areas cool during the summer.
 The filling of a fruit pie has a larger heat
capacity than the crust. This means that
fruit filling will retain heat better and the
crust will cool much quicker. As a result,
eating the fruit filling can cause burns (even
though it may appear that the pie is cool).
 Determine
the energy (heat) needed to
increase the temperature of 10.0 mol of Hg
by 7.5 K. The value of C for Hg is 27.8
J/K۰mol.
 Determine
the energy (heat) needed to
increase the temperature of 10.0 mol of Hg
by 7.5 K. The value of C for Hg is 27.8
J/K۰mol.
q=?
n = 10.0 mol
C = 27.8 J/K۰mol
∆T = 7.5 K
q = nC∆T
q = (10.0 mol)(27.8 J/K۰mol)(7.5 K)
q= 2.1 x 103 J
 The
molar heat capacity of tungsten is 24.2
J/K۰ mol. Calculate the energy as heat
needed to increase the temperature of 0.404
mol of W by 10.0 K.
 The
molar heat capacity of tungsten is 24.2
J/K۰ mol. Calculate the energy as heat
needed to increase the temperature of 0.404
mol of W by 10.0 K.
q=?
n = 0.404 mol
C = 24.2 J/K۰mol
∆T = 10.0 K
q = nC∆T
q = (0.404 mol)(24.2 J/K۰mol)(10.0 K)
q= 97.8J
 Suppose
a sample of NaCl increased in temperature
by 2.5 K when the sample absorbed 1.7 x 102 J
energy (heat). Calculate the number of moles of
NaCl if the molar heat capacity is 50.5 J/K۰mol.
 Suppose
a sample of NaCl increased in temperature
by 2.5 K when the sample absorbed 1.7 x 102 J
energy (heat). Calculate the number of moles of
NaCl if the molar heat capacity is 50.5 J/K۰mol.
q = 1.7 x 102 J
n=?
C = 50.5 J/K۰mol
∆T = 2.5 K
q = nC∆T
1.7 x 102 J = n(50.5 J/K۰mol)(2.5 K)
n= 1.3 mol
 Calculate
the energy as heat needed to increase
the temperature of 0.80 mol of nitrogen, N2, by 9.5
K. The molar heat capacity of nitrogen is 29.1
J/K۰mol.
 Calculate
the energy as heat needed to increase
the temperature of 0.80 mol of nitrogen, N2, by 9.5
K. The molar heat capacity of nitrogen is 29.1
J/K۰mol.
q=?
n = 0.80 mol
C = 29.1 J/K۰mol
∆T = 9.5 K
q = nC∆T
q = (0.80 mol)(29.1 J/K۰mol)(9.5 K)
q= 2.2 x 102 J
A
0.07 mol sample of octane, C8H18, absorbed 3.5 x
103 J of energy. Calculate the temperature
increase of octane if the molar heat capacity of
octane is 254.0 J/K۰mol.
A
0.07 mol sample of octane, C8H18, absorbed 3.5 x
103 J of energy. Calculate the temperature
increase of octane if the molar heat capacity of
octane is 254.0 J/K۰mol.
q = 3.5 x 103 J
n = 0.07 mol
C = 254.0 J/K۰mol
∆T = ?
q = nC∆T
3.5 x 103 J = (0.07 mol)(254.0 J/K۰mol) ∆T
∆T = 2.0 x 102 K
 Objective:
enthalpy.
To calculate the change in
 The
total energy of a system is impossible to
measure.
 However, we can measure the change in
enthalpy of a system.
 Enthalpy: The total energy content of a
sample.
H
= Enthalpy
 ∆H = Change in Enthalpy
 ∆H
A
can be measured with a calorimeter.
calorimeter is used to measure the heat
absorbed or released in a chemical or
physical change.
 Enthalpy
changes can be used to determine if a
process is endothermic or exothermic.
 Exothermic
Reaction: Negative Enthalpy
Change
 Endothermic
Changes
Reaction: Positive Enthalpy
 The
standard enthalpy of formation (  H f ) is
the enthalpy change in forming 1 mol of a
substance from elements in their standard
states.
0
 Note:
The value for the standard enthalpy of
formation for an element is 0.
 The
values for the standard enthalpies of
formation can be found using a table.
ΔHreaction = ΔH products - ΔHreactants
 Step
1: Determine
for each compound
using enthalpy table.
 Step 2: Multiply by the coefficients from the
balanced equation (# of moles).
 Step 3: Set up ΔH equation.
 Step 4: calculate.
H
0
f
ΔHreaction = ΔH products - ΔHreactants
Calculate ΔH for the following reaction
and determine if the reaction is
exothermic or endothermic.
SO2(g) + NO2(g)  SO3(g) + NO(g)
ΔHreaction = ΔH products - ΔHreactants
SO2(g) + NO2(g)  SO3(g) + NO(g)
-296.8(1)
33.1(1)
kJ/mol
kJ/mol
-395.8(1)
kJ/mol
90.3(1)
kJ/mol
ΔH= [(-395.8 kJ/mol)(1) + (90.3 kJ/mol)(1)] –
[(296.8 kJ/mol)(1) + (33.1 kJ/mol)(1)]
Δ H = -305.5 kJ – 329.9 kJ
Δ H = -635.4 kJ The reaction is exothermic.
 Calculate
the enthalpy change for the
following reaction:
2C2H6(g) + 7O2(g)  4 CO2(g) + 6H2O(g)
 Calculate
ΔH for the following reaction:
CaO(s) + H2O(l)  Ca(OH)2(s)
 Calculate
the enthalpy change for the
combustion of methane gas.
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
 Objective:
entropy.
To calculate the change in
A
reaction is more likely to occur if enthalpy
(ΔH) is negative.
 However, some endothermic reactions can
occur easily. Why? Entropy!
 Entropy (ΔS): A measure of the randomness
or disorder of a system.
 A process if more likely to occur if there is an
increase in entropy (or if ΔS is positive).
 Entropy







is increased by the following factors:
Diffusion (process of dispersion)
Dilution of a solution
Decreasing the pressure of a gas
Increasing temperature
The number of moles of product is greater than
the number of moles of reactant
Increasing the total number of particles in a
system
When a reaction produces more gas particles
(opposed to liquid or solids)
ΔSreaction = ΔSproducts - ΔSreactants
 Find
the change in entropy for the following
reaction:
2Na (s) + 2HCl (g)  2NaCl (s) + H2(g)
 Find
the change in entropy for the following
reaction:
2Na (s) + 2H2O (l)  2NaOH (s) + H2(g)
 Objectives:
(1)
(2)
To calculate the change in Gibbs energy.
To determine if a reaction is spontaneous
or nonspontaneous.
 Gibbs
Energy: the energy in a system that is
available to do useful work.
 Gibbs
 ΔG
Energy is also called Free Energy.
= Change in Gibbs Energy
ΔGreaction = ΔGproducts - ΔGreactants
 Calculate
ΔG for the following water-gas
reaction:
C(s) + H2O (g)  CO (g) + H2 (g)
 Calculate
the Gibbs energy change that
accompanies the following reaction:
C(s) + O2 (g)  CO2 (g)
 Calculate
the Gibbs energy change that
accompanies the following reaction:
CaCO3 (s)  CaO (s) + CO2 (g)
 If
ΔG is negative, the forward reaction is
spontaneous.
 If
ΔG is 0, the system is at equilibrium.
 If
ΔG is positive, the forward reaction is
nonspontaneous.

NOTE: In this case, the reaction is spontaneous in the
reverse direction.
ΔG = ΔH - TΔS
Step
1: Organize the information.
Step 2: Change the units.
Step 3: Step up ΔG equation.
Step 4: Calculate.
 Given
the changes in enthalpy and entropy are -139
kJ and 277 J/K respectively for a reaction at 25⁰C,
calculate the change in Gibbs energy.
 Given
the changes in enthalpy and entropy are -139
kJ and 277 J/K respectively for a reaction at 25⁰C,
calculate the change in Gibbs energy.
ΔH = -139 kJ
ΔS = 277 J/K / (1000 J/kJ) = 0.277 kJ/K
T = 25 ⁰C + 273 = 298 K
ΔG = ?
 Given
the changes in enthalpy and entropy are -139
kJ and 277 J/K respectively for a reaction at 25⁰C,
calculate the change in Gibbs energy.
ΔH = -139 kJ
ΔS = 277 J/K / (1000 J/kJ) = 0.277 kJ/K
T = 25 ⁰C + 273 = 298 K
ΔG = ?
ΔG = ΔH - T ΔS
ΔG = (-139 kJ) – (298K)(0.277 kJ/K)
ΔG = (-139 kJ) – (82.546 kJ)
ΔG = -221.55 kJ The reaction is spontaneous.
A
reaction has a ΔH of -76 kJ and a ΔS of -117
J/K. Calculate the ΔG at 298 K. Is the reaction
spontaneous?
A
reaction has a ΔH of 11 kJ and a ΔS of 49 J/K.
Calculate ΔG at 298 K. Is the reaction
spontaneous?
 Objective:
To calculate the melting and
boiling point of a substance.
T mp 
 Tmp:
H
fus
 S fus
T bp 
 H vap
 S vap
melting point temperature
 Tbp: boiling point temperature
 ΔHfus: molar enthalpy of fusion
 ΔSfus: molar entropy of fusion
 ΔHvap: molar enthalpy of vaporization
 ΔSvap: molar entropy of vaporization
 Step
1: Organize the information.
 Step 2: Change the units.
 Step 3: Step up the equation.
 Step 4: Calculate.

The enthalpy of fusion of Hg is 2.295 kJ/mol, and the molar
entropy of fusion is 9.79 J/mol·K. The enthalpy of vaporization at
the boiling point if 59.2 kJ/mol, and the molar entropy of
vaporization is 93.8 J/mol·K. Calculate the melting point and boiling
point of Hg.

The enthalpy of fusion of Hg is 2.295 kJ/mol, and the molar
entropy of fusion is 9.79 J/mol·K. The enthalpy of vaporization at
the boiling point if 59.2 kJ/mol, and the molar entropy of
vaporization is 93.8 J/mol·K. Calculate the melting point and boiling
point of Hg.
Tmp = ?
ΔHfus = 2.295 kJ/mol
ΔSfus = 9.79 J/mol·K = 0.00979 kJ/mol·K

The enthalpy of fusion of Hg is 2.295 kJ/mol, and the molar
entropy of fusion is 9.79 J/mol·K. The enthalpy of vaporization at
the boiling point if 59.2 kJ/mol, and the molar entropy of
vaporization is 93.8 J/mol·K. Calculate the melting point and boiling
point of Hg.
Tmp = ?
ΔHfus = 2.295 kJ/mol
ΔSfus = 9.79 J/mol·K = 0.00979 kJ/mol·K
Tmp = (2.295 kJ/mol) / (0.00979 kJ/mol·K)
Tmp = 234 K

The enthalpy of fusion of Hg is 2.295 kJ/mol, and the molar
entropy of fusion is 9.79 J/mol·K. The enthalpy of vaporization at
the boiling point if 59.2 kJ/mol, and the molar entropy of
vaporization is 93.8 J/mol·K. Calculate the melting point and boiling
point of Hg.
Tmp = ?
ΔHfus = 2.295 kJ/mol
ΔSfus = 9.79 J/mol·K = 0.00979 kJ/mol·K
Tmp = (2.295 kJ/mol) / (0.00979 kJ/mol·K)
Tmp = 234 K
Tbp = ?
ΔHvap = 59.2 kJ/mol
ΔSvap = 93.8 J/mol·K = 0.0938 kJ/mol·K

The enthalpy of fusion of Hg is 2.295 kJ/mol, and the molar
entropy of fusion is 9.79 J/mol·K. The enthalpy of vaporization at
the boiling point if 59.2 kJ/mol, and the molar entropy of
vaporization is 93.8 J/mol·K. Calculate the melting point and boiling
point of Hg.
Tmp = ?
ΔHfus = 2.295 kJ/mol
ΔSfus = 9.79 J/mol·K = 0.00979 kJ/mol·K
Tmp = (2.295 kJ/mol) / (0.00979 kJ/mol·K)
Tmp = 234 K
Tbp = ?
ΔHvap = 59.2 kJ/mol
ΔSvap = 93.8 J/mol·K = 0.0938 kJ/mol·K
Tbp = (59.2 kJ/mol) / (0.0938 kJ/mol·K )
T = 631 K
 For
ethanol, the molar enthalpy of fusion is
4.931 kJ/mol and the molar entropy of fusion
is 31.6 J/mol·K. The molar enthalpy of
vaporization at the boiling point is 42.32
kJ/mol and the molar entropy of
vaporization is 109.9 J/mol·K. Calculate the
melting and boiling points for ethanol.
 For
sulfur dioxide, the molar enthalpy of
fusion is 8.62 kJ/mol and the molar entropy
of fusion is 43.1 J/mol·K. The enthalpy of
vaporization at the boiling point is 24.9
kJ/mol and the molar entropy of
vaporization is 94.5 J/mol·K. Calculate the
melting and boiling points for sulfur dioxide.
 For
ammonia, ΔHfus is 5.66 kJ/mol and Δsfus is
29.0 J/mol·K. ΔHvap is 23.33 kJ/mol and ΔSvap is
97.2 J/mol·K. Calculate the melting and
boiling points for ammonia.