Chapter 17
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
Define key terms and concepts
Explain the three laws of thermodynamics.
Calculate the entropy for a system.
Determine if a reaction is spontaneous or
nonspontaneous.
Calculate the Gibbs Free Energy for a
reaction.

Energy cannot be created or destroyed, only
converted from one form to another.
ΔH = ΔU + PΔV
Δ H°rxn = ΣΔH°f(products) - ΣΔH°f(reactants)
Spontaneous
processes/reactions are
favored over nonspontaneous processes.
 Entropy

 How dispersed the energy
of a system is
 The more dispersed, the
greater the entropy



Solids have less entropy than liquids
Liquids have less entropy than gases
As a solute dissolves in solution, entropy increases.
ΔS = Sf - Si
S = k ln W
ΔS = k ln Wf - k ln Wi


Where k = 1.38x10-23 J/K
W = the number of microstates

Melting aluminum

Iodine crystals subliming

A car accident

Water freezing

Dissolving sodium chloride in water

The entropy of the universe increases in a
spontaneous process and remains unchanged
in an equilibrium process.
ΔSuniv = ΔSsys + ΔSsurr
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
For a spontaneous process, ΔSuniv > 0
For an equilibrium process, ΔSuniv = 0
For the general reaction:
aA + bB  cC + dD
ΔS°rxn =[cS°(C) + dS°(D)] – [aS°(A) + bS°(B)]
ΔS°rxn = ΣnS°(products) - ΣnS°(reactants)
 If the entropy for a system increases, ΔS° is positive.
 If the entropy for a system decreases, ΔS° is negative.
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

If a reaction produces more gas molecules
than it consumes, ∆S° is positive.
If the total number of gas molecules
diminishes, ∆S° is negative.
If there is no net change in the total number
of gas molecules, then ∆S° may be negative
or positive, but the numerical value will be
small.
C6H12O6(s)  C2H5OH(l) + CO2(g)
NH3(g) + CO2(g)  NH2CONH2(aq) + H2O(l)
CO(g) + H2O(g)  CO2(g) + H2(g)
Calculate the standard entropy of reaction for the following reaction at 25°C.
C6H12O6(s)  C2H5OH(l) + CO2(g)
Calculate the standard entropy of reaction for the following reaction at 25°C.
NH3(g) + CO2(g)  NH2CONH2(aq) + H2O(l)
Calculate the standard entropy of reaction for the following reaction at 25°C.
CO(g) + H2O(g)  CO2(g) + H2(g)
System 1
Surroundings
H
e
a
t
System 2
Entropy
Entropy
H
e
a
t

If the temperature of the surroundings is low,
the energy released by a system will increase
the entropy.
∆Ssurr = -∆Hsys
T

The Third Law of Thermodynamics
 The Entropy of a perfect crystalline substance is
zero at absolute zero temperature.
The heat of vaporization, ∆Hvap, of carbon tetrachloride at 25°C is 43.0 kJ/mole.
If 1 mole of liquid carbon tetrachloride at 25°C has an entropy of 214 J/K, what
is the entropy of 1 mole of the vapor in equilibrium with the liquid at this
temperature?
Liquid ethanol at 25°C has an entropy of 161 J/mole•K. If the heat of
vaporization, ∆Hvap, at 25°C is 42.3kJ.mole, what is the entropy of the vapor in
equilibrium with the liquid at 25°C?

The energy available to do work.
∆G= ∆H-T∆S
 If ∆G<0, the reaction is spontaneous in the
forward direction.
 If ∆G >0, the reaction is nonspontaneous in the
forward direction.
 If ∆G=0, the system is at equilibrium and no net
change in energy will be observed.
Using ∆H° and ∆S° values from Appendix 3, calculate the ∆G° for the following
reaction at 25°C.
N2(g) + H2(g)  NH3(g)
Using ∆H° and ∆S° values from Appendix 3, calculate the ∆G° for the following
reaction at 25°C.
CaCO3(s) ↔ CaO(s) + CO2(g)
Using ∆H° and ∆S° values from Appendix 3, calculate the ∆G° for the following
reaction at 25°C.
KClO3(s) ↔ KCl(aq) + O2(g)
∆H
∆S
∆G
+
+
Reaction is spontaneous in the forward direction at high
temperatures, but is spontaneous in the reverse direction at low
temperatures.
+
-
∆G is always positive. Non-spontaneous reaction.
-
+
∆G is always negative. Spontaneous reaction.
-
Reaction is spontaneous in the forward direction at low temperatures,
but is spontaneous in the reverse direction at high temperatures.
-

Standard Free Energy of a Reaction (∆G°rxn)
 The change in free energy when 1 mole of a
compound is synthesized from its elements in their
standard state.

For the general reaction:
aA + bB  cC + dD
ΔS°rxn =[cS°(C) + dS°(D)] – [aS°(A) + bS°(B)]
ΔG°rxn = ΣnG°f(products) - ΣnG°f(reactants)
Determine if the following reaction is spontaneous using Gibbs Free Energy if the
reaction is conducted at 25°C.
CaCO3(s)  CaO(s) + CO2(g)
Determine if the following reaction is spontaneous using Gibbs Free Energy if
the reaction is conducted at 25°C.
KClO3(s)  KCl(aq) + O2(g)
Determine if the following reaction is spontaneous using Gibbs Free Energy
if the reaction is conducted at 25°C.
NH3(g) + CO2(g)  NH2CONH2(aq) + H2O(l)
ΔG° = -RTlnK

For calculating reactions where not all
reactants and products are in their standard
states.
 Where R= 8.314 J/K•mole
 T is temperature in Kelvin
 K is the equilibrium constant.
K
InK
ΔG°
Affect at Equilibrium
>1
+
-
Products are favored
=1
0
0
Neither side of the reaction is favored
<1
-
+
Reactants are favored
What is the value of the equilibrium constant K at 25°C for the following reaction?
Predict whether the products or reactants are favored at equilibrium.
CaCO3(s) ↔ CaO(s) + CO2(g)
What is the value of the equilibrium constant K at 25°C for the following
reaction? Predict whether the products or reactants are favored at
equilibrium.
KClO3(s) ↔ KCl(aq) + O2(g)
What is the value of ΔG° at 25°C for the following reaction? The Ksp of
magnesium hydroxide is 1.8x10-11. Predict whether the products or
reactants are favored at equilibrium.
Mg(OH)2(s) ↔ Mg2+(aq) + OH-(aq)
What is the value of the equilibrium constant K at 25°C for the following
reaction? The Ksp of lead (II) iodide is 6.5x10-9. Predict whether the
products or reactants are favored at equilibrium.
PbI2(s) ↔ Pb2+(aq) + I2-(aq)