Science SS1
Thermochemistry
4th Class
Entropy and Free Energy
Thermodynamic
s vs. Kinetics
• Domain of Kinetics
– Rate of a reaction depends
on the pathway from
reactants to products.
• Thermodynamics tells us
whether a reaction is
spontaneous based only
on the properties of
reactants and products.
Copyright © Cengage Learning. All rights reserved
2
Spontaneous Processes and Entropy
• Thermodynamics lets us predict the direction in
which a process will occur but gives no information
about the speed of the process.
• A spontaneous process is one that occurs without
outside intervention.
Copyright © Cengage Learning. All rights reserved
3
Entropy, ΔS
• The driving force for a spontaneous process is an
increase in the entropy of the universe.
• A measure of molecular randomness or disorder.
Copyright © Cengage Learning. All rights reserved
4
The Expansion of An Ideal Gas Into an
Evacuated Bulb
Copyright © Cengage Learning. All rights reserved
5
Positional Entropy
• A gas expands into a vacuum to give a uniform
distribution because the expanded state has the
highest positional probability of states available to
the system.
• Therefore: Ssolid < Sliquid << Sgas
Entropy
ΔS°reaction = ΣnpS°products – ΣnrS°reactants
Entropy
ΔG=ΔH-tΔS
Entropy ΔS
J/t×mole of Reaction
(Watch Units)
• Negative ----Going to order (Products more
ordered than reactants)
• Positive -----Going to disorder (Product more
disorder than reactants)
ΔSsurr
• ΔSsurr = +; entropy of the universe increases
• ΔSsurr = -; process is spontaneous in opposite
direction
• ΔSsurr = 0; process has no tendency to occur
Copyright © Cengage Learning. All rights reserved
9
CONCEPT CHECK!
Predict the sign of ΔS for each of the following,
and explain:
+ a) The evaporation of alcohol
– b) The freezing of water
– c) Compressing an ideal gas at constant
temperature
+ d) Heating an ideal gas at constant
pressure
+ e) Dissolving NaCl in water
Copyright © Cengage Learning. All
rights reserved
10
ΔSsurr
• The sign of ΔSsurr depends on the direction of the
heat flow.
• The magnitude of ΔSsurr depends on the
temperature.
Copyright © Cengage Learning. All rights reserved
11
ΔSsurr
Heat flow (constant P) = change in enthalpy = ΔH
Ssurr
Copyright © Cengage Learning. All rights reserved
H
= 
T
12
Free Energy
ΔG°reaction = ΣnpG°products – ΣnrG°reactants
Electro
ΔG° = -nfe
Free
Energy
Equil
ΔG° = -RTlnk
ΔG° = ΔH° –
TΔS°
Free Energy ΔG
KJ/mole of Reaction
• Negative = spontaneous reaction
• Positive = non-spontaneous reaction (will not
occur at current conditions)
• Zero = System is in equilibrium
CONCEPT CHECK!
Describe the following as spontaneous/non-spontaneous/cannot tell,
and explain.
A reaction that is:
a) Exothermic and becomes more positionally random
Spontaneous
b) Exothermic and becomes less positionally random
Cannot tell
a) Endothermic and becomes more positionally random
Cannot tell
a) Endothermic and becomes less positionally random
Not spontaneous
Explain how temperature affects your answers.
Free Energy (G)
Suniv
G
= 
(at constant T and P )
T
• A process (at constant T and P) is spontaneous in the
direction in which the free energy decreases.
– Negative ΔG means positive ΔSuniv.
Copyright © Cengage Learning. All rights reserved
16
Relation between ΔG, ΔH, and ΔS
Based on the formula ΔG=ΔH-TΔS
ΔG
ΔH
ΔS
Spontaneous
at high temp
positve
positive
Always
spontaneous
negative
positive
Spontaneous
at low temp
negative
negative
Never
spontaneous
Positive
negative
∆S
Always
spontaneous
Spontaneous
High temp
-∆H
∆H
Spontaneous
Low temp
Never spontaneous
-∆S
CONCEPT CHECK!
A liquid is vaporized at its boiling point. Predict the signs of:
w
–
q
+
+
ΔH
+
ΔS
ΔSsurr –
0
ΔG
Explain your answers.
Copyright © Cengage Learning. All rights reserved
18
Third Law of Thermodynamics
• The entropy of a perfect crystal at 0 K is zero.
• The entropy of a substance increases with
temperature.
Copyright © Cengage Learning. All rights reserved
19
CONCEPT CHECK!
A stable diatomic molecule spontaneously
forms from its atoms.
Predict the signs of:
ΔH°
ΔS°
–
–
ΔG°
–
Explain.
Copyright © Cengage Learning. All rights reserved
20
The Meaning of ΔG for a Chemical
Reaction
• A system can achieve the lowest possible free
energy by going to equilibrium, not by going to
completion.
Copyright © Cengage Learning. All rights reserved
21
• The equilibrium point occurs at the lowest value of
free energy available to the reaction system.
ΔG = 0 = ΔG° + RT ln(K)
ΔG° = –RT ln(K)
Copyright © Cengage Learning. All rights reserved
22
Change in Free Energy to Reach
Equilibrium
Copyright © Cengage Learning. All rights reserved
23
Copyright © Cengage Learning. All rights reserved
24
Question 1
Which of the following is a graph that describes
the pathway of reaction that is exothermic and
has high activation energy?
A.
B.
C.
D.
Question 1 Breakdown
•
•
•
•
•
•
Correct answer is C
What can you say about A?
What can you say about B?
What can you say about D?
Label Activation energy on each graph
Label ∆H for each graph
Question 2
When solid NH4SCN is mixed with solid Ba(OH)2 in a
closed container, the temperature drops and a gas
is produced. Which of the following indicates the
correct signs for ΔG, ΔH, and ΔS for the process?
ΔG
ΔH ΔS
A) –
–
–
B) –
C) –
D) +
+
+
–
–
+
+
Question 2 break down
• Correct answer is C
• G is negative because it happens
spontaneously
• S is positive since going from solid to gas
• H is positive because it feels cold
(endothermic)
Question 3
N2(g) + 3 H2(g) → 2 NH3(g)
The reaction indicated above is thermodynamically
spontaneous at 298 K, but becomes
nonspontaneous at higher temperatures. Which of
the following is true at 298 K?
A) ΔG, ΔH, and ΔS are all positive.
B) ΔG, ΔH, and ΔS are all negative.
C) ΔG and ΔH are negative, but ΔS is positive.
D) ΔG and ΔS are negative, but ΔH is positive.
E) ΔG and ΔH are positive, but ΔS is negative.
Question 3 breakdown
• G = H -TS
• Spontaneous means negative G.
• S is negative since you are going from 4 gases
to 2 gases (disorder is leaving)
• H must be negative if G is negative.
• Correct answer is B.
Alternate process Question 3
• G is negative since it is spontaneous
• G is becoming more positive as Temp
increases. Only S is affected by the temp. S
must be negative since TS is positive.
• Since TS is positive H must be negative if G is
negative.
Question 4
Of the following reactions, which involves the
largest decrease in entropy?
A) 2 CO(g) + O2(g) → 2 CO2(g)
B) Pb(NO3)3(s) + 2 KI(s) → PbI2(s) + 2 KNO3(s)
C) C3H8(g) + O2(g) → 3 CO2(g) + 4 H2O(g)
D) 4 La(s) + 3 O2(g) → 2 La2O3(s)
Question 4 breakdown
• Correct answer D. Going from gas to solid
• Which choice is largest increase? Why?
Question 5
Assume the data graphed was collected at a
constant pressure of 0.97 atm and represents
four different temperature samples of pure
neon gas. Which of the following
temperatures most likely corresponds to the
data graphed for sample “D”?
A) 273 K
B) 298 K
C) 305 K
D) 338 K
Question 5 breakdown
• KE= ½ mv2
• Small molecules move faster than larger
molecules. The high point of each curve hits
the x axis in order from A to D. A is the
slowest and biggest. D is the smallest and
fastest.
• Temperature means molecules are moving
faster therefore D is the highest temperature
Question 6
• At 298 K, as the salt MX dissolves spontaneously to
form an aqueous solution, ∆S and ∆H are positive.
Which describes the value of ∆G and the absolute
values of its components, T∆S and ∆H?
• A) ∆G < 0; |T∆S| > |∆H|
• B)∆G < 0; |T∆S| < |∆H|
• C) ∆G > 0; |T∆S| > |∆H|
• D) ∆G > 0; |T∆S| < |∆H|
Breakdown 6
• Spontaneous means G is negative or less than
zero
• If both H and S are positive then S is the
driving force of the spontaneity therefore if
the reaction is spontaneous it must mean that
TS is greater than H.
• Correct answer is A.
Free response 1
• C6H5OH(s) + 7 O2(g) → 6 CO2(g) + 3 H2O(l)
• When a 2.000-gram sample of pure phenol,
C6H5OH(s), is completely burned according to
the equation above, 64.98 kilojoules of heat is
released. Use the information in the table
below to answer the questions that follow.
substance
Delta H
KJ/mol@25
Delta S Joules/mol
CO2
−393.5
213.6
H2
0.00
130.6
H2O
−285.85
69.91
O2
0.00
205.0
C6H5OH
?
144.0
• (a) Calculate the molar heat of combustion of
phenol in kilojoules per mole at 25°C.
• 2.00/94.12=.021 moles of phenol
• .021moles/64.98 KJ= 1mole/X
• X= -3058 KJ/mole = delta H
• (b) Calculate the standard heat of formation,
ΔH°f, of phenol in kilojoules per mole at 25°C.
• H= prod- react = -3058=(6(−393.5)+3(−285.85))-X
• X=-161 J/K*mol
• (c) Calculate the value of the standard free-energy
change, ΔG° for the combustion of phenol at
25°C.
• ∆ S= prod-react=(6(213.6)+3(69.91)) – (7(205.0) +
144.0)
• S= -87.67 J/mol
• G= H-TS = -3058 – 298( -.08767 )
• G = -3032 KJ/mol
Free Response 2
• O3(g) + NO(g) → O2(g) + NO2(g)
• Consider the reaction represented above.
• (a) Referring to the data in the table below,
calculate the standard enthalpy change, ΔH,
for the reaction atv25°C. Be sure to show your
work.
KJ/mol
O3
NO
NO2
Std. enthalpy
of formation
∆Hf at 25◦C
143
90
33
• A. H= 33- (90+143)= -200 KJ/mol
• (b) Make a qualitative prediction about the
magnitude of the standard entropy change, ΔS°,
for the reaction at25°C. Justify your answer.
• 2 moles of gas to 2 moles of gas very small
magnitude (very close to zero)
• (c) On the basis of your answers to parts (a) and
(b), predict the sign of the standard free-energy
change, ΔG°, for the reaction at 25°C. Explain
your reasoning.
• H is negative and TS is very small after converting
to KJ therefore G is negative.
Free Response 3
• C2H2(g) + 2 H2(g) ---> C2H6(g)
Substance
S° (J/mol K)
ΔH°f (kJ/mol)
C2H2(g)
200.9
226.7
H2(g)
130.7
0
C2H6(g)
--------
-84.7
• a) If the value of the standard entropy change, ΔS°, for the
reaction is -232.7 joules per mole Kelvin, calculate the
standard molar entropy, S°, of C2H6 gas.
• (b) Calculate the value of the standard free-energy change,
ΔG°, for the reaction. What does the sign of ΔG° indicate
about the reaction above?
• (c) Calculate the value of the equilibrium constant, K, for
the reaction at 298 K.
•
•
•
•
•
S = prod –react = -237.2= X – (200.9 + 2(130.7))
X = 229.6 J/mol
B. H = prod –react= -84.7–226.7= -311.4 KJ/mol
G = H – TS = -311.4 – 298(-.2372)
G = -242.1 KJ/mol
•
•
•
•
C. G = -RTlnK
-242000= -(8.31)(298)lnK
ln K = 97.7
K = 3x 1042
Formulas tell you what to do.
• ∆E = q (heat) + w(work)
• q=mC∆T (water’s specific heat is 4.184 J/g °C)
• ∆H, S, G = (sum of products) – (sum of reactants)
• ∆H bond energy = (sum of reactants) – (sum of products)
•
•
•
•
∆G = H-TS (watch for units on S)
∆G=-RTlnK (watch for units on G)
∆G=-nFE
Hess’s law match the equation and add up ∆Hrxn
(changes affect equation and ∆H)