Tuesday, Oct. 29th: “A” Day
Wednesday, Oct. 30th: “B” Day
Agenda
Homework questions/quick review
Sec. 10.3 quiz: “Changes in Enthalpy During
Chemical Reactions”
Section 10.4: “Order and Spontaneity”
Entropy, Standard Entropy, Gibbs energy
Homework:
Sec. 10.4 review, pg. 367: #3-5, 7-11
Sec 10.4 concept review
*Quiz next time over section 10.4*
Homework
Pg. 15 practice worksheet
Sec. 10.3 review, pg. 357: #1-5
Calculating a Reaction’s Change in Enthalpy
Sample Prob. E, pg.356
 Calculate the change in enthalpy for the reaction below
using data from Table 2 on pg 355.
2 H2(g) + 2 CO2(g)
2 H2O(g) + 2 CO(g)
State whether the reaction is exothermic or endothermic.
ΔHreaction = ΔH f0products - ΔH f0reactants
ΔHf0prod = [(2 mol)(-241.8 kJ/mol) + (2 mol)(-110.5 kJ/mol)]
= -704.6 kJ
ΔHf0reactants = [(2 mol)(0 kJ/mol) + (2 mol)(-393.5 kJ/mol)]
= -787 kJ
ΔHreaction= (-704.6 kJ) – (-787 kJ) = 82.4 kJ
*Reaction is endothermic because ΔH is positive.*
Sec. 10.3 Quiz:
“Changes in Enthalpy During Chemical
Reactions”
You can use your notes and your book to
complete the quiz with a partner of your choice…
Hint:
#6: ΔHf° reactants = -612 kJ/mol
ΔHf° products = -671 kJ/mol
Good Luck!
Entropy
 Entropy: a measure of the randomness
or disorder of a system
 Symbol: S
 Units: J/K
 A process is more likely to occur if it is
accompanied by an increase in entropy
( ΔS is positive)
Your room WANTS to be messy!
Factors that Affect Entropy
 Entropy increases as molecules or ions become
dispersed. (Diffusion)
 Entropy increases as solutions become more dilute
or when the pressure of a gas is reduced.
 Mixtures of gases have more entropy than a single
gas.
 Entropy increases when
 total # moles product > total # moles reactant
 Entropy increases when a reaction produces more
gas particles, because gases are more disordered
than liquid or solids.
Hess’s Law Also Applies to Entropy
Standard Entropy, So: the entropy of 1
mole of a substance at a standard
temperature, 298.15 K.
The entropy change of a reaction can be
calculated by:
ΔSreaction = S˚products
- S˚reactants
Elements can have standard entropies of
formation that have values other than zero.
Practice #1, pg. 361
Find the change in entropy for the reaction below by
using Table 4 and that S˚ for CH3OH(l) is
126.8 J/K·mol
CO(g) + 2 H2(g)
CH3OH (l)
ΔSreaction = S˚product - S˚reactants
S˚product = (1mol)(126.8 J/K·mol)
= 126.8 J/K
S˚reactant=[(1mol)(197.6 J/K·mol)+(2mol)(130.7J/K·mol)]
= 459 J/K
ΔSreaction = 126.8 J/K – 459 J/K = -332.2 J/K
Example
 Calculate the change in entropy for the following
reaction using Table A-11 starting on pg. 833.
2 Na(s) + 2 HCl(g)
2 NaCl(s) + H2(g)
ΔSreaction = S˚product - S˚reactants
S˚product= [(2 mol)(72.1 J/mol∙K)+(1 mol)(130.7 J/mol∙K)]
= 274.9 J/K
S˚reactants= [(2 mol)(51.5 J/mol∙K)+(2 mol)(186.8 J/mol∙K)]
= 476.6 J/K
Δsreaction= 274.9 J/K – 476.6 J/K = -201.7 J/K
Gibbs Energy
If ΔH is negative and ΔS is positive for a
reaction, the reaction will likely occur.
If ΔH is positive and ΔS is negative for a
reaction, the reaction will NOT occur.
How can you predict what will happen if ΔH
and ΔS are both positive or negative?
Gibbs Energy
Gibbs Energy: the energy in a system
that is available for work.
(also called free energy)
Symbol: G
G = H – TS
OR
ΔG = ΔH – TΔS
H = enthalpy (kJ or J)
S = entropy (J/K)
T = temperature (K)
Gibbs Energy Determines
Spontaneity
Spontaneous reaction: a reaction that does
occur or is likely to occur without continuous
outside assistance, such as the input of
energy.
Non-spontaneous reaction: a reaction that
will never occur without assistance.
Spontaneous vs. non-spontaneous
 On a snow-covered mountain in winter, an
avalanche is a spontaneous process because it
may or may not occur, but it always CAN occur.
The return of snow from the bottom of the
mountain to the mountaintop is a nonspontaneous process, because it can NEVER
happen without assistance.
Gibbs Energy Determines
Spontaneity
If ΔG is negative, reaction is spontaneous
If ΔG is greater than 0, reaction is nonspontaneous
If ΔG is exactly 0, reaction is at equilibrium
Entropy and Enthalpy Determine Gibbs
Energy
Standard Gibbs energy of formation: the
change in energy that accompanies the
formation of 1 mole of the substance from its
elements at 298.15 K.
Symbol: ΔGfo
Unit: kJ/mol
ΔGreaction = ΔGf˚products – ΔGf˚reactants
Sample Problem G, pg. 364
Given that the change in enthalpy and entropy are
-139 kJ and 277 J/K respectively for the reaction given below,
calculate the change in Gibbs energy. Then, state whether
the reaction is spontaneous at 25˚C.
C6H12O6(aq)
2 C2H5OH(aq) + 2 CO2(g)
ΔG = ΔH – TΔS
ΔH = -139 kJ
ΔS = 277 J/K (Change to kJ
0.277 kJ)
T = 25°C + 273 = 298 K
ΔG = (-139 kJ) – [(298K) (0.277 kJ/K)] = -222 kJ
Reaction is spontaneous because ΔG is negative.
Sample Problem H, pg. 365
Use Table 5 to calculate ΔG for the following
water-gas reaction with graphite.
C(s) + H2O(g)
CO(g) + H2 (g)
ΔGreaction = ΔGf˚ products - ΔGf˚ reactants
ΔGf˚ products = [(1mol)(-137.2 kJ/mol) + (1mol)(0)]
= -137.2 kJ
ΔGf˚ reactants= [(1mol)(0) + (1mol)(-228.6 kJ/mol)]
= -228.6 kJ
ΔGreaction = -137.2 kJ – (- 228.6 kJ) = 91.4 kJ
Predicting Spontaneity
ΔH
ΔS
ΔG
Spontaneous
?
Negative
Positive
Negative
Yes, at all
Temps
Negative
Negative
Either
Positive or
Negative
Only if
T < ΔH/ΔS
Positive
Positive
Either
Positive or
Negative
Only if
T > ΔH/ΔS
Positive
Negative
Positive
Never
Predicting Spontaneity
Since ΔG = ΔH – TΔS, temperature may greatly
affect ΔG.
Increasing the temperature of a reaction can
make a non-spontaneous reaction
spontaneous.
Homework
 Sec 10.4 review, pg. 367: #3-5, 7-11
 #10 and #11: C(s) use GRAPHITE
 Concept Review: “Order and Spontaneity”
Next Time:
 Quiz over section 10.4/Lab Write-up
 Mon/Tues: Calorimetry Lab
Chapter 10 test/concept review due:
Thursday, 11-14: “A” Day
Friday, 11-15: “B” Day