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Wednesday, Oct. 31st: “A” DayThursday, Nov. 1st: “B” Day
AgendaHomework questions/quick reviewSec. 10.3 quiz: “Changes in Enthalpy During
Chemical Reactions”Section 10.4: “Order and Spontaneity”
Entropy, Standard Entropy, Gibbs energyHomework:
Sec. 10.4 review, pg. 367: #3-5, 7-11Sec 10.4 concept review*Quiz next time over section 10.4*
Homework
Pg. 15 practice worksheetSec. 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 f0
products - ΔH f0
reactants
ΔHf0
prod = [(2 mol)(-241.8 kJ/mol) + (2 mol)(-110.5 kJ/mol)]
= -704.6 kJΔHf
0reactants = [(2 mol)(0 kJ/mol) + (2 mol)(-393.5 kJ/mol)]
= -787 kJΔHreaction= (-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…
Good Luck!
EntropyEntropy: a measure of the randomness
or disorder of a system
Symbol: SUnits: J/K mol∙
A process is more likely to occur if it is accompanied by an increase in entropy( ΔS is positive)
Factors that Affect EntropyEntropy 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 reactantEntropy increases when a reaction produces more
gas particles, because gases are more disordered than liquid or solids.
Hess’s Law Also Applies to EntropyStandard 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/KS˚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
ExampleCalculate the change in entropy for the following
reaction using Table A-11 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/KS˚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 EnergyGibbs Energy: the energy in a system
that is available for work. (also called free energy)
Symbol: GG = H – TS
ORΔG = 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 non-spontaneous 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 non-spontaneous
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. 364Given 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 – TΔSΔH = -139 kJΔS = 277 J/K (Change to kJ 0.277 kJ)T = 25°C + 273 = 298 KΔG = (-139 kJ) – [(298K) (0.277 kJ/K)] = -222 kJ
Reaction is spontaneous because ΔG is negative.
Sample Problem H, pg. 365Use 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 SpontaneityΔH 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, 7-11Concept Review: “Order and Spontaneity”
Next Time:Quiz over section 10.4/Lab Write-
upTues/Wed: Calorimetry Lab
Chapter 10 test/concept review due: Friday, 11-16: “A” Day
Monday, 11-19: “B”: Day
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