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AP C19 A 1415.notebook
March 27, 2015
Is it thermodynamic favored at 0°C?
NO!­ it is at equilibrium so no shift
NO!­ it is at equilibrium so no shift
Is it thermodynamically favored at +10°C?
YES!
YES!
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AP C19 A 1415.notebook
March 27, 2015
When ∆Suniv ° is 0, will ∆G° always be 0 too?
How do the signs of ∆Suniv ° and ∆G° relate?
Enthalpy and Entropy
When H and S are in opposition, the favorablity depends on T
(∆G = ∆ H ­ T ∆ S)
In opposition: +∆S and +∆H OR
–∆S and ­∆H
Exothermic direction is thermodynamically favored at low T
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AP C19 A 1415.notebook
March 27, 2015
Free Energy and Temperature
Occasionally, you are asked to calculate one of the following:
1) the specific temperature at which a reaction changes from being spontaneous to non­spontaneous
or 2) the specific temperature at which a reaction changes from being non­spontaneous to spontaneous
You can calculate the specific temperature at which ∆G changes sign by setting ∆G = 0.
Example 2
At what T is this reaction thermodynamically favored at 1 atm of pressure?
Br 2 (l) Br 2 (g)
∆H=31.0 kJ/mol, ∆S=93.0 J/molK
Looking for boiling point: ∆G=0
+∆H: endothermic
T = 333k
+∆S: increasing entropy
T>333K: spontaneous
T<333K: reverse is spont
T=333K: boiling pt.; equilibrium
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AP C19 A 1415.notebook
March 27, 2015
Third Law of Thermodynamics
The entropy of a perfect crystal at 0K is zero. As the temperature is raised, entropy increases.
Assigning S values
Normally deal with changes in S but can assign actual S values to substances
3rd law of thermodynamics
• S = 0 for a perfect crystal at 0 K
• A standard value to help us set S values
Appendix C contains S° for common substances at 298 K and 1 atm
Since S is state function, can find ∆S by subtracting final ­ initial
all values must be looked up in a chart, the ∆S°, for elements in their standard state is NOT zero
(for ∆Η, elements have values of zero)
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AP C19 A 1415.notebook
March 27, 2015
Standard Entropies
Larger and more complex molecules have greater entropies.
H
H
H
H
C
H
H
C
H
C
H
C
H
H
H
H
Methane, CH4
ΔS = 186.3 J/mol K Propane, C3 H8
ΔS = 270.3 J/mol K
17 The value of ΔSo for the catalytic hydrogenation of acetylene to ethane is _____ J/K∙ mol.
C2H2(g) + H2(g) à C2H4(g)
A
+18.6
B
+550.8 C
+112.0 D
­112.0 E
­18.6
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AP C19 A 1415.notebook
March 27, 2015
Example 3
Find the ∆S° (J/K mol) at 25°C for
:
2NiS(s) + 3O2 (g) 2SO2 (g) + 2NiO(s)
S° 53 205 248 38
Standard Free Energy Change
standard free energies of formation, ∆Gf °
• similar to standard enthalpies of formation
• change when reactants and products are both in standard states
• Cannot measure directly but can calculate it from other measured values
• Standard states: chart
• The more negative ∆G° is, the further a reaction’s equilibrium position lies to the right
∆G° = Σn∆Gf ° (products) ­ Σn∆Gf ° (reactants)
As it was for ΔHf o (but not ΔS), the ΔGf o for elements in their standard state is zero.
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AP C19 A 1415.notebook
March 27, 2015
3 Ways to Calculate ∆G°
Use equation: ∆G° = ∆ H ° ­ T ∆ S ° • Can calculate ∆ H ° and ∆ S ° using appendix and products – reactants
• Then put in the equation to find ∆G ° Use Hess’ Law
• Rearrange equations to get the goal equation
• Add the ∆G° values for the equations
Use ∆Gf° : standard free energy of formation
18 The value of ΔGo at 373 K for the oxidation of solid elemental sulfur to gaseous sulfur dioxide, as shown below, is ______ kJ/
mol. The ΔHo for this reaction is ­269.9 kJ/mol, and ΔSo is +11.6 J/K.
­300.4
B
+300.4
C
­4,597
D
+4,597
S(s, rhombic) + O2(g) à SO2(g)
answer
A
∆ G° = ∆ H° ­ T∆ S°
∆ G° = Σn∆ Gf ° (products ) ­ Σm ∆ Gf ° (reactants
E
E
)
­274.2
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AP C19 A 1415.notebook
March 27, 2015
Dependence on Pressure
?
• ∆ H is not dependent on the pressure of the system but ∆S is­ Why
pressure changes volume which changes the number microstates possible
∆ G is dependent on pressure ­ Why?
because dependent on ∆S can calculate the ∆G of a substance at a pressure other than 1 atm •
using:
Can also determine the ∆G for a reaction using the reaction quotient (Q)
• where R is gas law constant (8.314 J/molK) and Q is the reaction quotient in pressures or concentrations
Example
Find the ∆G for the following reaction at 25°C and with CO at 5.0 atm and H2 at 3.0 atm. CO(g) + 2H2 (g) CH3 OH(l)
­137
0
­166 ∆G°= ­166 ­ [­137 + (2x0)] = ­29 kJ
J/molK ­usually given in a table
∆G= ∆G° +RTlnQ
∆G= (­29x10 3 J) + (8.314)(298K)(ln2.2x10 ­2 )
∆G = ­3.8x10 4 J/mol
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AP C19 A 1415.notebook
March 27, 2015
What does ∆G mean?
a negative ∆G means thermodynamically favored, equilibrium lies to right
Equilibrium
Free Energy and Equilibrium
∆G° = ­RT lnK
If we examine this equation, let us consider the relationship between the standard free­energy change, ∆ G, and the equilibrium constant, K. ∆ G K
ln K K < 1
­ + , non­thermodynamically favored process,
reactants favored
G°prod > G°react
K = 1
0 0, equilibrium,
G°prod = G°react
neither products or reactants favored
K > 1 +
­, thermodynamically favored process
G°react > G°prod products favored
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AP C19 A 1415.notebook
March 27, 2015
A
K = 0
B
K = 1
C
K > 1
D
K < 1
E
K < 0
answer
19 If for a reaction ∆G is greater than zero, then __________. D
N2 (g) + 3H 2 (g) à 2NH3 (g)
For the reaction above, ∆G° =‐33.2734 kJ/mol. For each of the mixtures below at 25°C, predict the direction in which the system will shift to reach equilibrium
PNH3 = 1.00 atm, P N2 = 1.47, P H2 = 0.0100
∆G = (‐3.32734x10 4 J) + (8.314)(298)(ln6.80x10 5 )
∆G=0 already at equilibrium‐ no shift
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AP C19 A 1415.notebook
March 27, 2015
Example 2
PNH3 = 1.00 atm, P N2 = 1.00, P H2 = 1.00
∆G = (‐3.33x10 4 J) + (8.314)(298)(ln 1.00)
∆G= (‐3.33x10 4 J) + 0 = ‐3.33x10 4 J
since ∆G is negative, K > 1 so need to create more products, shift to right
Work
• Can calculate the maximum amount of work that can be done by a process
• wmax = ∆G
• if the ∆G is +, then w is the amount of work expended to make a process occur
• impossible to achieve maximum work because of wasted energy
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AP C19 A 1415.notebook
March 27, 2015
The AP test commonly uses the units of kJ/ mol rxn linked to a balanced chemical reaction.
Most college texts use the units kJ and assume the given number of kJ is directly proportional to the number of mores of each reactant of product in a balanced equation.
if you see kJ/mol rxn on AP test, be sure to translate it to kJ per the given reaction
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