chapter 17: free energy and thermodynamics (again) free energy case 1: ∆h negative, ∆s positive...
TRANSCRIPT
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Chapter 17: Free Energy and
Thermodynamics (again)
Mrs. Brayfield
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17.2: Spontaneous and Nonspontaneous
Processes
A spontaneous process is one that occurs without
outside intervention
For example, when you drop a book it falls to the floor
In chemistry, we want to know the chemical potential
that will predict the direction of a chemical system
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Spontaneous vs Nonspontaneous
DO NOT confuse spontaneity of a reaction with its
speed
Thermodynamics is the direction and the extent to which a
chemical reaction will proceed
Kinetics is the speed – how fast the reaction takes place
Thermodynamic ≠ Kinetic
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Spontaneous vs Nonspontaneous
The conversion of diamond to graphite is
thermodynamically spontaneous, but the process is VERY
slow – so diamonds won’t turn to graphite quickly
Also, the rate of a reaction can be increased with a catalyst, but
a nonspontaneous process still will not go with one
A nonspontaneous process is not impossible, it can
happen
You just need to supply enough energy to it
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17.3: Entropy and the 2nd Law
Remember entropy back from chapter 6…
Entropy is a function that increases with the number of
energetically equivalent ways to arrange the components
of a system to achieve a particular state (aka disorder)
This is melting ice (endothermic) happens spontaneously…
As the ice melts, the molecules become more “disordered” (or they
have more energy states), so entropy has increased (∴ spontaneous)
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Entropy
The definition of entropy can be mathematically defined:
𝑆 = 𝑘𝑙𝑛𝑊 [Ludwig Boltzmann]
Entropy, like enthalpy (H) is a state function:
∆𝑆 = 𝑆𝑓𝑖𝑛𝑎𝑙 − 𝑆𝑖𝑛𝑖𝑡𝑖𝑎𝑙
A chemical system proceeds in the direction that
increases the entropy of the universe (or towards the
largest number of ways to arrange the components)
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Entropy
How many ways can we arrange the following gas
molecules?
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2nd Law of Thermodynamics
The second law of thermodynamics states:
For any spontaneous process, the entropy of the universe
increases (∆Suniv > 0)
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Entropy Changes with States
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Prediction of ∆S
In general, entropy increases (∆S > 0) for each of the
following:
Phase transition from solid to liquid
Phase transition from solid to gas
Phase transition from liquid to gas
An increase in the number of moles of gas during a chemical
reaction
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Predicting
Predict the sign of ∆S for each of the following processes:
1. The boiling of water
2. I2(g) → I2(s)
3. CaCO3(s) → CaO s + CO2(g)
positive
positive
negative
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Homework Problems: #2, 3, 6, 8, 10, 12
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17.4: Heat Transfer and Changes in Entropy
If going from a liquid to a solid is spontaneous, why does
entropy decrease?
Well it doesn’t, even though you would think that.
Why not?
∆𝑆𝑢𝑛𝑖𝑣 = ∆𝑆𝑠𝑦𝑠 + ∆𝑆𝑠𝑢𝑟𝑟
So the entropy of the system CAN decrease, as long as the
entropy of the surroundings increases by a greater amount
So that the entropy of the universe still increases
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Spontaneity of Water Freezing
If we look at water freezing at a low temperature:
But if we looked at a high temperature:
This is like giving a rich verses a poor man $1000
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Entropy Changes
In general…
A process that emits heat into the surroundings (qsys negative)
increases the entropy of the surroundings (positive ∆Ssurr)
A process that absorbs heat from the surroundings (qsys
positive) decreases the entropy of the surroundings (negative
∆Ssurr)
The magnitude of the change in entropy of the surroundings is
proportional to the magnitude of qsys
We then get the relationship:
∆𝑆𝑠𝑢𝑟𝑟 =−∆𝐻𝑠𝑦𝑠
𝑇 (𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑃, 𝑇)
If at constant pressure qsys = ∆Hsys
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Calculating Entropy Changes Example
Consider the following reaction:
2𝑁2 𝑔 + 𝑂2 𝑔 → 2𝑁2𝑂 𝑔 ∆𝐻𝑟𝑥𝑛 = +163.2𝑘𝐽
a. Calculate the entropy change in the surroundings
associated with this reaction occurring at 25˚C.
∆𝑆𝑠𝑢𝑟𝑟 =−∆𝐻
𝑇=−(163.2𝑘𝐽)
298𝐾= −548𝐽/𝐾
b. Determine the sign of the entropy change for the
system.
ΔSsys is negative (less moles on product side)
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c. Determine the sign of the entropy change for the
universe. Will the reaction be spontaneous?
∆𝑆𝑢𝑛𝑖𝑣 = ∆𝑆𝑠𝑦𝑠 + ∆𝑆𝑠𝑢𝑟𝑟
∆𝑆𝑢𝑛𝑖𝑣 = − + −
Therefore ΔSuniv is negative -> process is not spontaneous
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17.5: Gibbs Free Energy
Using the equations for this chapter, we can solve for a
new thermodynamic function: (see page 653)
∆𝐺 = ∆𝐻 − 𝑇∆𝑆
Where G is Gibbs free energy
If you know the sign of G (positive or negative), you can
tell if the process is spontaneous or not
If ∆G is negative, the process is spontaneous
If ∆G is positive, the process is nonspontaneous
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Gibbs Free Energy
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Gibbs Free Energy
Case 1: ∆H negative, ∆S positive
G will be negative at all temperatures (∴ spontaneous)
Case 2: ∆H positive, ∆S negative
G will be positive at all temperature (∴ nonspontaneous)
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Gibbs Free Energy
Case 3: ∆H negative, ∆S negative
Case 4: ∆H negative, ∆S positive
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Predicting Spontaneity Example
Consider the following reaction:
𝐶2𝐻4 𝑔 + 𝐻2 𝑔 → 𝐶2𝐻6 𝑔 ∆𝐻 = −137.5𝑘𝐽; ∆𝑆 = −120.5𝐽/𝐾
Calculate ∆G at 25˚C and determine whether the
reaction is spontaneous. Does ∆G become more negative
or more positive as the temperature increases?
∆𝐺 = ∆𝐻 − 𝑇∆𝑆 = −317 × 103𝐽 − 298𝐾 −120.5𝐽
𝐾
∆𝐺 = −101.6𝑘𝐽
Reaction is spontaneous. As temperature increase, G
becomes more positive
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Homework Problems: #14, 15, 17, 22, 24, 26, 28, 32, 38
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17.6: Entropy Changes in Reactions
To find the change in entropy for any reaction, use an
equation that is similar to one we’ve already seen:
∆𝑆𝑟𝑥𝑛° = ∆𝑆𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠
° + ∆𝑆𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠°
Just use the standard molar entropy values (pg. 657 and
appendix IIB)
The third law of thermodynamics states:
The entropy of a perfect crystal at absolute zero (0K) is zero.
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Entropy Changes Example
Calculate ∆Srxn˚ for the following reaction:
2𝐻2𝑆 𝑔 + 3𝑂2 𝑔 → 2𝐻2𝑂 𝑔 + 2𝑆𝑂2(𝑔)
∆𝑆𝑟𝑥𝑛°= 𝑛𝑝𝑆
°(𝑝𝑟𝑜𝑑) − 𝑛𝑟𝑆° 𝑟𝑒𝑎𝑐𝑡
∆𝑆𝑟𝑥𝑛°= 2𝑚𝑜𝑙 188.8
𝐽
𝑚𝑜𝑙𝐾+ 2𝑚𝑜𝑙 248.2
𝐽
𝑚𝑜𝑙𝐾
− 2𝑚𝑜𝑙 205.8𝐽
𝑚𝑜𝑙𝐾+ 3𝑚𝑜𝑙 205.2
𝐽
𝑚𝑜𝑙𝐾
∆𝑆𝑟𝑥𝑛°= 874
𝐽
𝐾− 1027
𝐽
𝐾= −153.2
𝐽
𝐾
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17.7: Free Energy Changes
Using all of the thermodynamics information (and tables),
we can calculate the free energy change of a reaction:
∆𝐺𝑟𝑥𝑛° = ∆𝐻𝑟𝑥𝑛
° − 𝑇∆𝑆𝑟𝑥𝑛°
Example:
For the following reaction:
𝑁𝑂 𝑔 +1
2𝑂2(𝑔) → 𝑁𝑂2(𝑔)
Compute ∆Grxn˚ at 25˚C and determine if the process is
spontaneous.
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∆𝐻𝑟𝑥𝑛°= 33.2
𝑘𝐽𝑚𝑜𝑙 − 91.3
𝑘𝐽𝑚𝑜𝑙 +
1
20
∆𝐻𝑟𝑥𝑛°= −58.1𝑘𝐽
∆𝑆𝑟𝑥𝑛°= 240.1
𝐽𝑚𝑜𝑙𝐾
− 210.8𝐽𝑚𝑜𝑙𝐾 +
1
2205.2𝐽𝑚𝑜𝑙𝐾
∆𝑆𝑟𝑥𝑛°= −73.2
𝐽𝐾
∆𝐺𝑟𝑥𝑛°= ∆𝐻𝑟𝑥𝑛
° − 𝑇∆𝑆𝑟𝑥𝑛°
∆𝐺𝑟𝑥𝑛°= −58.1 × 103𝐽 − 298𝐾 −73.2
𝐽𝐾
= −36.3𝑘𝐽 ∴ 𝑟𝑥𝑛 𝑖𝑠 𝑠𝑝𝑜𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠
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17.8: Free Energy Changes with
Nonstandard States
Just use the equation:
∆𝐺𝑟𝑥𝑛 = ∆𝐺𝑟𝑥𝑛° + 𝑅𝑇𝑙𝑛𝑄
Where Q is the reaction quotient (equilibrium)
T in K
R is gas constant
8.314 J/mol•K
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Example
Consider the following reaction at 298K:
2𝐻2𝑆 𝑔 + 𝑆𝑂2 𝑔 → 2𝑆 𝑠 + 2𝐻2𝑂 𝑔 ∆𝐺𝑟𝑥𝑛
°= −102𝑘𝐽 Compute ∆Grxn under the following conditions:
𝑃𝐻2𝑆 = 2.00𝑎𝑡𝑚 𝑃𝑆𝑂
2= 1.50𝑎𝑡𝑚 𝑃𝐻
2𝑂 = 0.0100𝑎𝑡𝑚
Is the reaction more or less spontaneous under these conditions than under standard states?
𝑄 =𝑃𝐻2𝑂
2
𝑃𝐻2𝑆2𝑃𝑆𝑂2=(0.1)2
2 2(1.5)= 1.67 × 10−5
∆𝐺𝑟𝑥𝑛 = ∆𝐺° + 𝑅𝑇𝑙𝑛𝑄
= −102kJ + (8.314𝐽
𝑚𝑜𝑙𝐾)(298𝐾)(ln 1.67 × 10−5
= −129𝑘𝐽 𝑡ℎ𝑒 𝑟𝑥𝑛 𝑖𝑠 𝑚𝑜𝑟𝑒 𝑠𝑝𝑜𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠
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17.9: Free Energy and Equilibrium
If we know that at equilibrium, ∆Grxn = 0 and Q = K,
∆𝐺𝑟𝑥𝑛° = −𝑅𝑇𝑙𝑛𝐾
When K < 1, lnK is negative, ∆Grxn˚ is positive
Reaction is spontaneous in reverse direction
When K > 1, lnK is positive, ∆Grxn˚ is negative
Reaction is spontaneous in forward direction
When K = 1, lnK is zero, ∆Grxn˚ is zero
Reaction is at equilibrium
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Example
Compute ∆Grxn˚ at 298K for the following reaction:
𝐼2 𝑔 + 𝐶𝑙2 𝑔 ↔ 2𝐼𝐶𝑙 𝑔 𝐾𝑝 = 81.9
∆𝐺𝑟𝑥𝑛°= −𝑅𝑇𝑙𝑛𝐾
∆𝐺𝑟𝑥𝑛°= −(8.314
𝐽
𝑚𝑜𝑙𝐾)(298𝐾)(ln 81.9
= −10.9𝑘𝐽𝑚𝑜𝑙
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Homework Problems: #40, 42, 46
Review: #50, 51, 53
Entropy:
http://www.youtube.com/watch?v=ZsY4WcQOrfk&index
=21&list=PL8dPuuaLjXtPHzzYuWy6fYEaX9mQQ8oGr
Electrochemistry:
http://www.youtube.com/watch?v=IV4IUsholjg&index=38
&list=PL8dPuuaLjXtPHzzYuWy6fYEaX9mQQ8oGr