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What is a spontaneous reaction?
One, that given the necessary activation energy, proceeds without continuous outside assistance
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Why do some reactions occur spontaneously & others do not?
• Atoms react to achieve greater stability
• Therefore products are generally more energetically stable than reactants
• In general, exothermic reactions (-) tend to proceed spontaneously
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EXCEPTIONS
• Some endothermic reactions and those that produce less energetically stable products proceed spontaneously
EXAMPLES: • Ba(OH)2(aq) + 2 NH4NO3(aq) Ba(NO3)2(aq) + 2 NH4OH(l)
• NH4NO3(s) NH4 +
(aq) + NO3 -(aq)
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Entropy, S
- a measure of the disorder of a system or the surroundings
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Entropy of “The Universe”
the systemthesurroundings
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The Universe
The System
The Surroundings
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1st law of thermodynamics:
The total energy of the universe is constant
(The best you can do is break even)
2nd law of thermodynamics:
The entropy of the universe is increasing
(You can’t break even)
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Low entropy is less probable
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SuniverseSsystemSsurroundings
If Suniverse0, reaction is spontaneous
If Suniverse0, reaction is nonspontaneous
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heat
heat
Ssurr increases!
H < 0
How does the system impacts the Ssurr?
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Entropy is a State FunctionS = Sfinal - Sinitial
path taken is irrelevant
rate of change is irrelevant
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S > 0 for:
- melting
- vaporizing
- making a solution
- a reaction that produces
an increased number of moles
- heating a substance
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H2O(s) H2O(l)
ordered, low S
less ordered,
high SS > 0
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S > 0 for:
- melting
- vaporizing
- making a solution
- a reaction that produces
an increased number of moles
- heating a substance
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H2O(l) H2O(g)
high entropy
low entropy
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S > 0 for:
- melting
- vaporizing
- making a solution
- a reaction that produces
an increased number of moles
- heating a substance
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low entropy
high entropy
Very unlikely!
More likely!
Benzene Toluene
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S > 0 for:
- melting
- vaporizing
- making a solution
- a reaction that produces
an increased number of moles
- heating a substance
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Ba(OH)28H2O(s) + 2 NH4NO3(s)
Ba(NO3)2(aq) + 2 NH3(aq) + 10 H2O(l)
H = +80.3 kJ (unfavorable)
3 moles 13 moles
S > 0 (favorable)
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S > 0 for:
- melting
- vaporizing
- making a solution
- a reaction that produces
an increased number of moles
- heating a substance
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Temperature
Ent
ropy
S
L
G
Sfusion
Svaporization
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Entropy tends to increaseIn general, a system will increase in entropy (S > 0) if:
•the volume of a gaseous system increases
•the temperature of a system increases
•the physical state of a system changes from solid to liquid to gas
• the number of moles in a system increases
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Calculating S for a reaction
Srxn =np Soproducts - nr So
reactants
standard entropy in J/K i.e. (@ SATP)
stoichiometriccoefficient
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for example,C8H18(g) + 12.5 O2(g)8 CO2(g) + 9 H2O(g)
13.5 moles 17 moles
(expect S > 0)
Srxn=n Soproducts - n So
reactants
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for example,
= [8(213.6) + 9(188.6)] – [463.2 + 12.5(204.8)]
= +383.0 J K-1 mol-1
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• Temperature and pressure are strongly connected to ideas of enthalpy and entropy. (Remember that -∆H and +∆S are favourable).
• Consider the following three examples:For each reaction, identify the sign of ∆H and ∆S. Indicate whether the reaction is likely to be spontaneous.
1. Zn (s) + 2 HCl (aq) ↔ ZnCl2 (aq) + H2 (g)
2. 3 C (s) + 3 H2 (g) C3H6 (g)
3. 2 Pb(NO3)2 (s) 2 PbO (s) + 4 NO2 (g) + O2 (g)
• In a case where both ∆H and ∆S are favourable, we consider the reaction to be spontaneous and very likely to occur. What about in cases where only one is favoured?
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Gibbs Free EnergySunivSsysSsurr
and, Ssurr = -Hsys
T
thus, SunivSsys-Hsys
T
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-TSuniv= -TSsys+Hsys
now multiply through by -T
-TSuniv= Hsys-TSsysor,
or, Gsys= Hsys-TSsys
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G= H-TS
Gibbs energy change
or the “free energy change”
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G and spontaneity
recall that Gsys = -TSuniv
since Suniv > 0 for a spontaneous change,
Gsys < 0 for a spontaneous change
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What’s “free” about free energy?
G= H-TS
the energy transferred as heat
the energy used up creating “disorder”
the “free” energyleft over
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Ho So Go Spontaneous?
- + - always
+ - + never
- - + or - at lower T
+ + + or - at higher T
When is G < 0?
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G is a state functionG = Gfinal - Ginitial
path is irrelevant
rate of reaction is irrelevant
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How do we find G values?
1. Calculate H,S values, then use G = H - TS
2. Look up Gof values
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for example,
Will this reaction proceed at 25oC?
4 KClO3(s) 3 KClO4(s) + KCl(s)
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4 KClO3(s) 3 KClO4(s) + KCl(s)
rxn =npHoproducts - nrHo
reactants
= 3 Hof (KClO4(s)) + Ho
f (KCl (s))
- 4 Hof (KClO3(s))
= 3(-432.8) + (-436.7) - 4(-397.7)
= -144.3 kJ mol-1
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4 KClO3(s) 3 KClO4(s) + KCl(s)
Srxn =np Soproducts - nr So
reactants
= 3 So(KClO4(s)) + So(KCl (s)) - 4 So(KClO3(s))
= 3(151.0) + (82.6) - 4(143.1)
= - 36.8 J K-1 mol-1
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4 KClO3(s) 3 KClO4(s) + KCl(s)
G = H - TS
-144.3 kJ mol-1 - 298 K (-0.0368 kJ K-
1mol-1)
= -133.3 kJ mol-1
G < 0, thus reaction proceeds
spontaneously
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4 KClO3(s) 3 KClO4(s) + KCl(s)
G = H - TS
-144.3 kJ mol-1 - 298 K (-0.0368 kJ K-1mol-1)
= -133.3 kJ mol-1
N.B. conversion to kJ!
25oC
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How do we find G values?
1. Calculate H,S values, then use G = H - TS
2. Look up Gof values (standard
free energies of formation)
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4 KClO3(s) 3 KClO4(s) + KCl(s)
rxn =npGoproducts) - nr Go
reactants
= 3 Go(KClO4(s)) + Go (KCl(s))
- 4 Go (KClO3(s))
= 3(-303.2) + (-409.2) - 4(-296.3)
= - 133.6 kJ
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Homework:
p.g. 512: 1 - 14