What is a spontaneous reaction?
One, that given the necessary activation energy, proceeds without continuous outside assistance
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
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)
Entropy, S
- a measure of the disorder of a system or the surroundings
Entropy of “The Universe”
the systemthesurroundings
The Universe
The System
The Surroundings
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)
Low entropy is less probable
SuniverseSsystemSsurroundings
If Suniverse0, reaction is spontaneous
If Suniverse0, reaction is nonspontaneous
heat
heat
Ssurr increases!
H < 0
How does the system impacts the Ssurr?
Entropy is a State FunctionS = Sfinal - Sinitial
path taken is irrelevant
rate of change is irrelevant
S > 0 for:
- melting
- vaporizing
- making a solution
- a reaction that produces
an increased number of moles
- heating a substance
H2O(s) H2O(l)
ordered, low S
less ordered,
high SS > 0
S > 0 for:
- melting
- vaporizing
- making a solution
- a reaction that produces
an increased number of moles
- heating a substance
H2O(l) H2O(g)
high entropy
low entropy
S > 0 for:
- melting
- vaporizing
- making a solution
- a reaction that produces
an increased number of moles
- heating a substance
low entropy
high entropy
Very unlikely!
More likely!
Benzene Toluene
S > 0 for:
- melting
- vaporizing
- making a solution
- a reaction that produces
an increased number of moles
- heating a substance
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)
S > 0 for:
- melting
- vaporizing
- making a solution
- a reaction that produces
an increased number of moles
- heating a substance
Temperature
Ent
ropy
S
L
G
Sfusion
Svaporization
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
Calculating S for a reaction
Srxn =np Soproducts - nr So
reactants
standard entropy in J/K i.e. (@ SATP)
stoichiometriccoefficient
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
for example,
= [8(213.6) + 9(188.6)] – [463.2 + 12.5(204.8)]
= +383.0 J K-1 mol-1
• 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?
Gibbs Free EnergySunivSsysSsurr
and, Ssurr = -Hsys
T
thus, SunivSsys-Hsys
T
-TSuniv= -TSsys+Hsys
now multiply through by -T
-TSuniv= Hsys-TSsysor,
or, Gsys= Hsys-TSsys
G= H-TS
Gibbs energy change
or the “free energy change”
G and spontaneity
recall that Gsys = -TSuniv
since Suniv > 0 for a spontaneous change,
Gsys < 0 for a spontaneous change
What’s “free” about free energy?
G= H-TS
the energy transferred as heat
the energy used up creating “disorder”
the “free” energyleft over
Ho So Go Spontaneous?
- + - always
+ - + never
- - + or - at lower T
+ + + or - at higher T
When is G < 0?
G is a state functionG = Gfinal - Ginitial
path is irrelevant
rate of reaction is irrelevant
How do we find G values?
1. Calculate H,S values, then use G = H - TS
2. Look up Gof values
for example,
Will this reaction proceed at 25oC?
4 KClO3(s) 3 KClO4(s) + KCl(s)
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
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
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
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
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)
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
Homework:
p.g. 512: 1 - 14