what is a spontaneous reaction? one, that given the necessary activation energy, proceeds without...

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