Spontaneous Processes

The Second Law:

S 0The entropy of a closed system can only increase.

If a process will decrease entropy in a closed system, then it does not occur spontaneously. Its opposite will occur spontaneously.

But we very rarely work with closed systems...

Spontaneous Processes

…except for the universe as a whole!

The entropy of a system can spontaneously decrease, as long as the entropy of the surroundings increases by at least as much.

Ssys + Ssurr 0

Do we have to keep calculating Ssurr ?

Not necessarily!

Spontaneous Processes

Let’s stay at constant T and P:

ΔSsurr=−qP

T=−

ΔHT

ΔStot=ΔSsys−ΔHT

ΔStot≥0⇔ ΔSsys−ΔHT

≥0

Now everything is in terms of the system. The criterion for spontaneity given by the Second Law becomes:

Gibbs Free Energy

The Gibbs Free Energy is a new state function, defined as:

G ≡H −TS

ΔG =ΔH −TΔS

ΔStot≥0⇔ ΔG ≤0

Josiah Willard Gibbs(1839-1903)

At constant temperature and pressure, G is

From the previous slide, we end up with

Gibbs Free Energy

The condition of constant T and P is very important when using G. Otherwise, the entropy change of the surroundings might be different leading to a different result.

G is extremely useful for chemistry and biochemistry, since so much takes place at constant temperature and pressure.

At constant T and P, consideration of G will answer the question

“Will a given reaction be spontaneous?”

• G is still defined and can be calculated for any change of state, including changing P and T.

• We can also define another state variable, Helmholtz Free Energy (A = E - TS), which has similar characteristics as G, but relates more directly to constant T and V processes.

Gibbs Free Energy Summary

The Gibbs Free Energy is a direct measure of spontaneity:

G = H - TS

It sums up, in a way, the competition between energy considerations and “configurational” barriers.

We have also learned that a process is spontaneous if

G < 0

Thus, ifH < 0 the process is exothermic (downhill) S > 0 the process is increases disorder

So H dominates spontaneity at low temperaturesS dominates spontaneity at high temperatures

Calculation of G

In many cases, we can build on calculations we have already done in order to get G.

ΔG =ΔH −TΔS

In other cases, like chemical reactions, standard values have been found. We need only to add them up properly.

Example: ice melting at 100° C

H = 6.75 kJ mol–1

S = 45.5 J K–1 mol–1

soG = 6750 – (373)(45.5) = –10.2 kJ mol–1.

Calculation of G

H2O(g) H2(g) + 1/2 O2(g)

Is So298 greater than, less than, or equal to zero?

So298= -(188.82) + 130.684 + 1/2 (205.14) J/(K mol)

= 44.4 J / (K mol)Spontaneous?

Ho298=-(-241.82) + 0 + 1/2 (0) kJ/mol = 241.82

Go298= Ho

298-T So298 = 241.82 kJ/mol - (298 K)*0.0444 kJ/(K mol)

= 228.56 kJ/mol

The process is non-spontaneous!

Calculation of G

Example: CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

G°f –50.7 0 –394.36 –228.6

G°r = –800 kJ mol–1

Calculation of G

We can perform calculus on G directly:

dG = dH – TdS – SdT = dE + pdV + Vdp – TdS – SdT

but dE = –pdV +TdS

(dw = -pdV and dqrev = TdS)

In other words, constant TG(p2)−G(p1) = V(p)dpp1

p2

∫

G(T2)−G(T1) = S(T)dTT1

T2

∫and constant P

Calculation of GA puzzle:

ΔG =ΔE +pΔV −TΔS

ΔS=q T ΔE =w+q

ΔG =w+pΔV

w=−pΔV

Hence, ΔG =0

so

but

Assume everything is reversible.

at constant T and p,

???

According to this, we can’t ever have G < 0 if everything is reversible at constant T and p. But what about all those chemical reactions? Surely they can be run reversibly! But ΔG ≠0

Where is the mistake?

Calculation of G

A puzzle:

ΔG =ΔE +pΔV −TΔS

ΔS=q T ΔE =w+q

ΔG =w+pΔV

w=−pΔV

Hence, ΔG =0

so

but

Assume everything is reversible.

at constant T and p,

???

Not all work is PV work! For example, electrochemical, mechanical, etc.

“Free” means free to do non-PV work!

Temperature Dependence

Simplest approximation:

G = H - T S

assume both H and S do not change for moderate changes of T

G(T) = G(298 K) - (T - 298 K)S

If we are assuming H and S independent of T, why not just ignoreG dependence?

Explicit dependence on T in G, whereas only implicit for H and S

Temperature DependenceGeneral expression (reversible path, only PV work):

dG = dH - TdS - SdT = dE +PdV + VdP - TdS - SdT

but dE = –PdV +TdS since dw = -PdV and dqrev = TdSand dE = dw + dq (we are on a specified path so this is ok)so …

dG = (-PdV + PdV) + (TdS - TdS) + VdP - SdT = VdP - SdT

At constant pressure:dG = -SdT

dGT1

T2

∫ =− S(T)dTT1

T2

∫

G(T2)−G(T1) =− S(T)dTT1

T2

∫

ΔG T1( )=ΔG(298)− ΔS T( )dT298

T

∫

Temperature Dependence

Gibbs-Helmholtz equation

ΔG T2( )

T2

−ΔG T1( )

T1

=−ΔH T( )

T2 dTT1

T2

∫

If H changes little with temperature

ΔG T2( )

T2

−ΔG T1( )

T1

=ΔH1T2

−1T1

⎛

⎝ ⎜

⎞

⎠ ⎟

See text p. 93 for mathematical derivation.

Pressure Dependence

For the pressure dependence we hold T constant:

dG = VdP-SdT=VdPThus,

For solids and liquids, V does not vary with temperature so,

and for an ideal gas:

dG=G P1( )−G P2( ) =P1

P2

∫ VdPP1

P2

∫

G P2( )−G P1( )=V P2 −P1( )

G P2( )−G P1( )=nRTPP1

P2

∫ dP=nRTlnP2

P1

⎛

⎝ ⎜

⎞

⎠ ⎟

Pressure Dependence ExampleCan we force graphite to diamond by increasing the pressure? We will use:

and the fact that molar volumes of graphite and diamond are known:

V graphite=5.33cm3 /mol

V diamond=3.42cm3 /mol

ΔG(P) =ΔG(1atm)+ΔV*(P −1)

ΔG(P) =2.84−1.935∗10−4*(P −1)

Where we have used a conversion factor to convert from cm3 atm to kJ.

G P2( )−G P1( )=V P2 −P1( )

Now, we want the pressure that makes G = 0: Why?

0 = 2.84 - 1.935 10-4 (P-1) kJ/mol P = 15,000 atm

Experimentally, the required pressure is more! Why?

Example: Reversible Process

H2O (l) H2O (g)100 °C

What is the free energy change?

Well, this is at constant T and P. Its reversible. So,

G°vap= 0 = H°vap-T S°vap

Note that this implies:

oo

vapvap S

T

H=

G of Mixing

Consider the isobaric, isothermal mixing of two gases:

Gas Aat

1 Atm

Gas Bat

1 Atm

Gas A+Bat

1 Atm (total)

Is this reaction spontaneous?

ΔGmixing=ΔGexpansion of A+ΔGexpansion of B

ΔGmixing=nARTlnP2,A

1atm⎛ ⎝ ⎜ ⎞

⎠ ⎟ +nBRTln

P2,B

1atm⎛ ⎝ ⎜ ⎞

⎠ ⎟

Well, both P2,A and P2,A are less than 1 atm…so yes, this process is spontaneous.

Protein Unfolding

Proteins have a native state. (Really, they tend to have a tight cluster of native states.)

Denaturation occurs when heat or denaturants such as guanidine, urea or detergent are added to solution. Also, the pH can affect folding.

When performing a denaturation process non-covalent interactions are broken.

Ionic, van der-Waals, dipolar, hydrogen bonding, etc.Solvent is reorganized.

Protein Unfolding

Let’s consider denaturation with heat. We can determine a great deal about the nature of the protein from such a consideration.

The experimental technique we use for measuring thermodynamic changes here is the differential scanning calorimeter.

Basic experiment: Add heat to sample, measure its temperature change.

heat

Protein Unfolding

In differential scanning calorimetry you have two samples: Your material of interestControl

You put in an amount of heat to raise the temperature of the control at a constant rate, then measure the rate of change in temperature of the other sample as a function of the input heat.

This is a measure of the heat capacity!

Protein+

SolventSolvent

HeatT1

T2

T1-T2

Protein Unfolding

Data for glyceraldehyde-3-phosphate dehydrogenase.

pH8.0

pH6.0

Bacillus stearothermophilus

E. coliRabbit

Is the protein more stable at pH 8 or 6? Why is B. stear. more stable?

Protein Unfolding

We are given the following data for the denaturation of lysozyme:

10 25 60 100 °C

G° kJ/mol 67.4 60.7 27.8 -41.4H° kJ/mol 137 236 469 732S° J/ K mol 297 586 1318 2067TS° kJ/mol 69.9 175 439 771

Where is the denaturation temperature?

What then is special about the temperature at which the denaturation is spontaneous?