Chemistry 2402 - Thermodynamics

Lecture 11 : Phase Diagrams and Solubility

Lecture 12 : Kinetic Coefficients for the Relaxation to Equilibrium

Lecture 13: Non Ideal Solutions and Activity

Entropy and the Driving Forces Towards Equilibrium

Remember this basic problem regarding spontaneous change?

UI VI NI UII VII NII

Before

? ?

After

In an isolated system, we remove an internal constraint. Previously we asked, what happens?

Here we want to expand this question to also include “.. and how fast does it happen?”

Connecting Entropy Change to Flows of Extensive Properties

Now we can write the most general expression for a change in entropy dS in terms of changes is U, V and N as

or, making use of the intensive parameters,

If the expressions in the brackets corresponds to a non-zero value then the extensive quantity associated with that expression will spontaneously flow between the two systems (if permitted) in order to increase the total entropy

I

VSII

II

VSI

II

NSII

II

NSI

II

VNII

II

VNI

IIII dN

N

S

N

SdV

V

S

V

SdU

U

S

U

SSSd

,,,,,,

)(

III

II

I

II

II

II

I

II

IIIIII dN

TTdV

T

P

T

PdU

TTSSd

11)(

Introducing Affinities – the Thermodynamic Driving Forces

III

II

I

II

II

II

I

II

IIIIII dN

TTdV

T

P

T

PdU

TTSSd

11)(

Given the importance of the terms in the brackets we will give them a name – affinity.

There are as many affinities as there are independent variables in the entropy. In the system considered here we have 3 affinities associated with energy, volume and particle number changes.

The three affinities Fi

FU = 1/TI-1/TII FV = PI/TI-PII/TII FN = -μI/TI+μII/TII

Connecting Affinities to Kinetics:

Markoff Processes

What we want to do is connect the affinities to the rate of change of our extensive variables, i.e. dUI/dt, dVI/dt and dNI/dt.

Two assumptions are required to do this.

1) The Markoff (‘no memory’) Assumption

The time derivatives of properties depends only on the instantaneous state of the system and does not depend on the system’s state at previous times.

This means that the time derivative of, for example, the energy at a given moment is only a function of properties of the system at that moment and their affinities, i.e.

dUI/dt = f(FU,FV,FN,UI,NI,VI)

What we neglect here is values of properties or affinities at earlier times i.e. a ‘memory’ of previous states.

Connecting Affinities to Kinetics:

Linear Response

The second assumption:

2) The Linear (‘small perturbation’) Assumption

Since the rate of change must equal zero when the affinities vanish we can imagine expanding the right hand side of

dUI/dt = f(FU,FV,FN,UI,NI,VI)

in powers of the affinities, i.e.

dUI/dt = ΣiLUiFi + ΣiΣk LUikFiFk + .... (where i and k can be U, N or V)

and then keep only the linear terms by assuming that the affinities are small, i.e.

dUI/dt = ΣiLUiFi

where the coefficients Lik are called the kinetic coefficients .

Applying the Linear Response Theory

e.g. Heat flow

I II

UIIUIheat flow only

UUUI FL

dt

dU

IIIU TTF

11

IIIUU

I

TTL

dt

dU 11 where the temperatures will change as heat flows between the two systems.

LUU is directly related to the thermal conductivity.

Notice that the linear response equations allow for non-zero kinetic coefficients that allow an affinity in, say, N to drive a change in a different quantity, say U, i.e.

NUNUUUI FLFL

dt

dU

An important simplification: Lik = Lki

(proving this in 1931 won Lars Onsager a Nobel prize in 1968)

Examples of Kinetic Coefficients

Many familiar kinetic processes can be described with our linear response equations.

Process Kinetic Coefficient

Affinity

Heat flow Thermal conductivity

Temperature gradient

Electric current Electrical conductivity

Voltage

Diffusion Diffusion constant

Concentration gradient

How about chemical reaction kinetics? How does it fit into this linear response treatment of relaxation towards equilibrium?

Chemical Reactions Kinetics: A Revision

A generic chemical reaction is specified by an expression such as νAA + νBB ⇋νCC + νDD

where νi is the stoichiometric coefficient for the ith component.

Note that it is the relative values of the stoichiometric coefficients that matter, not their absolute values since we can multiply both sides of the reaction by some constant factor without changing the chemical description of the reaction.

The kinetics of such a reaction is described by a set of rate laws that, typically, cannot be trivially obtained from the expression above which simply identifies the reactants and products and their relative proportions.

Here is an example of the experimentally determined rate laws responsible for the overall reaction O3+O→2O2

i) d[O3]/dt = - k1[O3][Cl] arising from O3 + Cl → ClO + O2

ii) d[O2]/dt = k2[ClO][O] arising from ClO + O → Cl + O2

In fact for any elementary mechanistic step, X+yY→ products, the rate of reaction is d[X]/dt = -k[X][Y]y

Multi-Process Reactions Can Have Complex Overall Rate Laws

Consider the apparently simple overall reaction H2 + Br2 → 2HBrThe overall rate law was found by Bodenstein in 1916 to be

Max Bodenstein

1871-1942This complicated rate law must be the result of multiple elementary processes – in this case including chemical chain reactions.

(Bodenstein was the first to describe a chain reaction.)This is an example of kinetics for which the rate is clearly not proportional to a concentration and, hence, is unlikely to be described by linear response. The linear response approximation, which works quite well for heat flow, diffusion and electrical conduction, will not be able to describe most chemical kinetics except very close to equilibrium.

Remember, our thermodynamic treatment of kinetics knows nothing of the specific elementary mechanisms expressed in the empirical rate laws.

Kinetics and Chemical Equilibrium

Seeing that the kinetics of a reaction can involve some complicated sequence of elementary processes, possibly involving chemical species other than those identified as reactants or products, how do we connect the kinetics with the equilibrium state?

At equilibrium, the forward and backward rate of each elementary process must be equal. This requirement is known as the principle of detailed balance.

i.e. k1[NO]2eq = k-1[N2O2]eq and k2[N2O2]eq[H2]eq=k-2[N2O]eq[H2O]eq and k3[N2O]eq[H2]eq=k-

3[N2]eq[H2O]eq

For example: 2NO + 2H2 ⇋ N2 + 2H2O

is believed to occur as i) NO + NO ⇋ N2O2

ii) N2O2+ H2 ⇋ N2O + H2O

iii) N2O + H2 ⇋ N2 + H2Ok3

k1

k-1

k2

k-2

k-3

]][[

]][[

]][[

]][[

][

][

22

22

222

22222

321

321

HON

OHN

HON

OHON

NO

ON

kkk

kkk

KHNO

OHN

22

2

222

][][

]][[=

the equilibrium constant

What Are the Affinities for Chemical Reactions?

The rate laws we typically use to describe chemical reaction kinetics are found empirically through the comparison of the observed rate with the various concentrations.

Thermodynamics provides us with the definition of the affinities that directly connect the spontaneous direction of a reaction to the increase in entropy.

Consider the general reactionνAA + νBB ⇋ νCC + νDD

The change in entropy S when the reaction proceeds from left to right is

where dξ measures how far we let the reaction proceed.

The chemical affinity that drives the reaction towards equilibrium is

dS/dξ = Fξ= - (νCμC+νDμD-νAμA-νBμB)/T

dT

dS BBAADDCC )(1

Rate Constants and Kinetic Coefficients: What’s Their Connection?

So, finally, let’s compare the description of chemical kinetics provided by the empirical rate law and by the linear response approach.

For our example, let’s consider (once again) the isomerization A ⇋ BLinear Response

dNB/dt = -L(μB-μA)/T

d[B]/dt = -(L/V)(μB-μA)/T

Empirical Rate Law

d[B]/dt = k1[A]-k-1[B]

k1

k-1

Rate Constants and Kinetic Coefficients: What’s Their Connection?

So, finally, let’s compare the description of chemical kinetics provided by the empirical rate law and by the linear response approach.

For our example, let’s consider (once again) the isomerization A ⇋ BLinear Response

dNB/dt = -L(μB-μA)/T

d[B]/dt = -(L/V)(μB-μA)/T

Empirical Rate Law

d[B]/dt = k1[A]-k-1[B]

If we assume ideal solutions then TkBABeAkdtBd /)(

1 1][/][

Rate Constants and Kinetic Coefficients: What’s Their Connection?

So, finally, let’s compare the description of chemical kinetics provided by the empirical rate law and by the linear response approach.

For our example, let’s consider (once again) the isomerization A ⇋ BLinear Response

dNB/dt = -L(μB-μA)/T

d[B]/dt = -(L/V)(μB-μA)/T

Empirical Rate Law

d[B]/dt = k1[A]-k-1[B]

If we assume ideal solutions then TkBABeAkdtBd /)(

1 1][/][ This result shows that the empirical rate law i) includes the affinity in a nonlinear form and ii) it does predict equilibrium when the affinity goes to zero

Rate Constants and Kinetic Coefficients: What’s Their Connection?

So, finally, let’s compare the description of chemical kinetics provided by the empirical rate law and by the linear response approach.

For our example, let’s consider (once again) the isomerization A ⇋ BLinear Response

dNB/dt = -L(μB-μA)/T

d[B]/dt = -(L/V)(μB-μA)/T

Empirical Rate Law

d[B]/dt = k1[A]-k-1[B]

If we assume ideal solutions then

Finally, assume (μB-μA)/kBT is small, so

d[B]/dt ≈ -k1[A]eq(μB-μA)/kBT

TkBABeAkdtBd /)(1 1][/][

Rate Constants and Kinetic Coefficients: What’s Their Connection?

So, finally, let’s compare the description of chemical kinetics provided by the empirical rate law and by the linear response approach.

For our example, let’s consider (once again) the isomerization A ⇋ BLinear Response

dNB/dt = -L(μB-μA)/T

d[B]/dt = -(L/V)(μB-μA)/T

Empirical Rate Law

d[B]/dt = k1[A]-k-1[B]

If we assume ideal solutions then

Finally, assume (μB-μA)/kBT is small, so

d[B]/dt ≈ -k1[A]eq(μB-μA)/kBT

TkBABeAkdtBd /)(1 1][/][

Comparing the linear response equation with the linear version of the rate law we find

L = k1[A]eqV/kB

Rate Laws Represent the Most Complete Description of Chemical

KineticsTo summarise:

• The linear response approach provides a completely general description of the kinetics of relaxation to equilibrium for any quantity – sufficiently close to equilibrium.

• Unlike heat conduction of diffusion, chemical reactions often involve large excursions away from equilibrium (e.g. combustion) and so the linear response approach is of limited use in describing chemical kinetics.

• Empirical rate laws satisfy the thermodynamic requirement that the direction of reaction is determined by the affinity.

• Empirical rate laws and their rate coefficients represent the most complete description of chemical kinetics. In Lecture 15 we shall return to consider what factors influence the size and temperature dependence of these rate coefficients.

Summary

You should now• Be able to explain the terms affinity, and kinetic coefficient• Understand the assumptions behind the linear response

approach to relaxation kinetics, and the situations when those assumptions break down

• Understand the empirical nature of overall rate laws, and their connection to underlying elementary mechanisms

Next Lecture• Non Ideal Solutions and Activity