Chemistry 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? U I V I N I U II V II N II 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

Introducing Affinities – the Thermodynamic Driving Forces 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 F i F U = 1/T I -1/T II F V = P I /T I -P II /T II F N = -μ I /T I +μ II /T II

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. dU I /dt, dV I /dt and dN I /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. dU I /dt = f( F U, F V, F N,U I,N I,V I ) 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 dU I /dt = f( F U, F V, F N,U I,N I,V I ) in powers of the affinities, i.e. dU I /dt = Σ i L Ui F i + Σ i Σ k L Uik F i F k (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. dU I /dt = Σ i L Ui F i where the coefficients L ik are called the kinetic coefficients.

Applying the Linear Response Theory e.g. Heat flow I II U II UIUI heat flow only where the temperatures will change as heat flows between the two systems. L UU 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. An important simplification: L ik = L ki (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. ProcessKinetic CoefficientAffinity Heat flowThermal conductivity Temperature gradient Electric currentElectrical conductivity Voltage DiffusionDiffusion constantConcentration 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 ν A A + ν B B ⇋ ν C C + ν D D where ν i 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 O 3 +O→2O 2 i) d[O 3 ]/dt = - k 1 [O 3 ][Cl] arising from O 3 + Cl → ClO + O 2 ii) d[O 2 ]/dt = k 2 [ClO][O] arising from ClO + O → Cl + O 2 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 H 2 + Br 2 → 2HBr The overall rate law was found by Bodenstein in 1916 to be Max Bodenstein This 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. k 1 [NO] 2 eq = k -1 [N 2 O 2 ] eq and k 2 [N 2 O 2 ] eq [H 2 ] eq =k -2 [N 2 O] eq [H 2 O] eq and k 3 [N 2 O] eq [H 2 ] eq =k -3 [N 2 ] eq [H 2 O] eq For example:2NO + 2H 2 ⇋ N 2 + 2H 2 O is believed to occur asi) NO + NO ⇋ N 2 O 2 ii) N 2 O 2 + H 2 ⇋ N 2 O + H 2 O iii) N 2 O + H 2 ⇋ N 2 + H 2 O k3k3 k1k1 k -1 k2k2 k -2 k -3 = 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 reaction ν A A + ν B B ⇋ ν C C + ν D D The change in entropy S when the reaction proceeds from left to right is where d ξ measures how far we let the reaction proceed. The chemical affinity that drives the reaction towards equilibrium is dS/d ξ = F ξ = - ( ν C μ C +ν D μ D -ν A μ A -ν B μ B )/ T

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 ⇋ B Linear Response dN B /dt = -L(μ B -μ A )/T d[B]/dt = -(L/V)(μ B -μ A )/T Empirical Rate Law d[B]/dt = k 1 [A]-k -1 [B] k1k1 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 ⇋ B Linear Response dN B /dt = -L(μ B -μ A )/T d[B]/dt = -(L/V)(μ B -μ A )/T Empirical Rate Law d[B]/dt = k 1 [A]-k -1 [B] If we assume ideal solutions then

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 ⇋ B Linear Response dN B /dt = -L(μ B -μ A )/T d[B]/dt = -(L/V)(μ B -μ A )/T Empirical Rate Law d[B]/dt = k 1 [A]-k -1 [B] If we assume ideal solutions then 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 ⇋ B Linear Response dN B /dt = -L(μ B -μ A )/T d[B]/dt = -(L/V)(μ B -μ A )/T Empirical Rate Law d[B]/dt = k 1 [A]-k -1 [B] If we assume ideal solutions then Finally, assume (μ B -μ A )/k B T is small, so d[B]/dt ≈ -k 1 [A] eq (μ B -μ A )/k B T

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 ⇋ B Linear Response dN B /dt = -L(μ B -μ A )/T d[B]/dt = -(L/V)(μ B -μ A )/T Empirical Rate Law d[B]/dt = k 1 [A]-k -1 [B] If we assume ideal solutions then Finally, assume (μ B -μ A )/k B T is small, so d[B]/dt ≈ -k 1 [A] eq (μ B -μ A )/k B T Comparing the linear response equation with the linear version of the rate law we find L = k 1 [A] eq V/k B

Rate Laws Represent the Most Complete Description of Chemical Kinetics To 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