(A) Energy Relationships, Temperature Dependence for an Equilibrium Wher

Home , Arrhenius equation, Eyring equation, Transition state theory

3. Rate and equilibrium constants and free energies. (a) Energy relationships, temperature dependence For an equilibrium K A B []B K = []A

∆°=−GRTKln( ) where ∆G° is the standard free energy of the reaction and

∆G° = ∆H° - T∆S°.

∆G° can be positive (unfavorable equilibrium) or negative (favorable equilibrium.

This means that ∆G° and K are readily translated: they describe the same essential idea but in very different languages. At 25°C, 1.366 kcal or 4.72 kJ in ∆G° corresponds to a factor of 10 change in K. − RT ln(K) = ∆H o − T∆S o ∆S o ∆H o ln(K) = − R RT 1 ln(K) ∝ T (van’t Hoff equation)

Rate constants also vary with temperature; it is common to raise the temperature of a reaction to speed it up, or lower the temperature to slow it down. A useful (though approximate) rule of thumb is that for reactions which occur in humanly reasonable times at humanly reasonable temperatures (minutes to hours near 25°C) the rate constant doubles for a 10° increase in temperature and increases 10 fold for a 25°C increase intemperatures. Rate constants may also be related to energy changes, either by the Arrhenius equation (an empirical relation by analogy to the van’t Hoff equation) kAe= −ERTa /

where Ea = (Arrhenius) activation energy and A "the pre-exponential factor",

or the Eyring equation (from transition state theory):

kT ‡ ke= B −∆GRT/ h where ∆G‡ is the free energy activation and ∆G‡ = ∆H‡ -T∆S‡ 18

∆G‡ is always positive because there is always a barrier (sometimes small) to reaction.

∆G‡ is, of course, calculated directly from k and the Eyring equation.

‡ kTB ∆=GRTln − RTk ln( ) h  That is to say that one can readily translate between the two ways of describing the rate of a reaction: ∆G‡ or k.

By rearranging the Eyring equation and taking logarithms:

‡‡ kkSH B ∆∆ 1 ln=+− ln    ThRRT    we obtain a strictly linear realtionship which can be plotted. The Eyring equation as normally written

‡‡ kTB ∆∆ S H 1 lnk =+− ln   hRRT 

is not strictly linear in (1/T)

Plotting ln(k/T) vs l/T, the slope of the straight line is -∆H‡/R and hence ∆H‡ = -(slope)R. From the same plot, the intercept is: ‡ kSB ∆ intercept= ln + hR 1 1 1 1 10 1 1 (R = 1.987 cal K- mol- = 8.314 J K- mol- , (kB/h) = 2.083x10 K- s- , ln(kB/h) = 23.760)

∆S‡ is readily calculated either from

‡ kSB ∆ intercept= ln + hR ≠ ∆S = R(intercept – ln(kB/h) ‡ or from ∆H and kexp at a particular temperature

‡ ‡ kexp kHB ∆ ∆=SRln − ln  + ThT ‡ ‡kexp ∆H ‡ -1 -1 ‡ -1 ∆=S 4.576 log − 10.319 + ,∆SincalKmol,∆Hincalmol TT ‡ ‡kexp ∆H ‡ -1 -1 ‡ -1 ∆=S 19.146 log − 43.17 + ,∆SinJKmol,∆HinJmol TT

Alternatively, from the Arrhenius equation

Ea 1 ln(kA )=− ln( )  Ea = activation energy R T A = Arrhenius prexponential factor the slope of a plot of ln k vs l/T gives Ea/R and the intercept gives ln (A)

We can translate between the Eyring and Arrhenius languages, but this is a bit tricky because the Arrhenius equation makes ln (k) linear in (1/T) while the Eyring equation does not. At some temperature T in the middle of the range studied we determine the instantaneous slope ∂ ln(k )

1 ∂  T for both equations: ∂ ln(k ) E Arrhenius: =− a 1 R ∂  T Eyring: here we have to cast the equation in a form with (1/T) as the variable throughout ‡‡ kTB ∆∆ S H 1 lnk =+− ln   hRRT 

becomes ‡‡ kSHB  11∆∆  lnk =−+− ln  ln   hTRRT    and then ∂∆∆+∆ln(kHHRTH ) 1 ‡‡ ‡ =− − =−T − =− 1 1 T RRR ∂  T and by comparison we can say ‡ ‡ Ea = ∆H + RT ∆H = Ea – RT (with T in the middle of the range studied)

‡ and ∆H is readily obtained from Ea (or vice versa). ≠ ‡ Then to translate A into ∆S we substitute the expression for ∆H as a function of Ea into the Eyring equation ‡ kTB ∆ S ERTa − 1 lnk =+− ln   hR RT 

‡ kTB ∆ S Ea =++−ln 1 hR RT ‡ ekB T∆ S Ea =+−ln  hRRT and by comparison ‡ ekB T∆ S A = exp hR

‡ ekB T ∆=SRARln − ln  h ∆=SA‡4.58(log − 13.2),∆S ‡ in cal K -1 mol -1 ∆=SA‡19.16(log − 55.23),∆S ‡ in J K -1 mol -1 (at 25°C)

(b) Reaction coordinate diagrams; the rate-determining step

Armed with rate and equilibrium data we may now draw the familiar "reaction coordinate diagrams", i.e. plots of energy (commonly free energy, ∆G) vs the "extent of the reaction".

Reaction coordinate

The "extent of the reaction" can, if necessary, be defined in terms of something quantitative, e.g. internuclear distances between reacting atoms, but for any but the simples processes this usually requires a switch from one set of distances to another as the reaction proceeds, and hence may merely be clumsy without adding much in the way of clarity or precision. Reaction coordinate 21

diagrams are particularly useful for reactions with more than one step. Recall the expression for a reaction of the type

k1 k2 A B C k -1

A number of situations may arise depending on the relative magnitudes of the k’s.

In (i) kobs = k1k2/k-1 = K1 k2 and in (iv) kobs = k1

In (i) and (iv) we may point to a rate-determining (or rate-limiting or rate- controlling) step. This is step 2 in (i) and step 1 in (iv).

In (ii) and (iii) both steps are partially rate-determining, and the full steady state expression would have to be used.

kk12 kobs = kk−12+

The terms "rate-determining", "rate-limiting", and "rate-controlling" are not always defined the same way by all users. One consistent usage is to define the "rate-determining step" as that with the highest energy transition. state. With such usage, however, one 22 must recognize that other steps may be "partially rate-determining" (e.g. step 1 in (ii) or step 2 in (iii)) . With cases like (ii) and (iii) it is also reasonable to state simply that the reactions have no single rate - determining step. The term has its uses but the user should be aware of the possibility of easily confusing others (and even himself!).

One point:

Though step 2 in (i) is rate-determining kobs ≠ k2, i.e. the rate constant for the overall reaction is often not the same as the rate constant for the rate-determining step.

The mechanism of a reversible reaction is the same, in microscopic detail (except for the direction of the reaction), for the reaction in one direction as in the other under a given set of conditions. In other words, if the mechanism of a reaction in one direction is determined, the mechanism of the reverse reaction is known immediately.

A B

C E.g.

If, for example the process A → B → C is n-times faster than the alternative route A → C (directly) then C → B → A is also n-times faster than C → A.

direct route, A → C

indirect route, A → B → C

the same thing can be shown in a projection onto 23 the A → C reaction coordinate

With transition state theory (and hence with reaction coordinate diagrams) the principle is virtually axiomatic. The relative rates of alternative pathways are decided by the ∆∆G‡’s, the differences in energy between transition states; ∆∆G‡’s are the same regardless of the direction of the reaction. Hence if, for example, 90% of the reaction goes by one route ("X") and 10% by another ("Y"), then the reverse reaction will also proceed 90% by "X" and 10% by "Y".

Note that the principle of microscopic reversibility excludes such processes as

A B or ABor AB

C C C

(where the single arrow implies irreversibility.) In principle all reactions are reversible, but in practical terms if the equilibrium constant for the forward reaction is large and favorable the forward reaction is much faster than the reverse, and the reaction is in effect irreversible. Try to draw an energy diagram for the three schemes shown above; you will find it is impossible (unless perhaps you are Maurits Escher; see his drawing “Ascending and Descending” at: http://home.comcase.net/~eschermc/ {or google on escher})

4. Information from the rate law

(a) Normally, the rate law indicates the composition of the transition state. More specifically, the numerator shows all of the species that must be assembled to form the rate-limiting transition state (and the denominator any species that are discarded before that transition state is reached).

E.g. if Rate = k[A]a[B]b[C]c.... then the empirical formula of the transition state is aA.bB.cC.... another example k [CH] 1 k2 A + B I P k -1[C] under pseudo first order conditions, with [B] >> [A], and CH or C a catalyst, also at fixed concentration,

dP[] k12 [][ B CHk ] rate=== k2[] I [] A = kψ [] A dt k−12[] C+ k

kBCHk12[][ ] IF k-1[C] >> k2, then kψ =  kC−1[] The transition state contains A + B + H H is what is left from (CH - C) C is discarded before the transition state for the k2 step.

IF k-1[C] << k2, then kkBCHψ = 1[][ ] and the transition state for the k1 step includes CH.

Two points:

(i) Exceptions: a) - chain reactions, for instance the polymerization of styrene with AIBN (azoisobutyronitrile) as initiator. The rate law is: ½ dstyrene[]ki ½ rate=− = kp[][] AIBN styrene dt kt

b) - species with unchanged concentrations in the experiments (e.g. solvent, walls of vessel, an unsuspected adventitious catalyst, etc.) do not show up in the rate law.

(ii) The rate law in itself says nothing about the structure of the transition state, only its composition. 25

(b) The rate law in itself gives no information about the mechanism of any pre- equilibrium process(es) (i..e. any step(s) before the rate determining step). It doesn’t tell you how it got there.

(d) A multiple term rate law indicates parallel reactions with transition states of different composition.

dA[] E.g. −=+kA[] kAB [][] dt 12

The two terms refer to separate reactions with different transition states.

(e) If there are fewer components in the rate law (and hence in transition state) than are shown by the stoichiometry, then there is at least one step after the rate-determining step. i.e. if aA + bB + ... → product shows the rate law Rate = k[A]a-1[B]b... there is a further step (at least) after the rate determining step in which the other molecule of A reacts.

E.g. Under some conditions O O CH C CH 3 3 + Br2 CH3 C CH2Br + HBr shows the following rate law

Rate = k CH3 C CH3 i. e. is zero order in Br2, (also I2 reacts at the same rate.)

The rate determining step is enolization O OH CH C CH 3 3 CH3 C CH2

followed by rapid reaction of the enol with Br2 (or I2)-

Another example: 26

H Cl H OH acetone +H2O H3CCH3 H3CCH3

Addition of NaN3 (0.05 M) gives a product consisting of 40% of the alcohol (above) plus 60% of the azide (Ar-CH(N3)-Ar). Except for a small salt effect, the rate is unaffected by the NaN3 in spite of the fact that it is incorporated into the major part of the products. Addition of any salt makes the solvent more polar and this speeds up the reaction. For example if 0.05M NaClO4 is added the rate increases by the same small amount, but there is no change in products.

Mechanism

Cl OH slow H2O Ar CH Ar Ar CH Ar Ar CH Ar fast fast NaN3

N3 Ar CH Ar

Alteration in product - ratios (or composition) without altering the rate is sometimes referred to as

"the rate-product discrepancy" or "the rate-product criterion", and requires a mechanism with an intermediate consumed after the rate-determining step.

Hammond Postulate (George Hammond) “If two states occur consecutively during a reaction process and have nearly the same energy content, their interconversion will involve only a small reorganization of molecular structure.” Corollary: The transition state will be more similar in structure to that species (reactant, product, or intermediate) that is closest in energy to it.

≠≠ The first transition The second state will resemble transition state the corresponding will resemble the final state, B. B corresponding initial state, B

When there is only one step and the overall free energy change is small, all one can say is that there is a central transition state with a structure between that of starting material and product.

If one could get some information about the position of the transition state (more or less than half way along, one could say something about the likely overall free energy change. (Uphill reactions have TS higher and later; downhill reactions have TS lower and earlier.)