Chem Ch 28/#3 Today’s To Do List Relaxation Methods & Fast Reactions Temperature Dependence Transition-State Theory

Reversible Reactions l A = B k 1 = forward reaction k -1 = reverse reaction l At equilibrium: -d[A]/dt = d[B]/dt = 0 Rate forward = k 1 [A] Rate reverse = k -1 [B] k 1 [A] eq = k -1 [B] eq k 1 /k -1 = [B] eq /[A] eq = K eq

Reaching Equilibrium

The Mixing Problem with Fast Reactions l Consider: H + (aq) + OH - (aq)  H 2 O(aq) with k 1 = 1.4 x dm 3 /mol-s at 298 K l Calculate t 1/2 when [H + ] 0 = [OH - ] 0 = 1 x mol/dm 3 Recall: K w = [H + ] eq [OH - ] eq = 1 x nd -order Reaction: t 1/2 = 1/(k 1 [A] 0 ) = 1/(1.4x10 11 )(1x10 -5 ) t 1/2 = 1 x s << s (mixing time)

Relaxation Methods l Start with a system at equilibrium. l Perturb the system to knock it out of equilibr. T-jump P-jump pH- and pOH-jump l Measure time necessary to relax to new equilibr. state. l k 1 and k -1 are related to this relaxa. time (  )

T-Jump Relaxation l Relaxation processes tend to decay exponentially with time:  x =  x 0 e -t/  where  = relaxation time = 1/e of the time for a system to decay to its new equilibrium state after a “shock” such as a sudden  T. If x = [B], then  [B] = the change in [B] as a reaction approaches its new equilibr.  [B] =  [B] 0 e -t/ 

l  is uniquely related to k 1 and k -1 For A  B  = 1/(k 1 + k -1 ) For A + B  P  = 1/{k 1 ([A] e + [B] e ) + k -1 } l Plot ln [B] vs t & measure slope to find k’s.

Relaxation for A  B

Some Examples l Ionic aqueous reactions are fast! l H + + Ac -  Hac k 1 =3.5 x dm 3 /mol-s l H + + NH 3 +  NH 4 + k 1 = 4.3 x 10 10

T-Dependence of k: The Arrhenius Equation l k carries the T-dependence of the rate law. l Most common is an exponential growth: k = A e -Ea/RT (The Arrhenius Eq.) ln k = ln A – E a /RT A = pre-exponential factor E a = Activation Energy Plot of ln k vs 1/T will be linear with slope –E a /R and intercept ln A.

Reaction Energy Diagram

2HI(g)  H 2 (g) + I 2 (g) E a = 184 kJ/mol

Transition-State Theory l A + B  P dP/dt = k[A][B] l Assume an initial equilibr - l A + B  AB ‡  P AB ‡ = activated complex l K ‡ = [AB ‡ ]/[A][B]

A + B  AB ‡  P l An alternate rate in terms of 2 nd step: dP/dt = c [AB ‡ ] c = freq. with which complex crosses barrier max. l Combining: dP/dt = k[A][B] = c [AB ‡ ] = c [A][B] K ‡ k = c K ‡ Let c  = {k B T/(2  m ‡ )} 1/2

Continued l Substituting: k = c K ‡ = (k B T/h) K ‡ l From thermo:  ‡ G o = -RT ln K ‡ l K ‡ = e -  ‡Gº/RT l k = (k B T/h) e -  ‡Gº/RT l But  ‡ G o =  ‡ H o – T  ‡ S o l k = (k B T/h) e  ‡Sº/R e -  ‡Hº/RT

Relating to E a l Comparing with experimental: k = A e -Ea/RT E a =  ‡ H o + RT Thus  ‡ H o can be obtained from empirical data, then  ‡ S o from  ‡ G o =  ‡ H o – T  ‡ S o A = (e 2 k B T/h) e  ‡Sº/R l Thus A (through  ‡Sº) indicates relative structures of reactants & activated complex.

Next Time Start Chapter 29: Reaction Mechanisms Elementary Reactions Molecularity Detailed Balance