Chem 300 - Ch 28/#3 Today’s To Do List

• Relaxation Methods & Fast Reactions• Temperature Dependence• Transition-State Theory

Reversible Reactions

A = B • k1 = forward reaction• k-1 = reverse reaction

At equilibrium: -d[A]/dt = d[B]/dt = 0• Rate forward = k1[A] • Rate reverse = k-1[B]• k1[A]eq = k-1[B]eq

• k1/k-1 = [B]eq/[A]eq = Keq

Reaching Equilibrium

The Mixing Problem with Fast Reactions

Consider: H+(aq) + OH-(aq) H2O(aq)

• with k1 = 1.4 x 1011 dm3/mol-s at 298 K

Calculate t1/2

• when [H+]0 = [OH-]0 = 1 x 10-7 mol/dm3

• Recall: Kw = [H+]eq[OH-]eq = 1 x 10-14

• 2nd-order Reaction:

• t1/2 = 1/(k1[A]0) = 1/(1.4x1011)(1x10-5)

• t1/2 = 1 x 10-5 s << 10-3 s (mixing time)

Relaxation Methods

Start with a system at equilibrium. Perturb the system to knock it out of equilibr.

• T-jump• P-jump• pH- and pOH-jump

Measure time necessary to relax to new equilibr. state.

k1 and k-1 are related to this relaxa. time ()

T-Jump Relaxation

Relaxation processes tend to decay exponentially with time:• x = x0 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/

[B] = [B]0 e-t/

is uniquely related to k1 and k-1

• For A B = 1/(k1 + k-1)

• For A + B P = 1/{k1([A]e + [B]e) + k-1}

Plot ln [B] vs t & measure slope to find k’s.

Relaxation for A B

Some Examples

Ionic aqueous reactions are fast! H+ + Ac-Hac k1=3.5 x 1010 dm3/mol-s

H+ + NH3+ NH4

+ k1 = 4.3 x 1010

T-Dependence of k:The Arrhenius Equation

k carries the T-dependence of the rate law. Most common is an exponential growth:

• k = A e-Ea/RT (The Arrhenius Eq.)

• ln k = ln A – Ea/RT• A = pre-exponential factor

• Ea = Activation Energy

• Plot of ln k vs 1/T will be linear with slope –Ea/R and intercept ln A.

Reaction Energy Diagram

2HI(g) H2(g) + I2(g)Ea = 184 kJ/mol

Transition-State Theory

A + B P• dP/dt = k[A][B]

Assume an initial equilibr - A + B AB‡ P

• AB‡ = activated complex K‡ = [AB‡]/[A][B]

A + B AB‡ P

An alternate rate in terms of 2nd step:• dP/dt = c[AB‡]

c = freq. with which complex crosses barrier max.

Combining:• dP/dt = k[A][B] = c[AB‡] = c[A][B] K‡

• k = c K‡

• Let c <uac> = {kBT/(2m‡)}1/2

Continued

Substituting:• k = c K‡ = (kBT/h) K‡

From thermo: ‡Go = -RT ln K‡

K‡ = e-‡Gº/RT

k = (kBT/h) e-‡Gº/RT

But ‡Go = ‡Ho – T ‡So k = (kBT/h) e‡Sº/R e-‡Hº/RT

Relating to Ea

Comparing with experimental:• k = A e-Ea/RT

• Ea = ‡Ho + RT• Thus ‡Ho can be obtained from empirical data,

then ‡So from ‡Go = ‡Ho – T ‡So

• A = (e2kBT/h) e‡Sº/R

Thus A (through ‡Sº) indicates relative structures of reactants & activated complex.

Next Time

• Start Chapter 29: Reaction Mechanisms• Elementary Reactions• Molecularity• Detailed Balance