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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