Unformatted text preview: [ A]0
time
L Grondahl CHEM2056 1 k 1
k1 k 1 Perturbation such as change in temperature (Tjump
method), pressure, or electric field induced.
Remember, the equilibrium constant is temperature
dependent! H 0 1 S 0
ln K R T R Possible to raise temperature a few degrees in less than
107 s!
Even a small change in T allows for useful signal to be
obtained
L Grondahl CHEM2056 Relaxation Methods Relaxation Kinetics If perturbation is small, first order
kinetic to new equilibrium is
obtained. A T1 (initial temperature) The rate constants obtained are
for T2 A k1 Initially:
[A]eq, T1
After perturbation: [A]eq,T2 + At new equilibrium:
[A]eq, T2 k1 Relaxation time: = k1([A]eq + ) – k1([P]eq  ) = k1[A]eq + k1 – k1[P]eq + k1 = (k1 + k1) [P]eq, T1
[P]eq, T2  [P]eq, T2 A k1
k1 P The differential rate law is a small perturbation! Relaxation kinetics The integrated rate law Obtain a first order rate equation. Plotting ln vs. time will give a linear
plot where the sum of the rate constants can be obtained from the
slope. A k1
k1 2P Reaction is perturbed such that A shifts by , at any time:
2
[A] = [A]eq + [ P]eq
k
K
1
k1[A]eq = k1[P]eq2
[P] = [P]eq  2 [ A]eq L Grondahl CHEM2056 d
d [ A] dt
dt L Grondahl CHEM2056 τ, 1 = k1 + k1 ln (k1 k1 )t 0 Because [A] = [A]eq + = k1[A] – k1[P] P Relaxation Kinetics
d
= (k1 + k1) dt At equilibrium R is zero; k1 [A]eq = k1 [P]eq d
d [ A] dt
dt L Grondahl CHEM2056 P Degree of perturbation from new equilibrium: [A] = [A]eq + [P] = [P]eq  T2 (temperature after
perturbation) k1 k1 k 1 d
d [ A] = k1([A]eq + ) – k1([P]eq  2 )2
dt
dt
d  4k1 2 + 4k1[P]eq + k1 dt If we neglect of higherorder terms i.e. 2 we get: 1 = k1 + 4k1[P]eq
L Grondahl CHEM2056 is a small perturbation! Relationship for relaxation time expression
aA + bB k1
k1 pP + qQ
a) If we neglect of higherorder terms in we get: 1 = k1([A]eq)a([B]eq)b(…)(a2/[A]eq + b2/[B]eq + …) + k1([P]eq)p([Q]eq)q(…)(p2/[P]eq + q2/[Q]eq + …) A+B Determination of the individual rate
constants k1
k1 If the equilibrium constant is known, k1 and k‐1 can be found by solving two simultaneous equations • An example of such a use is for determination of the rate constants for water dissociation
Section 18.12 b) If the equilibrium constant in not known, values for the rate constants can be found can sometimes be found but it depends on the reaction stoichiometry P k1
K
k 1 L Grondahl CHEM2056 Problem a) L Grondahl CHEM2056 Solution The equilibrium A
B + C is subjected to a temperature
jump. The measured relaxation time is 3.0 s. The equilibrium
constant for the system is 2.0×1016 M. The equilibrium
concentrations of B and C are both 2.0×104 M. Calculate the
rate constants for the firstorder forward and secondorder
reverse reaction. Try problems 1+2 on ‘Practice problem sheet week 6’.
L Grondahl CHEM2056 L Grondahl CHEM2056 Problem b) Solution The kinetics for complexation of lutetium(III) with anthranilante ion was studied by the T‐jump method. Lu3+ + An k1
k1 LuAn2+ [Lu3+]/...
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 Three '14
 Mole, Kinetics, ea, Grondahl CHEM2056

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