HO_CHEM2056_kinetics_2013_L4-6

Eigen 1967 nobel prize in chemistry applicable only

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: [ A]0 time L Grondahl CHEM2056 1 k 1 k1 k 1 Perturbation such as change in temperature (T-jump 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 10-7 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 k-1 Initially: [A]eq, T1 After perturbation: [A]eq,T2 + At new equilibrium: [A]eq, T2 k-1 Relaxation time: = k1([A]eq + ) – k-1([P]eq - ) = k1[A]eq + k1 – k-1[P]eq + k-1 = (k1 + k-1) [P]eq, T1 [P]eq, T2 - [P]eq, T2 A k1 k-1 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 k-1 2P Reaction is perturbed such that A shifts by , at any time: 2 [A] = [A]eq + [ P]eq k K 1 k1[A]eq = k-1[P]eq2 [P] = [P]eq - 2 [ A]eq L Grondahl CHEM2056 d d [ A] dt dt L Grondahl CHEM2056 τ, --1 = k1 + k-1 ln (k1 k1 )t 0 Because [A] = [A]eq + = k1[A] – k-1[P] P Relaxation Kinetics d = (k1 + k-1) dt At equilibrium R is zero; k1 [A]eq = k-1 [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 + ) – k-1([P]eq - 2 )2 dt dt d - 4k-1 2 + 4k-1[P]eq + k1 dt If we neglect of higher-order terms i.e. 2 we get: --1 = k1 + 4k-1[P]eq L Grondahl CHEM2056 is a small perturbation! Relationship for relaxation time expression aA + bB k1 k-1 pP + qQ a) If we neglect of higher-order terms in we get: --1 = k1([A]eq)a([B]eq)b(…)(a2/[A]eq + b2/[B]eq + …) + k-1([P]eq)p([Q]eq)q(…)(p2/[P]eq + q2/[Q]eq + …) A+B Determination of the individual rate constants k1 k-1 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 2.0×10-16 M. The equilibrium concentrations of B and C are both 2.0×10-4 M. Calculate the rate constants for the first-order forward and second-order 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 k-1 LuAn2+ [Lu3+]/...
View Full Document

This document was uploaded on 04/10/2014.

Ask a homework question - tutors are online