Kinetics (Topics 6,7,9,11,13) Flashcards
Rate=
k [A]^n[B]^b = -d[A]/dt = -delta[A]/delta T = d[P]/dt
Concentration
Number of reactant molecules in a volume
Order of reaction
How the reaction rate depends on the concentration of each reactant
Overall- sum of orders of each
Molecularity
The number of molecules involved in the elementary step
Where aA + bB —> cC + dD
Rate =
-1/a d[A]/dt = -1/b d[B]/dt = 1/c dC/dt = 1/d d[D]/dt
Concentration [A] at given time is
[A]0 e^-kt
Integrated rate laws: 1st order
Differential equation: -d[A]/dt = k[A]
Separate variables: d[A]/[A] = -k dt
Integrate between [A]0 and [A] (1/[A] d[A] ) = integrate between 0 and t (-k dt)
Find antiderivatives: [ln[A]] between [A]0 and [A] = -k[t] between 0 and t
Evaluate antiderivatives at boundaries: ln [A] - ln [A]0 = -kt
Rearrange for [A] to find function f(t): [A] = f(t) : [A] = [A]0 e^-kt
1st order reaction graph plot
ln [A] vs time
k = -slope
Integrated rate laws: 2nd order
Differential equation: -1/2 d[A]/dt = k[A][A]
Separate variables: d[A]/[A]^2= -2k dt
Integrate between [A]0 and [A] (1/[A]^2 d[A] ) = -2k integrate between 0 and t (dt)
Find antiderivatives: [-1/[A]] between [A]0 and [A] = -2k[t] between 0 and t
Evaluate antiderivatives at boundaries: 1/[A] - 1[A]0 = 2kt
Rearrange for [A] to find function f(t): [A] = f(t) : [A] = [A]0 / (1 + 2[A]0 kt)
2nd order reaction graph
A + A —> B
A + B —> C
Plot 1/[A] vs time
k = slope/2
Plot ln([B][A]0/[A][B]0) vs time
k = slope/([B]0-[A]0)
Integrated rate laws: zero order
Differential equation: -d[A]/dt = k
Separate variables: d[A] = -k dt
Integrate between [A]0 and [A] ( d[A] ) = integrate between 0 and t (-k dt)
Find antiderivatives: [[A]] between [A]0 and [A] = -k[t] between 0 and t
Evaluate antiderivatives at boundaries: [A] - [A]0 = -kt
Rearrange for [A] to find function f(t): [A] = f(t) : [A] = -kt + [A]0
Zero order graph
Plot [A] vs time
k = -slope
Half life of 1st order reaction calculations
t1/2 / [A]0: 1/2 and 0 are subscript
1/2 [A]0 = [A]0 e^-kt1/2
ln(1/2) = -kt1/2
t1/2 = -ln(1/2)/k = ln2/k
Half life 2nd order
t1/2 / [A]0: 1/2 and 0 are subscript
1/2 [A]0 = [A]0 / (1+2[A]0 kt1/2)
1/2 = 1 / (2[A]0 kt1/2)
1 = [A]0 kt1/2
t1/2 = 1 / [A]0k
Half life zero order
t1/2 / [A]0: 1/2 and 0 are subscript
1/2 [A]0 = -kt1/2 + [A]0
kt1/2 = 1/2 [A]0
t1/2 = [A]0 / 2k
k =
Ae^-Ea/RT
lnk =
Graph plot
lnA- Ea/RT
-Ea/R (1/T) + lnA
lnk vs 1/T
Slope = -Ea/R
Kinetic Theory of Gases
Gas molecules/atoms are in constant random motion
They collide with each other and the wall of the container
Kinetic theory of gases assumptions
Total volume of gas particles negligible compared to volume of container
Collisions are elastic
No attractive or repulsive forces between particles
Number of particles vast so statistical treatment can be applied
Mu =
mAmB/mA+mB in kg
c bar rel =
Root 2 c bar
Sigma =
Pi d^2
Collision distance (d) =
r1 + r2
[J] =
NJ/NA
pJ / kB T NA
Rate = (2 equations)
Z (collisions of molecules/time/volume) / NA (avogardos constant 6.022 x 10^23 mol^-1)
dKt/dt = k’(k infinity - kt)^m
Integrate between k0 and k1 1/k infinity - kt dK = k’ integrate between 0 and t dt
ln (K infinity - Kt)/K infinity = -k’ t
Steric factor:p =
A (pre exponential factor) / Z (collision rate constant)
3 conditions of experimental determinations
Rapid mixing of reactants at time 0
Ability to measure concentration of reactants or products as function of time after initiation
Accurate control and determination of temperature
2 general types of experimental methods
Quenching (quench reaction + analyse reaction mixture
In situ (measure concentrations in real time)
What is quenching
Mix reactants rapidly
Extract aliquots from the reaction mixture at regular intervals + measure their composition
Ways to to achieve quenching
Large dilution of aliquot to reduce concentration + slow reaction
Add co-reagent which adds rapidly with one of the reactants to remove it completely and prevent further reaction
Rapid lowering of temperature to slow reaction
Mixing and quenching deficiencies
Mixing takes fairly long time
Rapid quenching is difficult and introduces error in the measurement of reactant or product concentrations
In Situ measurements
Composition measured in reaction mixture without extracting aliquot e.g. monitoring changes in pressure or use UV/Vis spectroscopy to monitor rate and progress of reaction
Polarimetry is an experimental method for
Optically active species
Conductometry is an experimental method for
Charged species
pH-metry is an experimental method for
Acidic/basic species
Electrode potentiometry is an experimental method for
Oxidation-reduction reactions
In situ limitations
Only works with reactions not too fast
Half life must be > ca. 50s / k’ < ca.0.014 s^-1
K infinity not straight forward to measure
Continuous flow + Stopped-flow methods
Continuous- Rapid turbulent mixing of 2 reagents in a flow system
Reactants slow rapidly down an outlet tube and reaction monitored at varying distances down the tube to progress the reaction at different times
Can be applied to gas or liquid phase
Stopped- uses advantage of flow method while avoiding large volumes of reagent issue
Reactants mixed rapidly and flow into observation chamber pushing a piston up to stop
Once stop is reached, reagent flows stopped and the reaction monitored within already mixed
volume
Both have light source
Beer-Lambert’s Law can be used to probe reaction (Stopped)
Continuous flow method disadvantage
Large volume of reactant used rapidly
Detectors in continuous flow and stopped-flow methods
Continuous- move detector down stream to observe time dependence
Stopped- detector stays in same position to observe time dependence
Flash photolysis is used to…
It works by…
It requires…
Study reactions involving highly reactive radicals
Creating radicals by short light impulse (fixes issue of initial mixing)
Spectrophotometer
Femtochemistry uses…
Test pulses of laser light of 10^-15 s
Cis-trans isomerism in rhodopsin in the retina
Can look at transition states- “activated complexes”
d[B]/dt =
kf[A] - kb[B]
Dynamic equilibrium
forwards and backwards reaction happen at equal rate. Molecules still reacting but no overall change in concentration of reactants or products
d[B]/dt = kf[A]eq - kb[B]eq = 0
kf[A]eq = kb[B]eq
[B]eq/[A]eq = kf/kb
(subscript f/b)
kf/kb =
(subscript f/b)
K
Differentiate lnK=ln(kf/kb)
What does this equal?
(subscript f/b)
=lnkf - lnkb
dlnK / dT = d lnkf / dT - d lnkb / dT
van’t Hoff equation = delta H/RT^2
(subscript f/b)
kf = Af e ^ -(Ef^act)/RT
kb = Ab e ^-(Eb^act)/RT
(subscript f/b)
Take natural logs and differentiate
lnkf = lnAf - Ef^act / RT
lnkb = lnAb - Eb^act / RT
d lnkf / dT = 0 + (Ef^act) / RT^2
d lnkb / dT = 0 + (Eb^act) / RT^2
(subscript f/b)
d lnK / dT = delta H / RT^2
So …
And delta H =
= lnkf / dT - d lnkb / dT
= (Ef^act) / RT^2 - (Eb^act) / RT^2
Ef^act - Eb^act
(subscript f/b)
lnK =
-delta H / RT + delta S/R
= ln (Af/Ab) + delta S/R - 1/RT (Ef^act - Eb^act)
(subscript f/b)
Elementary reactions
1 step
molecular level
molecularity is number of molecules involved in the step
rate law can be written from reaction stoichiometry
Complex reactions have
More than 1 elementary step through intermediates
In complex reactions, rate law can reflect
stoichiometry
Kinetic vs Thermodynamic product
K - formed quicker, lower activation energy
T - most energetically favourable, lowest energy at equilibrium
[A]t in first order of kinetics =
[A]0 e^-k1t
d[C] / dt = k1 [A]0 e^-k1t
Integrate this
[C]t = k1 [A]0 Integrate between 0 and t e^-k1t
= (1 - e^-k1t) [A]0
How to simplify rate expressions when the first step is slow
Steady State Approximation
How to simplify rate expressions when the first step is fast
Pre-Equilibrium Approximation
Where A + B form I in an equilibrium reaction (k1 / k-1) and I forms P (k2)
Use pre-equilibrium approximation to write an equation for K and rate
K = [I] / [A][B] = k1 / k-1
rate = d [P] / dt = k2 [I] =
k2 K [A][B]
Catalyst increases the
Rate constant
The catalyst causes the formation of…
Meaning a decrease in…
A transition state with lower energy
Activation energy
Ratio of catalysed to uncatalysed rate constant
k (catalyst) / k(no catalyst) = e^-Ea/RT (catalyst) / e^-Ea/RT (no catalyst)
3 types of catalyst
homogeneous
heterogeneous
enzymatic
2 spontaneous processes that occur with enzyme catalysis
Reverse binding: E + S <=> ES
ka / k’a : forwards + backwards rate constants
Enzyme, Substrate, Enzyme - Substrate complex
Catalysis: ES –> P + E
kb : rate constant
Product, Enzyme
E + S <=> ES
ka / k’a : forwards + backwards rate constants
Catalysis: ES –> P + E
kb : rate constant
Calculate d[ES] / dt
ka [E][S] - ka’ [ES] - kb [ES] = 0
[ES] = (ka / k’a + kb )[E][S]
KM = ([E]0 - [ES])[S] / [ES]
[ES] = [E]0[S] / KM + [S]
What is KM (Michaelis constant) in terms of [E],[S],[ES] and in k (2 equations) and rearrange to make [ES] the subject
[E][S] / [ES] = (k’a + kb / ka)
KM = ([E]0 - [ES])[S] / [ES]
[ES] = [E]0[S] / KM + [S]
[E]0 (initial enzyme molecules) =
[E] + [ES]
Michaelis-Menton rate law
kb [E]0 [S] / KM + [S]
= r max [S] / KM + [S]
KM «_space;[S]
so rate = r max
How would you plot a graph for Michaelis-Menton rate law
1 / r0 (y) vs 1 / [S]0 (x)
Slope = KM / r max
y intercept = 1 / r max
x intercept = -1 / KM
r can interchange with V
Lineweaver - Burk plot
The Maxwell–Boltzmann distribution describes
the speed distribution of gas molecules
