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Year 1 Physical Chemistry > Kinetics (Topics 6,7,9,11,13) > Flashcards

Kinetics (Topics 6,7,9,11,13) Flashcards

1
Q

Rate=

A

k [A]^n[B]^b = -d[A]/dt = -delta[A]/delta T = d[P]/dt

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2
Q

Concentration

A

Number of reactant molecules in a volume

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3
Q

Order of reaction

A

How the reaction rate depends on the concentration of each reactant
Overall- sum of orders of each

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4
Q

Molecularity

A

The number of molecules involved in the elementary step

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5
Q

Where aA + bB —> cC + dD
Rate =

A

-1/a d[A]/dt = -1/b d[B]/dt = 1/c dC/dt = 1/d d[D]/dt

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6
Q

Concentration [A] at given time is

A

[A]0 e^-kt

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7
Q

Integrated rate laws: 1st order

A

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

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8
Q

1st order reaction graph plot

A

ln [A] vs time
k = -slope

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9
Q

Integrated rate laws: 2nd order

A

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)

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10
Q

2nd order reaction graph
A + A —> B
A + B —> C

A

Plot 1/[A] vs time
k = slope/2

Plot ln([B][A]0/[A][B]0) vs time
k = slope/([B]0-[A]0)

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11
Q

Integrated rate laws: zero order

A

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

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12
Q

Zero order graph

A

Plot [A] vs time
k = -slope

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13
Q

Half life of 1st order reaction calculations

A

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

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14
Q

Half life 2nd order

A

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

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15
Q

Half life zero order

A

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

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16
Q

k =

A

Ae^-Ea/RT

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17
Q

lnk =
Graph plot

A

lnA- Ea/RT
-Ea/R (1/T) + lnA
lnk vs 1/T
Slope = -Ea/R

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18
Q

Kinetic Theory of Gases

A

Gas molecules/atoms are in constant random motion
They collide with each other and the wall of the container

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19
Q

Kinetic theory of gases assumptions

A

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

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20
Q

Mu =

A

mAmB/mA+mB in kg

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21
Q

c bar rel =

A

Root 2 c bar

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22
Q

Sigma =

A

Pi d^2

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23
Q

Collision distance (d) =

A

r1 + r2

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24
Q

[J] =

A

NJ/NA
pJ / kB T NA

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25
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
26
Steric factor:p =
A (pre exponential factor) / Z (collision rate constant)
27
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
28
2 general types of experimental methods
Quenching (quench reaction + analyse reaction mixture In situ (measure concentrations in real time)
29
What is quenching
Mix reactants rapidly Extract aliquots from the reaction mixture at regular intervals + measure their composition
30
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
31
Mixing and quenching deficiencies
Mixing takes fairly long time Rapid quenching is difficult and introduces error in the measurement of reactant or product concentrations
32
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
33
Polarimetry is an experimental method for
Optically active species
34
Conductometry is an experimental method for
Charged species
35
pH-metry is an experimental method for
Acidic/basic species
36
Electrode potentiometry is an experimental method for
Oxidation-reduction reactions
37
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
38
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)
39
Continuous flow method disadvantage
Large volume of reactant used rapidly
40
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
41
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
42
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”
43
d[B]/dt =
kf[A] - kb[B]
44
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)
45
kf/kb = (subscript f/b)
K
46
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)
47
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)
48
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)
49
lnK =
-delta H / RT + delta S/R = ln (Af/Ab) + delta S/R - 1/RT (Ef^act - Eb^act) (subscript f/b)
50
Elementary reactions
1 step molecular level molecularity is number of molecules involved in the step rate law can be written from reaction stoichiometry
51
Complex reactions have
More than 1 elementary step through intermediates
52
In complex reactions, rate law can reflect
stoichiometry
53
Kinetic vs Thermodynamic product
K - formed quicker, lower activation energy T - most energetically favourable, lowest energy at equilibrium
54
[A]t in first order of kinetics =
[A]0 e^-k1t
55
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
56
How to simplify rate expressions when the first step is slow
Steady State Approximation
57
How to simplify rate expressions when the first step is fast
Pre-Equilibrium Approximation
58
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]
59
Catalyst increases the
Rate constant
60
The catalyst causes the formation of... Meaning a decrease in...
A transition state with lower energy Activation energy
61
Ratio of catalysed to uncatalysed rate constant
k (catalyst) / k(no catalyst) = e^-Ea/RT (catalyst) / e^-Ea/RT (no catalyst)
62
3 types of catalyst
homogeneous heterogeneous enzymatic
63
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
64
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]
65
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]
66
[E]0 (initial enzyme molecules) =
[E] + [ES]
67
Michaelis-Menton rate law
kb [E]0 [S] / KM + [S] = r max [S] / KM + [S] KM << [S] so rate = r max
68
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
69
The Maxwell–Boltzmann distribution describes
the speed distribution of gas molecules

Year 1 Physical Chemistry (3 decks)

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