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

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

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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
Q

Rate = (2 equations)

A

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

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26
Q

Steric factor:p =

A

A (pre exponential factor) / Z (collision rate constant)

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27
Q

3 conditions of experimental determinations

A

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

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28
Q

2 general types of experimental methods

A

Quenching (quench reaction + analyse reaction mixture
In situ (measure concentrations in real time)

29
Q

What is quenching

A

Mix reactants rapidly
Extract aliquots from the reaction mixture at regular intervals + measure their composition

30
Q

Ways to to achieve quenching

A

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
Q

Mixing and quenching deficiencies

A

Mixing takes fairly long time
Rapid quenching is difficult and introduces error in the measurement of reactant or product concentrations

32
Q

In Situ measurements

A

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
Q

Polarimetry is an experimental method for

A

Optically active species

34
Q

Conductometry is an experimental method for

A

Charged species

35
Q

pH-metry is an experimental method for

A

Acidic/basic species

36
Q

Electrode potentiometry is an experimental method for

A

Oxidation-reduction reactions

37
Q

In situ limitations

A

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
Q

Continuous flow + Stopped-flow methods

A

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
Q

Continuous flow method disadvantage

A

Large volume of reactant used rapidly

40
Q

Detectors in continuous flow and stopped-flow methods

A

Continuous- move detector down stream to observe time dependence
Stopped- detector stays in same position to observe time dependence

41
Q

Flash photolysis is used to…
It works by…
It requires…

A

Study reactions involving highly reactive radicals
Creating radicals by short light impulse (fixes issue of initial mixing)
Spectrophotometer

42
Q

Femtochemistry uses…

A

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
Q

d[B]/dt =

A

kf[A] - kb[B]

44
Q

Dynamic equilibrium

A

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
Q

kf/kb =
(subscript f/b)

A

K

46
Q

Differentiate lnK=ln(kf/kb)
What does this equal?
(subscript f/b)

A

=lnkf - lnkb
dlnK / dT = d lnkf / dT - d lnkb / dT
van’t Hoff equation = delta H/RT^2
(subscript f/b)

47
Q

kf = Af e ^ -(Ef^act)/RT
kb = Ab e ^-(Eb^act)/RT
(subscript f/b)
Take natural logs and differentiate

A

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
Q

d lnK / dT = delta H / RT^2
So …
And delta H =

A

= lnkf / dT - d lnkb / dT
= (Ef^act) / RT^2 - (Eb^act) / RT^2
Ef^act - Eb^act
(subscript f/b)

49
Q

lnK =

A

-delta H / RT + delta S/R
= ln (Af/Ab) + delta S/R - 1/RT (Ef^act - Eb^act)
(subscript f/b)

50
Q

Elementary reactions

A

1 step
molecular level
molecularity is number of molecules involved in the step
rate law can be written from reaction stoichiometry

51
Q

Complex reactions have

A

More than 1 elementary step through intermediates

52
Q

In complex reactions, rate law can reflect

A

stoichiometry

53
Q

Kinetic vs Thermodynamic product

A

K - formed quicker, lower activation energy
T - most energetically favourable, lowest energy at equilibrium

54
Q

[A]t in first order of kinetics =

A

[A]0 e^-k1t

55
Q

d[C] / dt = k1 [A]0 e^-k1t
Integrate this

A

[C]t = k1 [A]0 Integrate between 0 and t e^-k1t
= (1 - e^-k1t) [A]0

56
Q

How to simplify rate expressions when the first step is slow

A

Steady State Approximation

57
Q

How to simplify rate expressions when the first step is fast

A

Pre-Equilibrium Approximation

58
Q

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

A

K = [I] / [A][B] = k1 / k-1
rate = d [P] / dt = k2 [I] =
k2 K [A][B]

59
Q

Catalyst increases the

A

Rate constant

60
Q

The catalyst causes the formation of…
Meaning a decrease in…

A

A transition state with lower energy
Activation energy

61
Q

Ratio of catalysed to uncatalysed rate constant

A

k (catalyst) / k(no catalyst) = e^-Ea/RT (catalyst) / e^-Ea/RT (no catalyst)

62
Q

3 types of catalyst

A

homogeneous
heterogeneous
enzymatic

63
Q

2 spontaneous processes that occur with enzyme catalysis

A

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
Q

E + S <=> ES
ka / k’a : forwards + backwards rate constants
Catalysis: ES –> P + E
kb : rate constant
Calculate d[ES] / dt

A

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
Q

What is KM (Michaelis constant) in terms of [E],[S],[ES] and in k (2 equations) and rearrange to make [ES] the subject

A

[E][S] / [ES] = (k’a + kb / ka)
KM = ([E]0 - [ES])[S] / [ES]
[ES] = [E]0[S] / KM + [S]

66
Q

[E]0 (initial enzyme molecules) =

A

[E] + [ES]

67
Q

Michaelis-Menton rate law

A

kb [E]0 [S] / KM + [S]
= r max [S] / KM + [S]
KM &laquo_space;[S]
so rate = r max

68
Q

How would you plot a graph for Michaelis-Menton rate law

A

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
Q

The Maxwell–Boltzmann distribution describes

A

the speed distribution of gas molecules

Year 1 Physical Chemistry (3 decks)

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