Kinetics

Question 1

Define the average rate of reaction

Answer 1

Avr. Rate = Change in conc/change in time = deltaC/deltaT

Question 2

what is the instantaneous rate of reaction

Answer 2

Rate = d(conc.)/dt = derivative of concentration wrt time

i.e. derivative of the line of a conc-time graph

Question 3

What does a +ve and what does a -ve ROR represent

Answer 3

+ve ROR means the conc. of the species is increasing with time

-ve ROR means the conc. of the species is reducing with time

Question 4

what are the units for ROR and what is the shorthand for concs

Answer 4

ROR has units of moldm^-3s^-1 usually

[A] is the shorthand for conc. of whatever is inside the square brackets

Question 5

what is the equation for rate of reaction in terms of the stoichiometric coeffs.
how are these different for products / reactants
use formation of H2O as example

Answer 5

2H2 + O2 —> 2H2O

ROR = r = (1/va) (d[A] / dt)

where va is the stoichiometric coeff
va has -ve sign for reagents, +ve sign for products

e.g. using example
ROR = r = -1/2 d[H2]/dt
d[H2O]/dt = -2 d[O2]/dt

Question 6

what is the general form of the rate equation (for our purposes), what is k, what do we call the dependence on different species

Answer 6

Rate = k[A]^a [B]^b

k = Rate constant or coefficient

a, b are constants, these are powers which give the ‘orders’ wrt different species

Question 7

what is a first order rate equation, what is a second order rate equation, give the units of K in each case

Answer 7

r = k[A]
is a first order rate equation, in this case k has units of s^-1

r = k[A]^2
or
r = k[A][B]
are examples of second order rate equations, the units of k are dm^3mol^-1s^-1

Question 8

give the main three reasons for the importance of rate laws

Answer 8

1) If we know the rate law and the constants in it then we can use it to predict the rate for any set of conditions

2) the form of the rate law can tell us something about the mechanism of the reaction

3) knowing the rate law enables us to sperate the conc. dependence from underlying fundamental effect which is size of rate const.

Question 9

are rate constants temperature dependent, what is usually the trend

Answer 9

Rate constants (and so rates overall) are heavily temperature dependent so ALWAYS state temperature

generally size of rate coefficient increases with temp

Question 10

what is the most common relationship between rate coefficient and temp, give the equation and explain what each of the parts give

Answer 10

• Many rate constants will follow the Arrhenius equation

k = A exp(-Ea / RT)

k = rate constant (variable units)
A = pre-expo factor (same units as k)
Ea = activation energy (Jmol^-1)
T = temp (T)
R = ideal gas const. (J K^-1 mol^-1)

Question 11

what does a graph of ln(k) against 1/T give, what are its key features

Answer 11

ln(k) against 1/T will give a straight line graph of
gradient = -Ea/R
Y-int = ln(A)

Question 12

derive the expression for the equilibrium constants in terms of rate constants for the reaction

Answer 12

Assume we have some equil. of

A+B —REV.—> C+D

assume the forwards direction follows rate law
rate(f) = kf [A] [B]
and the reverse follows
rate(r) = kr[A][B]

At equil. these have the same value as the forwards and backwards reactions have the same rate so

kf [A(eq)] [B(eq)] = kr [B(eq)] [C(eq)]

K = [C(eq)] [D(eq)] / [A(eq)] [B(eq)] = kf / kr

so

K = kf / kr

Question 13

What is the main thing that occurs in all reactions that we must understand to analyse RORs

Answer 13

MOs need to interact:
- this is usually HOMO-LUMO interactions
- Bonds are broken, Bonds are formed

Question 14

What is a potential energy surface, why are they difficult in practice

Answer 14

• a potential energy surface uses the fact that in theory, using MOs, we can calculate the energy of any arbitrary arrangement of molecules
• in principle we can calculate the energy of any given arrangement of A and B as they interact
• we end up with a potential energy surface (PES) which gives the PE of the system as a function of the positions of all the atoms in the system
• in reality this is monumentally difficult to form and usually requires multi-dimensional plots

Question 15

What are some key features of PESs, what can we use them to imagine

Answer 15

we can use the PE surface to imagine the atoms ‘moving’ over it, they start out in a position which corresponds to reactants then move along some path over the surface as they rearrange themselves and end up in a position corresponding to products

the stable molecules, products/reactants exist in potential energy minima

Question 16

which path over the PES is most favoured and why

Answer 16

• when going from reactants to products over the PES, some energy will always have to be put in because the reactants and products exist in potential energy minima
• the route which involves the least expenditure of energy is the most favoured route
• it’s the most favoured route because the energy E which has to be put in in order to reach the energy of the transition state is only present in some cases, the lower this energy is, the more reactions can occur
• we can safely assume as very few molecules have sufficient energy to react, that they all go via the favoured route

Question 17

what is the transition state of a reaction, what are the key points about it

Answer 17

• The transition state is the state of the molecule(s) at the highest energy point on the PES on the route between the reactants and products
• it is not a molecule in the conventional sense because it is very unstable as molecules exist in potential energy minima and intermediates exist on potential energy maxima (unstable equil.) so any slight change and there’s a driving force to remove the intermediate
• they exist for a very small amount of time, in the order of a single molecular vibration

Question 18

what is an intermediate, how is this different to a transition state

Answer 18

an intermediate is a standard molecule but one which exists in a metastable equilibrium between the reactants and products

a transition state is not stable

Question 19

what is the minimum energy pathway from reactants to products called, what can we plot with it, what is the key feature of this plot

Answer 19

• the minimum energy pathway from reactants to products is called the reaction coordinate
• it can be plotted (on X) against potential energy
• this means we have
  Reactants (low energy)
  rising up to maxima (transition state)
  lowering in energy again to products
• the activation energy is the key feature of this plot, Ea is the energy difference between the maxima and the reactants

Question 20

what is collision theory, what are the key concepts we’ll work with, is it accurate?

Answer 20

• Collision theory is a theory about the motions and energies of the particles of gases, its entirely classical
• the theory treats particles as objects where size «distance between them
• the particles have kinetic energy which causes them to move in random directions
• they collide elastically with each other and the walls of the container

Question 21

what is the mean speed of gas particles in collision theory given by

Answer 21

• the energies and speeds of the particles can be given by the Maxwell Distribution
• the mean relative speed (cbar(rel))

cbar(rel) = (8kbT / pi*mu)^1/2

kb = Boltzmann constant
T = temp
mu = reduced mass of system

Question 22

what is mu, the reduced mass of a system of two molecules A and B

Answer 22

mu = (mA*mB) / (mA + mB)

Question 23

what is the mean relative speed of two molecules A and B in collision theory

Answer 23

cbar(rel) = (8kbT / pi*μ)^1/2

Question 24

Assuming the molecules are structureless spheres, state the expressions for collision rate between A and B (according to collision theory), explain each of the variables

Answer 24

Collision rate = Z(AB), collisions m^-3 s^-1

Z(AB) = cAcBpi(rA + rB)^2 * cbar(rel)
= cAcBpi(rA+rB)^2 * (8kbT / pi*mu)^1/2

cA = conc. of A in molecules per m^3
cB = conc. of B in molecules per m^3
rA = radius of A
rB = radius of B
cbar(rel) = mean relative speed

this expression can be simplified by defining the collision cross-section, σ, as
pi*(rA+rB)^2, giving

Z(AB) = cAcBσ(8kbT / pimu)^1/2

Question 25

how can we calculate the number of successful collisions per unit volume per unit time, what assumption do we have to make to do this.

from this how do we deduce an alternative expression for rate of reaction

Answer 25

• if we assume that every molecule with sufficient energy undergoes a successful collision then we can say:
  the number of successful collisions per unit volume per unit time is Z(AB)exp(-Ea/RT)

so number of moles of product formed per unit time = above / Avagadros = rate of reaction, r

r = Z(AB)exp(-Ea/RT)/L

r = (1/L)cAcBσ(8kbT / pi*mu)^1/2 * exp(-Ea/RT)

Question 26

Assuming our reaction is first order wrt A and first order wrt B, how can we link our expression for rate of reaction in terms of successful collisions/Arrhenius to our standard rate eq to deduce k

Answer 26

r = k (cA/L) (cB/L)
AND
r = (1/L)cAcBσ(8kbT / pi*mu)^1/2 * exp(-Ea/RT)

so by comparing we get

k = σ(8kbT /pimu)^1/2 * L * exp(-Ea/RT)

this is our collision theory definition for k

Question 27

Link the collision theory definition for k to the Arrhenius definition for k to get an expression for A

Answer 27

Collision theory is
k = σ(8kbT /pimu)^1/2 * L * exp(-Ea/RT)

Arrhenius is
k = A exp(-Ea/RT)

This gives

A(coll.theory) = σ (8kbT/pimu)^1/2 * L

NOTE: although this does depend on T, the exponential term in the Arrhenius eq. dominates it

Question 28

what is the expression for A(coll.theory), is it accurate?

Answer 28

A(coll.theory) = σ (8kbT/pimu)^1/2 * L

this almost always overestimates A
especially for more complicated reactions
sometimes by many orders of magnitude

Question 29

why does collision theory predict A to be too large, how can we adjust for this

Answer 29

• we have not taken account for sterics
• by modelling the molecules as hard spheres, simple collision theory ignores orientational effects altogether

we can introduce a steric factor, p

p = A(experiment) / A(coll. theory)

Question 30

How are rate laws actually determined experimentally (just as a general method)

Answer 30

• we cannot look at concentration-time graphs and immediately suggest a rate law so instead we must suggest a rate law and see if it fits a graph of experimental data
• if it doesn’t then different forms are tried until one which works is found

Question 31

what is the way that we can do data fitting for a first order rate equation for a reaction of A —> products, what do we plot?

Answer 31

for a first order reaction of
A —-> products
we know

rate = k[A]
hence
d[A]/dt = -k[A]
so separating the variables and integrating gives
ln([A]) = -kt + ln([A]o)
[A] = [A]o exp(-kt)

So:
- plot ln([A]) against t, this will give a straight line graph of gradient -k and Y-int ln([A]o)

Question 32

what should we note about what we can plot/what alternative to conc. can we plot in data fitting for a first order reaction of the form A —-> products

Answer 32

• we don’t actually need to know absolute conc. to plot a suitable graph
• plotting a quantity that is proportional to conc also works

e.g.
I = b[A]
[A] = I/b
[A]o = Io/b

ln(I/b) = -kt + ln(Io/b)
so
ln(I) = -kt + ln(Io)

this still has the right slope

Question 33

what process can we follow if we want to express the rate in terms of appearance of product when data fitting for a 1st order reaction A —-> nB

Answer 33

A —-> nB
d[B]/dt = k[A]
also
[A] = [A]o - n[B]
so
[A]o - n[B] = [A]o exp(-kt)
so
[B] = 1/n [A]o (1-exp(-kt))

• as expected, this predicts that [B] rises from zero and tends towards [A]o over a very long period

Question 34

what is the way that we can do data fitting for a second order rate equation for a reaction of A —> products, what do we plot?

Answer 34

we know for a second order reaction (in A) that

rate = -d[A]/dt
rate = k[A]^2

so
d[A]/dt = -k[A]^2
so separating variables, integrating and multiplying through by -1 gives
1/[A] = kt + 1/[A]o

so plotting 1/[A] against t gives:
- straight line graph of gradient k and y-int 1/[A]o

Question 35

what is the way that we can do data fitting for a more complex second order rate equation for a reaction of A + B —> C, what trick is it useful to use?

Answer 35

we can assume our rate law is
rate = k[A][B]
and
rate = -d[A]/dt

so
d[A]/dt = -k[A][B]

• a trick we can use is that we have control over the starting conditions of the reaction
• noting that if we start the reaction where [A]o = [B]o, then we know (as they react 1:1) that at all times [A] = [B]
• this simplifies the equation to
  d[A]/dt = -k[A]^2
  as before

Question 36

What is the isolation method? Why is it useful?

Answer 36

• the isolation method is used when a reaction’s rate law is too complex to integrate or solve normally

e.g. we have A + B —> products
we assume rate = k[A]^2[B]
so
d[B]/dt = d[A]/dt = -k[A]^2 [B]

• the idea is to make one reagent (e.g. [A]) in a very large excess (e.g. 50 times)
• this means that over the course of the reaction the conc. of the large excess substance changes very little from its init. conc. but the other changes a lot
• this means the conc. of the large excess substance can be thought of as constant

Question 37

what is k(eff) in the isolation method, how can we analyse it

Answer 37

• because in the isolation method, the substance(s) that are in very large excess remain virtually constant in concentration, they can be incorporated into a rate constant to make and effective rate constant K(eff) or a pseudo first order rate constant, this also reduces the rate equation to a pseudo first order equation

e.g.
A+B —> P

d[B]/dt = -k[A]^2[B]
for example
We could make A in large excess meaning
[A] = [A]o approx. at all times
so
d[B]/dt = -k(eff) [B]
where k(eff) = k [A]o^2

this is then solvable as normal

Question 38

what are some of the limitations of the isolation method

Answer 38

• putting a reagent in a very large excess may make the rate inconveniently fast
• having a large excess could change the mechanism of a reaction

Question 39

describe the differential method of rate law data fitting, explain why its useful and when it is used, what are some disadvantages

Answer 39

• if the rate law is in the form

rate = k[A]^n

NOTE: more complex laws can be made into this form using the isolation method

then we can use the differential method
- this involves taking logs of both sides so
log(r) = n*log([A]) + log(k)
- this is in the form of a straight line graph of gradient n, y-int log(k)
- this means we don’t have to make any assumptions about n
- a disadvantage is that we have to plot rates, these are often much harder to measure than conc.’s

Question 40

define the half life, t(1/2) of a reaction

Answer 40

The half life of a reaction is defined as the time it takes for the conc. of a specified reagent to fall to half of its initial value

Question 41

for a first order reaction, does t(1/2) depend on initial conc., prove it

Answer 41

no, half life remains const. throughout the reaction

we know for a first order reaction wrt A

r = k[A] = -d[A]/dt
so
ln[A] = -k*t + ln[A]o

so
ln([A]o / 2) = -kt(1/2) + ln[A]o
-ln(2) = -kt(1/2)
t(1/2) = ln(2)/k

Question 42

Give the equation linking rate constant and half life for a first order reaction, is this an accurate way of determining k

Answer 42

t(1/2) = ln(2)/k

• this is not a particularly accurate way to find k, you are better to plot some sort of conc. against time graph as explained before

Question 43

give the equation linking rate constant and half life for a second order reaction

Answer 43

t(1/2) = (1/k)*(1/[A]o)

• DIFFERENT TO FIRST ORDER, this does depend on init. conc.

Question 44

why is light/UV used in studying the kinetics of chemical reactions, what are the most common wavelengths and molecules used for this sort of study

Answer 44

• we know molecules absorb light at characteristic frequencies, these are associated with energy level transitions
• the commonest kinds of transitions in kinetics are those in the visible or UV region (1000-200nm)
• All molecules can absorb UV to some extent but special groups cause some stronger absorptions

Question 45

what is the relationship between the light absorbed by a species (in a light absorbance analysis of kinetics) and its conc., how can this be analysed to give something useful

Answer 45

• the extent to which light is absorbed is related to conc. through the Beer-Lambert law
  I = Io exp(-εcl)
  Io = intensity entering medium
  I = intensity exiting medium
  ε = extinction coeff.
  c = conc.
  l = path length through which light passes

we can define

ln(I) = ln(Io) - εcl
so
ln(Io/I) = εcl = absorbance, A

• the absorbance, A is often read out directly from meters
• A can be plotted against a series of known conc.’s to give a straight line graph of grad εl, this can then be used to convert future readings of A of an unknown conc. substance to actual conc.’s
• Note: in first order reactions, A can be plotted directly as A is direct. prop. to conc.

Question 46

what are the advantages and disadvantages of using light absorbance as a method to determine conc.’s over time

Answer 46

Advs:
- Absorbance measurements are convenient, non-invasive, rapid and can be automated
- molecules which absorb at diff wavelengths can be studied simultaneously in a reaction mixture

Disadvs
- it is, however, possible that more than one species will absorb at a given wavelength

Question 47

how can proton NMR be used to monitor conc.’s over time in a kinetic investigation, what are its Advs and Disadvs

Answer 47

Proton NMR can be used because under the right conditions peak height is prop. to conc.

Advs.:
- individual chemical species can easily be followed

Disadvs.:
- not very sensitive so it’s not very fast

Question 48

How can conductivity measurements be used to analyse concentrations of solutions over time in kinetics investigations, what are its advs and disadvs

Answer 48

• ions in solution conduct electricity
• the conductance of a solution is directly proportional to the concentration of conducting ions
• even in reactions where both the products and reactants contain ions, this process can be used because the identity of the ions will change

Advs:
- effective and easy for first order reactions
- rapid and non-invasive

Disadvs:
- relationship between conc. and conductance can be quite involved for more complicated reactions so this becomes an inconvenient method

Question 49

what is the background behind using partial pressures of gases to help determine conc.s over time in a kinetics investigation

Answer 49

we know for ideal gases
pV = nRT
so
n/V = p/RT
so at a constant temp, pressure is directly prop. to conc.

we can redefine this using partial pressures of

ni/V = pi/RT

Question 50

how can we use our relationship between partial gas pressures and concentrations to measure suitable concentration readings for reaction where only gas is formed and more complex ones with gas in products and reactants

Answer 50

our relationship is
ni/V = pi/RT

• for simple reactions where solids/liquids form gases this is simple as we can just take the total pressure to be the partial pressure
• for more complex reactions we need to restrict some variables
  e.g. take the reaction
  2NO(g) + O2(g) —> 2NO2(g)
  we know Pt = P(NO) + P(O2) + P(NO2)
  as we can only record total pressure this means we have 3 unknowns and only one measurable quantity
• we can start by ensuring that at time = 0, there is no NO2 present and NO, O2 are in their stoichiometric ratios, our total initial pressure is p0
• we can consider what happens when a single stoichiometric reaction occurs and rearrange the numbers using changes in each P.P. and the p0 to find the P.P of certain reagents/products in terms of only the initial and measured pressures p0 and pt

Question 51

Give the advs and disadvs of using gas pressures to determine conc.’s over time in a reaction when doing a kinetics investigation

Answer 51

advs:
- many devices are capable of measuring pressure

disadvs:
- need restrictions on starting conditions and maths can be complicated

Question 52

How can electrochemistry be used to analyse concentrations of solutions over time in kinetics investigations, what are its advs and disadvs

Answer 52

• we know the EMF of an electrochemical cell can depend on the conc.’s of the species present (think nernst eq.)
• we can use a reference electrode and an appropriate electrode for the reaction to determine the voltage for the reaction, this if the right species are in a large excess is dependent only on the conc. of a certain species

e.g. take
Br2 + HCOOH —> 2Br(-) + 2H(+) + CO2
we can dip a reference calomel electrode and a platinum electrode into the solution, the reaction at the platinum electrode is
Br2 + 2e(-) —> 2Br(-)
we have an equation for this of
E = E” - RT/2F ln([Br-]^2 / [Br2])
if Br(-) is in a large excess then the change in EMF will be dependent only on the change in conc. of Br2

Question 53

Give an example of a classical method for determining the conc. of a species over time in a reaction for a kinetics investigation, why is this method limited

Answer 53

• take titrations at different points
• these methods are slow and the the reaction mixture taken at a certain time must be quenched to ensure no further reaction takes place whilst carrying out the titration

Question 54

Give examples of how reaction mixtures taken when doing the titration method of kinetics analysis can be quenched

Answer 54

• sudden cooling
• removing one of the reagents

Question 55

How can gas chromatograms (GC) and mass spectrometers (MS) be used to analyse concentrations of solutions over time in kinetics investigations

Answer 55

• a gas chromatogram will separate out phases of an inert gas by passing them through a long column in an inert gas (mobile) phase, this column contains the stationary phase (porous solid), each of the gas components spend a different amount of time in the stationary/mobile phase so are separated
• this can be used in conjunction with a mass spectrometer so that the different fractions of gas can be recorded and their relative concentrations determined
• samples of the gas can be taken throughout the reaction and run

Question 56

What is the continuous flow method of studying fast reactions, why is it important, how does it work, give an example of where it is used, give advs and disadvs

Answer 56

• if we have a reaction where the reactants take more time to mix than the reaction takes to occur then we have a problem
• the continuous flow method solves this by having two (or more) tubes containing each of the reactants flowing into a single tube
• different distances down the mixed tube correspond to different points in the reaction (dependent on flow rate), this means measurements can be taken at leisure

Advs:
- measurements can be taken at leisure i.e. doesn’t have to be in real time
- mixing can be completed very quickly

disadvs:
- need a constant (and large) supply of reactants

an example is the complexation reaction between Fe(2+) and SCN(-) in aqueous solution

• the main difficulties encountered using this method are how to measure the conc.’s, it is generally done using spectrophotometry

Question 57

What is the stopped flow method of studying fast reactions, how does it work, give an example of where it is used, give advs and disadvs

Answer 57

• the apparatus and uses are generally similar to the continuous flow method in the way that the reactants are mixed and travel down a tube
• the difference is that in the stopped flow system, the reaction mixture pushes a plunger back until it hits a stop
• at that point a measurement is made at a specific fixed place
• observations are made at a point where mixing is complete

advs:
- doesn’t require large amounts of reagents
- generally time resolution is slightly faster than continuous flow

disadvs:
- measurements must be made in real time

• It’s generally used in analysing enzyme catalysed reactions

Question 58

what is flash photolysis, when is it used in kinetics analysis of conc.’s of a substance over time

Answer 58

• a short intense flash of light (from laser) is used to generate a reactive species (usually free radicals), this is photolysis
• then the reaction of these reactive species with other reagents present is monitored, usually spectrophotometrically

Question 59

what are elementary steps, what are their features

Answer 59

• each of the reactions which form the mechanism is called an elementary step
• it is elementary because it takes place in a single chemical encounter between the species involved
• the elementary steps cannot be broken down any further

Question 60

how can kinetics be useful in determining mechanisms of reactions, what general process do we do this by

Answer 60

• we can use the kinetics of a reaction along with other methods such as spectroscopy of intermediates to determine a mechanism
• generally a mechanism is proposed and then analyse its kinetics, if this agrees with experiment then the mechanism is consistent

Question 61

what can we say about the rate laws of elementary reactions, why are they easy

Answer 61

• we can always write down the rate law for elementary reactions simply by observing the stoichiometric equation
• we believe they occur in single chemical encounters so we can straight away write down the rate of loss of reagents or gain of products
• this is different to full equations where we can’t be certain of the form of the rate equation

Question 62

what can we do to form rate laws for all steps of a reaction once we know the mechanism, why does it then become very complex to determine how the conc.’s depend exactly on time

Answer 62

• if we know a reaction mechanism, we have all the elementary steps so we know what the reaction laws will be for each step
• Hence, it is possible to consider the rate of chance of conc. of a single species by observing which steps form said species and which remove it
• this can be made into an expression of the form d[P]/dt = ….
• in this expression any term which is derived from an expression forming P has a positive coeff., any term which derived from an expression removing P has a negative coeff.
• this is good but we quickly end up with very complex coupled differential equations if we want to solve them explicitly for time dependence

Question 63

How can we use sequential reactions to help us to understand the differential equations which arise in complex mechanisms, what are some limiting cases

Answer 63

with sequential reactions such as
A —(k1)—> B
B —(k2)—->C

we can consider some limiting cases

1. k1&raquo_space; k2 i.e. rate of eq 1 is much greater than eq.2. In this case we can say that step two is the rate determining step. So we can say the formation of C depends on k2 only

2.
k2&raquo_space; k1 i.e. rate of eq 2 is much greater than eq 1. In this case we can say that as soon as B is formed it is made into C. This means the production of C mirrors the fall in A. the rate of formation of C is now only dependent on k1

Question 64

How do the limiting cases in sequential reactions simplify the rate equations, how can it be applied to even more complex sequential reactions

Answer 64

• if either of the two limiting cases apply then we can say that the overall rate of production is dependent only on the stage with the smallest rate const.
• this can just as well be applied to more complex sequential reactions, e.g. with 4 sequential reactions

Question 65

what is the pre-equilibrium hypothesis, where does it apply

Answer 65

• the pre-equilibrium hypothesis applies when the species directly involved in the RDS are in equilibrium with the reagents
• the pre-equil hypothesis is where we assume the rate of the RDS is sufficiently slow that it does not affect the equilibrium reaction
• as such, we can assume that the second (non-equil) step is the RDS and that the rates of the forwards and backwards equil reactions are always equal

Question 66

give an example of a reaction where it is useful to use the pre-equilibrium hypothesis, derive the expression for the rate of that reaction using it

Answer 66

we can use the reaction

E + H3O(+) –REV.–> EH(+) + H2O eq1
EH(+) + H2O —> other eq2

we know if the pre-equilibrium hypothesis applies then the rates of the two reactions forming the equil will be equal, this gives

rate(f) = k1[E][H3O+]
rate(b) = k-1 [EH+][H2O]

keq = k1/k-1 = [EH+][H2O]/[E][H3O+]
rearranging gives
[EH+] = [E][H3O+]/[H2O] keq

we can substitute this into our rate expression for eq 2 to obtain an overall expression

rate(eq2) = k2[EH+][H2O] = k2([E][H3O+]/[H2O]eq) [H2O]

rate =k2keq [E][H3O+]

Question 67

what can we do with the overall equation that we obtain after having applied the pre-equil hypothesis in terms of defining a new rate constant

Answer 67

we can ‘clean’ up our new equation e.g. from before we had

rate = k2keq [E][H3O+]

this can be made into a single equil const of kexp = k2keq = k1k2/k-1

rate = kexp[E][H3O+]

Question 68

in what general reactions is the pre-equil a fair assumption

Answer 68

equilibrium steps involving protonation or deprotonation

Question 69

what is the steady state approximation, when is it generally used, how is it expressed mathermatically

Answer 69

• it is used when there are sequential reaction where the second reaction has a much higher rate constant than the first reaction
• this means as soon as an intermediate is formed it is almost immediately reacted again in the second step
• as such, the conc. of the intermediate is constant and low

mathematically d[B]/dt = 0
B is intermediate

Question 70

do the calculations for the simple reaction scheme
A—k1—>B
B—k2—>C
k2»k1
using the steady state approximation, show how it massively simplifies the rate laws

Answer 70

the rate of change of B has two terms
d[B]/dt = k1[A] - k2[B]
assuming the steady state we obtain
[B]SS = k1[A]/k2

we know
d[C]/dt = k2[B] = k2k1[A]/k2 = k1[A]

this is as expected because we expect B to react the moment it is formed so C is only dependent on A

Question 71

when is the steady state approx. valid or not valid

Answer 71

• the steady state approx. can generally only be applied to high energy intermediates
• it should not be applied to reactants or products as these change massively over time
• it is also only valid once the reaction has reached a steady state, i.e. NOT in the initial or final stages of the reaction

Question 72

do an analysis of the acid catalysed hydrolysis of esters using the steady state hypothesis i.e.

E + H3O(+) –k1–> EH(+) + H2O [1]
EH(+) + H2O –k(-1)–> E + H3O(+) [-1]
EH(+) + H2O –k2–> rest of reaction [2]

consider three circumstances and compare to what is expected using knowledge and the pre-eq hypothesis

Answer 72

• identify the intermediate EH(+)
  d[EH+]/dt = k1[E][H3O+] - k(-1)[EH+][H2O] - k2[EH+][H2O] = 0

[EH+] = k1[E][H3O+] / (k(-1) + k2)[H2O]

so

d[Prod.]/dt = k2[H2O][EH+] = k1k2[E][H3O+] / (k(-1)+k2)

case 1:
rate of [2]»rate of [-1], this gives k(-1)+k2 = k2 approx.
so
d[Prod.] / dt = k1[E][H3O+] as expected

case 2:
rate of [2] «rate of [-1], this gives k(-1) + k2 = k(-1) approx.
so
d[Prod.] / dt = k1k2[E][H3O+]/k-1 = K(eq)k2[E][H3O+]
K(eq) = k1/k(-1)
note: this is identical to the expression where we assumed the pre-equil hypoth., step 2 is RDS

case 3:
rates of [-1] and [2] are comparable
no change

Question 73

give the reaction scheme which forms the background to the michaelis menten eq.

Answer 73

E + S —REV.—> ES —> E + P
the three elementary steps are

E + S –k1–> ES [1]
ES —k(-1)–> E + S [-1]
ES –kcat —> E + P [2]

Question 74

how can we use the steady state approximation and the suitable reaction scheme to deduce the michaelis menten eq. what other assumptions must we make

Answer 74

E + S –k1–> ES [1]
ES —k(-1)–> E+S [-1]
ES –kcat —> E + P [2]

we assume ES is a short lived intermediate so use the steady state assumption on it
d[ES]/dt = k1[E][S] - k(-1)[ES] - kcat[ES] = 0

we can also assume
[E] = [E]o - [ES]
substituting this into the above expression gives:

[ES] = (k1[E]o[S]) / (k1*[S]+k(-1)+kcat)

thus we can say rate of formation of products = velocity

V = kcat[ES] = (kcatk1[E]o[S]) / (k1[S]+k(-1)+kcat) = (kcat[E]o[S]) / ([S]+(k(-1)+kcat)/k1) = (kcat[E]o[S]) / ([S]+Km)

where Km is the michaelis menten eq.

Km = k(-1)+kcat / k1

Question 75

how can we interpret the michaelis menten equation, what happens where [S]«Km or [S]»Km

Answer 75

V =(kcat[E]o[S]) / ([S]+Km)

• from this we can say that where [S]«Km we can approximate the denominator to just Km and hence the expression becomes first order in both [S] and [E]o, this is the correct linear regression for the expected graph
• in this region kcat/Km acts as a second order rate constant and determines the rate where enzyme is in excess, typically kcat/Km is 10-10^8 moldm^-3s^-1
• where [S]»Km the expression simplifies to
  Vmax = kcat [E]o
• in this limit the velocity is independent of [S] and is at its max.

Question 76

how can the michaelis menten eq. be re-written using using Vmax and what does this imply/tell us about the meaning of the michaelis menten constant

Answer 76

Vmax = kcat [E]o
so

V = Vmax[S] / [S] + Km

consider the substrate conc S which gives a velocity of half the max. this gives

Vmax/2 = VmaxS / S + Km
so
S = Km

• typically large Km values imply no strong affinity between the enzyme and the substrate and small Km values imply the opposite

Question 77

how can we analyse the michaelis menten equation graphically

Answer 77

• the michaelis menten eq is
  V =(kcat[E]o[S]) / ([S]+Km)

inverting both sides gives

1/V = 1/kcat[E]o + (Km/kcat[Eo])*(1/[S])

i.e. a plot of 1/V on y and 1/[S] on x gives a straight line

Question 78

what is the overall rate equation for the reaction of H2 and Br2

Answer 78

d[HBr]/dt = (ka[H2][Br2]^3/2) / ([Br2] + kb[HBr])

Question 79

what occurs in a chain reaction, give the suitable terminology, what are the 5 different types of process

Answer 79

• generally these are free radical substitutions e.g.

initiation:
Br2 —UV—> 2Br*

propagation:
these are the processes that keep the reaction going
Br* + H2 —> HBr + H*
H* + Br2 —> HBr + Br*
the Br* formed in the second step can be reused in the first step causing a chain reaction, in this case Br* and H* are chain carriers

termination:
these are processes which destroy chain carriers e.g.
Br* + Br* + M —> Br2 + M
note: in this equation, M is simply another body which dissipates the energy released from the Br2 formed, it does not react itself

Inhibition:
these are processes which consume the product and lead to a reduction in the overall rate of reaction, e.g.
HBr + H* —> H2 + Br*

chain branching steps:
some reactions have processes where one chain carrier reacts in such a way as to give rise to two or more carriers e.g.
H* + O2 —> OH* + O*
these can lead to very fast increases in ROR

Question 80

using H2 + Br2 as an example derive the complex rate law of
d[HBr]/dt = ka[H2][Br2]^1/2
(i.e. the expression for when the reaction has not been going on for long)
use the steady state assumption

Answer 80

initiation step:
Br2 + M –k1–> 2Br* + M

propagation step:
Br* + H2 —k2—> HBr + H*
H* + Br2 —k3—> HBr + Br*

termination step:
2Br* + M —> Br2 + M

Note: there are other reactions possible but they are too slow to be significant in the reaction scheme

using the steady state assumption on Br* and H* gives

d[Br]/dt = 2k1[M][Br2] - k2[Br][H2] + k3[H*][Br2] - 2k4[M][Br]^2 = 0

d[H]/dt = k2[Br][H2] - k3[H][Br2] = 0

rearrange for expressions of [H] and [Br] (noting that the middle two terms = 0 using the second eq above) and substitute into the equation
d[HBr]/dt = k2[Br][H2] + k3[H][Br2]

this gives the answer

doing this yields the correct answer
where
ka = 2k2sqrt(k1/k4)

Question 81

how does adding an inhibition steps change the rate law derived for when the reaction of H2 and Br2 has been going for a long time to the actual rate law, what are the inhibition steps, what is the only one we should worry about and why

Answer 81

• adding inhibition steps removes product where we hadn’t accounted for hence making our rate law incorrect
• two possible inhibition steps are
  H* + HBr —k5—> H2 + Br*
  Br* + HBr —k6—> Br2 + H*
• the k5 reaction is much quicker than the k6 reaction so it can effectively compete with the rest of the processes but k6 can’t, so only the k5 reaction is used
• using the same analysis as above, this gives the required rate law of
  d[HBr]/dt = (ka[H2][Br2]^3/2) / ([Br2] + kb[HBr])

where
ka = k2sqrt(k1/k4)
kb = k5/k3

Question 82

what is chain length and how is it calculated

Answer 82

• chain length is the average number of times that the closed cycle of steps which produce products can be repeated per chain carrier

chain length, l = rate of overall reaction / rate of initiation