Kinetics

Question 1

Overall reaction rate

Answer 1

For a reaction: eE + fF –> gG + hH
Rate = -1/e d[E]/dt = -1/f d[F]/dt = 1/g d[G]/dt + …

Question 2

Rate law

Answer 2

Dependence of reaction rate on concentration of reactants
Experimentally determined equation
Not given by stoichiometric coefficients unless elementary

Question 3

Rate laws for 0th, 1st and 2nd order

Answer 3

Dependence of rate on conc.
0th: d[R]/dt = k[R]^0 = k
1st: -d[R]/dt = k[R]
2nd: -d[R]/dt = k[R]^2

Question 4

Integrated rate equations for 0th, 1st and 2nd order

Answer 4

Dependence of conc. with time
0th: [R] = [R]0 - kt
1st: [R] = [R]0e^-kt
2nd: 1/[R] - 1/[R]0 = kt

Question 5

Linear plots for 0th, 1st and 2nd order

Answer 5

Graphical determination of rate law
0th: [R] vs t
1st: ln[R] vs t
2nd: 1/[R] vs t

Question 6

Half lives for 0th, 1st and 2nd order

Answer 6

Measure of speed of reaction
0th: t1/2 = [R]0/2k
1st: t1/2 = ln2/k
2nd: t1/2 = 1/k[R]0

Question 7

Elementary reactions

Answer 7

Reactions occurring in one step
No way of establishing whether a reaction is elementary but by experimentally determining reaction mechanism
E.g. H2 + Cl2 –> 2HCl Includes initation, propagation and terminations steps

Question 8

Arrhenius equation

Answer 8

k(T) = Ae^-E(A)/RT
Smaller activation energy, greater rate.
Plotted by linear plot of 1/T vs lnk where -Ea/R is gradient and lnA is y-intercept

Question 9

Kinetic gas model

Answer 9

Molecules are hard spheres, motion predicted by classical mechanics (random motion)
Size is negligible and distance travelled between collisions&raquo_space; size
Only elastic collisions (so no energy lost)
Molecules don’t interact except during collisions
pV = nRT = 1/3nMv^2 where M = N(a)m

Question 10

Collision theory

Answer 10

Collision between A + B for reaction to occur
Reaction rate proportional to speed of molecules
Collision density (Z(AB)): Number of collisions per unit time and volume
Z(AB) = sigma(AB) x v(rel) x N(A)^2 x [A][B]
Z(AB) = dNc/dt = [d[C]/dt]N(A)
V = pi(rA + rB)^2 x vt where vt = d
Collision cross section: sigma(AB) = pi x r^2 = pi (rA + rB)^2
where rA = hard sphere radius
v(rel) = Mean relative speed = square root of 8kBT/pi x mu
where mu is reduced mass = (mAmB)/(mA + mB)
Number density = conc. x Avogadro’s constant = N(A)[A] x N(B)[B]

Question 11

Maxwell-Boltzmann distribution

Answer 11

f(v) is distribution function for speed of molecules
Derived by assuming Boltzmann distribution of kinetic energies for gas particles
Probability of finding particle within given kinetic energy proportional to e^-KE/kBT

Question 12

Energy requirement (f react)

Answer 12

f react is probability that molecule will have specific energy Ea
f react = e^-Ea/RT
Boltzmann distribution gives rise to functional form for collisional energy:
k calc = sigma AB = [the square root of 8kBT/(pi x mu)] x N(A)e^-Ea/RT

Question 13

Geometric/ Steric requirement

Answer 13

Molecules need to collide with correct conformation
Account for this using steric factor, P
P = k exp/k theory
Normally P < 1 as expected
Sometimes P &laquo_space;1 as highly specific conformation needed for reaction to occur (usually for larger molecules/more complicated reactions)
P > 1 when reaction more likely to occur than expected

Question 14

0th order

Answer 14

For 0th order reaction: R –> P
Rate of removal given by: v = -d[R]/dt = k[R]^0 = k
Rewritten as: d[R] = -k dt
Integrated between t and t = 0: [R] = - [kt]
R - R = -(kt - 0)
R = [R]0 - kt

Question 15

1st order

Answer 15

R –> P
v = -d[R]/dt = k[R]^1
d[R]/[R] = -k dt
Integrated between t and t = 0: ln(R) - ln([R]0) = -k(t-0)
ln(R/[R]0) = -kt
R = [R]0e^-kt

Question 16

2nd order

Answer 16

A + B –> C
reactant removal rate = -d[A]/dt = k[A][B]
Assume case [A]0 = [B]0 so A = B
-d[A]/dt = k[A]^2
d[A]/[A]^2 = -k dt
Integrate between A and [A]0: -1/A + 1/[A]0 = -kt

Question 17

Half life, t1/2

Answer 17

Time taken for con. of reactant to decrease by half its initial value
0th order: 1/2[R]0 = [R]0 - kt1/2
kt1/2 = [R]0 - 1/2[R]0
t1/2 = [R]0/2k
1st order: -kt1/2 = ln(1/2[R]0/[R]0) = ln(1/2)
t1/2 = ln2/k = 0.693/k
2nd order: kt1/2 = 1/0.5[R]0 - 1/[R]0 = 2/[R]0 - 1/[R]0 = 1/[R]0
t1/2 = 1/k[R]0

Question 18

Lifetime (Tau) - 1st order

Answer 18

Tau = 1/k
Time for conc. to drop to 1/e of initial value
Commonly used in photochem.
Only sensible for 1st order

Question 19

tmax in consecutive reactions

Answer 19

t max = 1/(k2 - k1) x ln(k2/k1)

Question 20

Measuring reaction kinetics

Answer 20

• Initiate reaction with short (fs/ns) laser pulse
• Use UV/Vis abs. spectroscopy to monitor how conc. of reactive intermediate changes with time
• High laser powers allow for generation of high conc. of intermediates

Question 21

Rate-determining step

Answer 21

• Slowest step of complex reaction scheme
• RDS dominates reaction (determines time scale of reaction)
• Holds for more complicated cases too
• Determines order of reaction

Question 22

Fractional yield

Answer 22

psi = actual moles of product formed/ theoretical max. moles of product
If reactant –> product then by stoichiometry:
psi = moles of product formed/ moles of reactant used

Question 23

Relaxation towards equilibrium

Answer 23

K new = [B]new/[A]new = kf/kr –> kf = K(new)kr
t1/2 = ln2/kf + kr = ln2/K(new)kr + kr
t1/2 = ln2/ (Knew + 1)kr
kr = ln2/(Knew + 1)t1/2
kf = Knewln2/(Knew + 1 )t1/2

Question 24

Pre-equilibrium reactions

Answer 24

Combine sequential reactions with an equilibrium
A <–> I –> P
Assume all steps elementary
k-1&raquo_space; k2
A and I in equilibrium at all times (reaction to form products slow, only slight “leakage” to form P)
K = [I]/[A] = k1/k-1
[I] = (k1/k-1)[A]
d[P]/dt = k2[I] = (k1k2/k-1)[A] = k obs x [A]

Question 25

Pre-equilibrium reactions with 2 reactants

Answer 25

A + B <--> I --> Instead of k1, k1' Assume all steps to be elementary k1' is second order rate constant Still require k-1>>k2 K = [I]c^-/[A][B] At eqm, forward rate = reverse rate so k1'[A][B] = k-1[I] [I] = k1'[A][B]/k-1 d[P]/dt = k2[I] = k1'k2[A][B]/k-1 = k obs[A][B] k1' is 2nd order so 2nd order rate equation

Question 26

Reactions in dilute solution

Answer 26

- Small amount of reactants in an inert solvent - Solvent acts as barrier - Limited by diffusion of A + B together to form "encounter complex" AB - Leads to reduced collision rate so expectation of reduced reaction rate - A + B <--> AB --> P

Question 27

Solvent cage/encounter complex

Answer 27

- A + B come into close contact, normally only a fraction react (energetic requirement) - Gas phase, unreactive collision, molecules separate - In solution, can be harder for molecules to separate (they're held together) - Large number of collisions with reactive partner and solvent (pick up energy), increased reaction probability

Question 28

Steady-state approximation

Answer 28

- Assume that after induction period, net rate of conc. change for all intermediates is very small - d[I]/dt ~ 0 - Can be better approx. than pre-eqm - Greatly simplifies studies of reaction schemes, can explain seeming complex systems - Best demonstrated with example - d[P]/dt = (k1'k2/k-1 + k2) [A][B]

Question 29

Diffusion controlled kinetics

Answer 29

- k1' is bimolecular diffusion coefficient, often labelled kd - kd determined by speed of diffusion of molecules, depends on size of molecules and viscosity (T dependent) of solvent - kd = 8RT/3 eta (n with tail) where eta = viscosity

Question 30

Enzyme-catalysed reactions

Answer 30

- E + S --> E + P - For given [S]0, initial rate of product formation (P) proportional to [E]0 - For given [E]0 and low relative [S]0, rate of product formation proportional to [S] - For given [E]0 and high [S]0, rate of product formation becomes independent of [S]0

Question 31

Michaelis-Menten mechanism

Answer 31

E + S <--> (ES) --> E + P Pre-eqm approx. bad because if k-1>>k2, poor enzyme Can use steady state approx. d[ES]/dt = 0 = k1[E][S] - (k-1 + k2)[ES] [ES] = (k1/k-1 + k2 )[E][S] Michaelis constant K(M) = k-1 + k2/k1 [E] = KM [ES]/[S] Since [E]0 = [E] + [ES], [E]0 = KM[ES]/[S] + [ES] When [S]0 >> KM, KM/[S] + 1 --> 1 d[P]/dt ~ k2[E]0 = v max

Question 32

Lineweaver-Burk plot

Answer 32

Can calculate v max, hence k2 from y-intercept

Question 33

Lindemann-Hinshelwood mechanism

Answer 33

- Transition from 1st to 2nd order quite common for unimolecular gas phase reactions - Activation by collision can occur followed by transformation - C + C <--> C + C* - C* --> P - Suggests use of steady-state approx. with excited state (C*) as intermediate - d[C*]/dt = k1[C]^2 - k-1[C][C*] - k2[C*] - d[C*]/dt = 0 - At high p, [C] is large, assume k-1 [C] >> k2, 1st order - At low p, [C] small, assume k-1[C] << k2, 2nd order

Question 34

Electronic excitations

Answer 34

- Singlet state, S0: molecule in ground state, all electrons paired in molecular orbitals - Singlet state, S1: If electron excited with light to another singlet state, its spin stays the same - Triplet state, T1: State with different numbers of up and down electrons

Question 35

Photochemistry - Jablonski diagram

Answer 35

- Simplified portrayal of relative positions of electronic energy levels and associated vibrational states - States arranged vertically by energy - Horizontal separation of diferent state bears no resemblance to nuclear separation - Fluorescence and phosphorescence point down from singlet and triplet states respectively to singlet state - Excitation (electronic) points up - Vibrational transition points across between S1 states - Intersystem crossing between S1

Question 36

Rate of excitation

Answer 36

transition rate (W12) from s0 to s1: -d[S]0/dt = -dn1/dt = W12 = Bp(lambda) x n1 = k exc x n1 where B = Einstein coefficient of stimulated absorption, p(lambda) = photon density at wavelength, n1 = population (conc.) of groud state S0 1st order kinetics

Question 37

Stern-Volner kinetics

Answer 37

Without quencher: k0 + kQ[Q]/k0 psi 0/psi Q = 1 + tau0kQ[Q]