Rate Processes Flashcards

Question

What is the approx value for q_vib?

Answer 1

q_vib = q_vⁿ n = No of vib modes 3N-5 for linear, 3N-6 is for non-linear

Answer 2

q_vib = q_vⁿ q_vib^~ = q_v^n-1

Answer 3

SCT is equal to linear TS in TST q_A⁰ = q_B⁰ = q_T³ q_AB‡⁰ = q_T³ x q_R² x q_V Simplifies to q_R² / q_T³ Then insert into eyring

Answer 4

q_AB⁰ = q_CD⁰ = q_T³ x q_R² x q_V q_ABCD‡^0,~ = q_T³ x q_R³ x q_V⁵ Simplifies to q_V³ / [q_T³ x q_R]

Answer 5

A_TST α q_R² / q_T³ A_SCT α q_V³ / [q_T³ q_R] P = (q_V/q_R)³

Answer 6

q_R = 10 q_V = 1 No of accessible states at T

Answer 7

Common assumption q_V = 1 q_T α Sqrt(T) q_R α Sqrt(T)

Answer 8

k_r α T^-1 exp[-ΔE₀^‡/RT]

Answer 9

Ea = ΔE₀^‡ + nRT Mostly ΔE₀^‡ >>> RT so Ea has little T-dependence

Answer 10

Low masses Such as H/D as largest % difference in mass

Answer 11

ln[kr^H/kr^H] = ln[A^H/A^D] - [ΔE_0,H^‡ - ΔE_0,D^‡]/RT 2nd term dominates ln() α 1/T

Answer 12

ΔE_0,H^‡ - ΔE_0,D^‡ = ZPE(D) - ZPE(H) Assumes TS has same E, assume m_H/D much smaller than other atom so μ_HB/D = m_H/D ΔE_0,H^‡ - ΔE_0,D^‡ = 0.5N_Ahbar x Sqrt[kf/m_H] x (1/Sqrt2 - 1)

Answer 13

kr^H/kr^D = 8 For typical diatomics @ 298K H transfer is faster

Answer 14

Mass change effects: q_trans α m^3/2 q_rot α I Means A larger when lighter A_H/A_D = 2.8

Answer 15

3 ways which causes this: * T-dependence of A * kr depends on reactant quantum states * Quantum mech tunneling at low T

Answer 16

If the reactant is in a higher energy state then the reaction is faster e.g. v=1 instead of v=0, then the rate is 100x faster

Answer 17

Structure & vib frequency of TS may not be known so need to use a simpler equation

Answer 18

Use ideal gas, Vm^θ = RT/P^θ so kr = κ(kT/h)(RT/P^θ)K~^‡ Where K~^‡ is partial pressure eqm const Then relate to ΔG^‡ kr = κ(kT/h)(RT/P^θ) exp(ΔS^‡/R) exp(-ΔH^‡/RT) And as Ea = RT² x dlnkr/dT, Ea = ΔH^‡ + 2RT kr = κ(kT/h)(RT/P^θ)e² exp(ΔS^‡/R) exp(-Ea/RT)

Answer 19

A is therefore entropy-based difference between reactants & TS Most important is difference in no of vib modes When greatest loss of rot energy (Δn most -ve) then smaller A

Answer 20

A = atom, D = diatomic, P = Polyatomic A + A -> TS, Δn=2 A + D -> non-linear TS, Δn=1 A + D -> linear TS, Δn=0 D + D -> non-linear TS, Δn=-1 D + D -> linear TS, Δn = -2 non-linear P x2 -> non-linear TS, Δn=-3

Answer 21

Similar rates in both phases Little solvent dependence

Answer 22

Gas - rare collisions with rel free motion, low density Solution - frequent solvent-solute collisions with rare solute-solute encounters

Answer 23

Gas - collisions between molecules occur singly Solutions - collisions occur in bunches with rel long intervals between them

Answer 24

R-N=N-R -> {R* R*} -> R*R* -> R-R Without scavenger then recombines to form R-R Low conc competition between scavenging and bulk recomb of free radicals

Answer 25

Diffusion -> RDS is formation of encounter pair Activation -> RDS is reaction pair, overall will depend on K_AB (eqm const for encounter formation)

Answer 26

`#`of moles passing through a surface area of 1m² per second

Answer 27

Rate = total flux of B molecules diffusing towards & colliding with A molecules [B]_r = (1-R`*/R`) [B] where encounter distance, R`*` = R_A + R_B and bulk conc is [B]

Answer 28

J_r = D_B(d[B]_r/dr) = D_B(R`*`/r²)[B] where J_r = flux of B down a conc gradient D_B = diffusion coefficient

Answer 29

Rate = 4πR`*`² J_r=R`*` = (4πR`*`²)(D_B[B]/R`*`) = 4πR`*`[B]

Answer 30

Rate = 4πDR`* [A][B]`= k_d `[A][B]` where kd = 4πDR`*` N_A

Answer 31

D_A = kT/6πηR_A Where viscocity, η = η₀ exp(Ea/RT)

Answer 32

D_A = kT/6πηR_A Where viscocity, η = η₀ exp(Ea/RT) so kd α η^-1 α exp(-Ea/RT)

Answer 33

Coulomb interaction adds drift term kd = 4πDR_eff N_A where R_eff= Rc/[exp(Rc/R`*`)-1] Rc = z_az_be² / 4πε₀εkT

Answer 34

In stokes-einstein equation and diffusion control Rc, separation @ which Coulomb interaction = kT

Answer 35

A+B <-> {AB} -> P ka is {AB} ->, and kd is forwards of eqm and kd' is backwards reaction ka << kd' so kr=ka(kd/kd') K_AB

Answer 36

Add coulomb interaction to ΔG lnkr = lnkr₀ - (1/RT)(z_az_be²N_A/4πεε₀R`*`) where lnkr₀ is rate without ionic interaction z_az_b > 0 then kr

Answer 37

As ε decreases, shielding of ions by solvents decreases Means less stable encounter complex and a slower reaction

Answer 38

Ionic can stabilise or destabilise the TS which effects the kr

Answer 39

Use dG = Vdp - SdT (dlnkr/dp)_T = (-1/RT)(dΔG‡/dp)_T = -ΔV‡/RT Where ΔV‡ is molar vol of reactants and TS

Answer 40

(dlnkr/dp)_T = -ΔV‡/RT lnkr = lnkr^θ - (ΔV‡/RT)(p-p^θ) where kr^θ is kr at p=p^θ For bimol reactions -> ΔV‡ dominated by ΔVm of reactants & encounter pair For ionic reactions -> dominated by ΔV occupied solvent, called electrostriction If TS more charged than reactants, solvent around TS tighter packed & ΔV‡ -ve

Answer 41

Ea = ΔH‡ + RT kr = κN_A(kT/h) exp[ΔS‡/R] exp[-Ea/RT]

Answer 42

z_az_b > 0 then ΔS‡ < 0, higher TS charge so solvent more ordered z_az_b < 0, then ΔS‡ > 0

Answer 43

K = a_AB/a_Aa_B = [{AB}]/`[A][B]` * (γ_AB/γ_Aγ_B) = K_AB (γ_AB/γ_Aγ_B) where a_J = activity of species J γ_J = activity of coefficient of J

Answer 44

log γ_J = -Az²_J Sqrt[I] Where I = 0.5 Σ_J (b_J/b^θ)z²_J Where b_J = molarity

Answer 45

kr = k_aK_AB = k_a K (γ_Aγ_B/γ_AB) = kr₀ (γ_Aγ_B/γ_AB) logkr = logkr₀+ logγ_A + logγ_B - γ_AB = logkr₀ - ASqrt[I](z²_A + z²_B - z²_AB simplifies to logkr = logkr₀ + 2Az_Az_BSqrt[I]

Answer 46

No bond breaking in condensed phases but still has an Ea

Answer 47

Effect similar to Franck-Condon factor e- transfer faster than change in bond length or orientation change of solvent

Answer 48

Absorption of light nearly instantaneous No time for change in position of nuclei

Answer 49

All occurs in aqueous G is in 1D, and treats it as harmonic

Answer 50

Where () is in one solvent sphere, [] is another and {} is a third (2+) + {3+}<-> [(2+) + {3+}] <-> [2+ 3+] <-> [3+ 2+] <->{3+} + (2+)

Answer 51

D + A -> D⁺ + A^- Before: solvent reorganisation & bond reorgansiation so reactants and products have same energy ET: Tunelling in rigid nuclear framework then relaxation

Answer 52

Depends on overlap of donor & acceptor elec wavefn decreases wrt r k_{et_{α exp(-βr)

r = edge-to-edge separation of D & A}}

Answer 53

When identical PE curves and ΔG_F(D+A) = ΔG_F(D⁺ + A^-) Then Δ‡G = ( Δ_rG^θ +λ)² / 4λ where λ = reorganisation energy

Answer 54

Reorganisation energy Gibbs energy to distort eqm nuclear framework of prods to that of reactants without electron transfer

Answer 55

kr α exp[-βr]exp[-Δ‡G/RT] plot of lnkr vs Δ_rG gives normal on left of λ, and inverted on RHS

Answer 56

-Δ_rG^θ < λ More -ve Δ_rG^θ is the driving force, so rate increases

Answer 57

Activation-less ET Δ_rG^θ = 0 so kr is at max Little change in bond length or molec geometry

Answer 58

Δ_rG^θ > λ More -ve Δ_rG^θ so larger driving force However, k decreases

Answer 59

φ = potential @ point is work done in moving unit +ve charge from infinity to a point Units of V

Answer 60

E = φ₂ - φ₁ Made by charge separation Measure by estabilishing eqm when charged species exchanged across a surface

Answer 61

Rate of flow of charge j = dQ/dT

Answer 62

When charge ze & subject to φ G = G(0) + zFφ Where F is faraday const

Answer 63

Charge on mole of e-

Answer 64

As φ increases, the reactant is destabilised But destab TS by smaller amount so reaction occurs faster

Answer 65

j = j₀( [Red]₀/[Red]_∞) exp[βFη/RT] - j₀( [Ox]₀/[Ox]_∞) exp[-αFη/RT] Where: []₀ is conc @ electrode surface []_∞ is conc in bulk η is activation overpotential

Answer 66

Activation overpotential η = E - E₀

Answer 67

β is transfer coefficient α = 1-β if β~1 then TS resembles reactants if β~0 then TS resembels products

Answer 68

for reaction in aq: Oxⁿ⁺ + e^- <-> OxOx^(n-1)+ Forward reaction is k_red & backwards is k_ox φ_m & φ_s for in metal and solutions respectively G_r = G_r(0) + (n-1)Fφ_s - F(φ_m-φ_s) G_p = G_p + (n-1)Fφ_s G^‡ = G^‡(0) + (n-1)Fφ_s - βF(φ_M-φ_S) gives ΔG^‡red = G^‡ - Gr ΔG^‡oxn = G^‡ - Gp j = F d[e-]/dt = F(k_ox[Red]₀ - k_red[Ox]₀] Include overpotential to give Butler-Volmer j = j₀( [Red]₀/[Red]_∞) exp[βFη/RT] - j₀( [Ox]₀/[Ox]_∞) exp[-αFη/RT]

Answer 69

j₀ = Fk₀ [Red]^α_∞ [Ox]^β_∞ where k0 is rate const for forward and backwards

Answer 70

When η large & +ve then v slow redn but fast oxn exp(βFη/RT)~0 j~j₀exp(βFη/RT) so lnj = lnj₀ + βFη/RT When η large & -ve then v slow oxn but fast redn exp(-αFη/RT)~0 j~ -j₀exp(-αFη/RT) ln-j = lnj₀ - αFη/RT)

Answer 71

Tafel plot of η vs ln|j| Get α&β from gradients j₀ from extrapolated intercept of two lines E₀ is from

Answer 72

Tafel plot of η vs ln|j| Get α&β from gradients j₀ from extrapolated intercept of two lines E₀ is from η-axis intercept

Answer 73

Time vs Measured current (function of potential) X <-> Y + e- More +ve potential gives oxn More -ve potential gives redn

Answer 74

When low, kox small so rate of reaction governed by rate of e- transfer When high, kox higher but electrolysis laready carried out depleted surface [X] So k determined by rate of diffusion of X to electrode