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Reaction Thermodynamics Review

DGrx – indicates direction of reaction given the current

distribution of reactants and products.

DG˚rx – indicates the free energy difference between reactants and products at their standard state concentrations).

The intrinsic‘favorability’ of a reaction.

Q – indicates the current distribution of reactants and

products. DG = DG˚ + RT ln Q

K – indicates the equilibrium distribution of reactants

and products. DG˚ = -RT ln K

r – The rate of the reaction – M s-1

k – The rate constant for the reaction – Indicates

the ‘intrinsic’ speed of a reaction.

Ea –The activation energy for a reaction – Indicates

the free energy difference between the reactants

and the transition state.

Rate law – Indicates the dependence of r on k and the

concentrations of reactants and any other

reagentthat influences the rate of a reaction.

aA + bBcC + dD

Rate (r) = 1/nidci/dt

= -(1/a) (d[A]/dt) or ..... + (1/c) (d[C]/dt) etc.

rate = k [A]a [B]b [L]l

k = rate constant (if you determine k using the change in a reagent

for which the stoichiometric coefficient ≠ 1

you must also adjust for this.

A and B are reactants. L = catalyst, intermediate

a/b/l = reaction orders with respect to A, B, L, respectively.

overall order, n = a + b + l.

A + B C + D and r = k [A] [B]

Partial orders of reactants = stoichiometric coefficient

i.e. a = a and b = b.

2. no catalysts or intermediates in rate law.

3. reverse reaction is also elementary

4. Represents the actual ‘collision’ that takes place resulting in the change in molecular arrangement.

5. Typically n will be 2 or less for an elementary reaction (and its reverse reaction).

I + B F + C

Mechanism – A reaction is represented as a series of

elementary steps that add up to overall

stoichiometry and represent the actual

collision order in the reaction.

slow

fast

stoichiometry: A + B D + F

I is intermediate and C is catalyst r = k[A][C]

Experimental Goals

Determine rate law – Find n and k

e.g. r = k [A] [C], n = 2, and k = r/([A][C]).

- Determine mechanism(s) consistent with rate expression

3. Determine Activation energy by T dependence of k.

1st order:

rate = k [A]

rate = k [A]2 or....

2nd order:

rate = k [A][B]

rate expression can include ......

orders > 2, half-integral orders,

inverse dependency – [X] in denominator

half-life

ln([A]/[Ao]) = -kt

r = -d[A]/dt = k[A]

if [A] = [Ao]/2

then t = t1/2 & .....

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

[Ao][A]d[A]/[A] = -k 0tdt

ln 0.5 = -kt½

t½= ln 2/k

ln [A] - ln[Ao] = ln([A]/[Ao]) = -kt

Linear: ln [A] = -kt + ln [Ao]

t½= 0.693/k

[A] = [Ao] e(-kt)

t½is independent of [Ao]

half-life

r = - d[A]/dt = k[A]2

1/[A] – 1/[Ao] = kt

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

2/[Ao] - 1/[Ao] = kt½

1/[Ao] = kt½

[Ao][A] [A]-2 d[A] = -k 0tdt

1/[A] – 1/[A0] = kt

linear 1/[A] = kt + 1/[A0]

t½= 1/(k[Ao])

[A] = [Ao]/(1 + kt[Ao])

as [Ao] t½

0 Order Reactions (rare – some free radical reactions)

r = -d[A]/dt = k

∫AoAd[A] = k ∫0tdt

[A] = kt + [Ao]

[Ao]

Plot [A] vs. t

slope = k

Yint = [Ao]

[A]

t

Half-life Method

1st order – t½ is constant regardless of [A]0.

2nd order – t½ doubles as [A]0↓ by ½.

plot lnt½vs. ln [Ao] slope = 1-n

115 (115) 230 (115) 345

167(333)500

Advantage: Single experiment

Disadvantage: Requires reaction integrity over multiple half-lives

unless ‘fraction’ < ½ used.

Initial rate method ― r = k [A]a [B]b

1. vary [Ao] while holding [Bo] etc. cst.

2. find initial rate r from plot of [A] vs. t

r2/r1 = ([A0,2]/[A0,1])a

log (r2/r1) = a log ([A0,2]/[A0,1])

two data points: a = log(r2/r1)/log([A0,2]/[A0,1])

Multiple data points: plot log (ro) vs. log [Ao]: slope = a

3. repeat for other reagents in rate expression

20.4 - p726

A + B

G

C + D

Transition State theory (collision theory) A + B → C + D

Boltzmann Distribution

N2/N1 =exp(-Ea/RT)

Ea

where N2 = # collisions leading to reaction &

N1= total # collisions

DG

Eadetermines rate - DG° determines Equil.

Determining the Rate Law

X = 1

rate = k [F2]x[ClO2]y

r2 = k•(2[F2])x•[ClO2]y = k•2x•[F2]x•[ClO2]y = 2x = 2

r1 k•[F2]x•[ClO2]y k• [F2]x•[ClO2]y

r2 = k•[F2]x•(4•[ClO2])y = k•[F2]x•4y•[ClO2]y = 4y = 4

r1 k•[F2]x•[ClO2]y k•[F2]x• [ClO2]y

y = 1

rate = k [F2][ClO2] ― Note that partial orders ≠ reaction coefficients.

Problem 13.70

The T dependence of reaction rates is due to the dependence of k on T. This in turn is due to the dependence of Ea on T. Through empirical observation Arrhenius determined that ….

k = A exp(-Ea/RT)

ln k = (-Ea/R)(1/T) + ln A

plot ln k vs. 1/T slope = -Ea/RYint = ln A

or .. ln (k2/k1)/(1/T2 – 1/T1) = -Ea/R

REACTION MECHANISM

1. List of Elementary Steps

2. Must add to overall stoichiometry

3. Must be “consistent” with Rate Law

1. Rate Determining Step Method (RDS)

2. Steady State Method (SS)

C+ EB+ F

A + ED + F

C is an intermediate

Product in an early step, reactant in a later step.

Doesn’t appear in stoichiometry.

May appear in rate law

B is a catalyst

Reactant in an early step, product in a later step.

Doesn’t appear in stoichiometry.

Must appear in the rate law.

Rate Law: r = k [H+] [HNO2] [Br-]

Br- must be catalyst or intermediate and must

show up in mechanism.

FNH2 is reactant that is not in the rate law.

It must show up in the mechanism in a later step.

If an RDS mechanism is sufficient to explain rate

law then it must be a reactant in a step after the

rate-determining step.

H+ + HNO2 H2NO2+ fast

k-1

k2

H2NO2+ + Br- ONBr + H2O slow

k3

ONBr + FNH2 FN2+ + H2O + Br- fast

H+ + HNO2 + FNH2 FN2+ + 2H2O

Rate Law: r = k [H+] [HNO2] [Br-]

r = k2 [H2NO2+] [Br-]

[H2NO2+] = k1/k-1 [H+][HNO2]

r = k [H+][HNO2][Br-] k = (k2k1/k-1)

H+ + HNO2 H2NO2+

k-1

k2

H2NO2+ + Br- ONBr + H2O

k3

ONBr + FNH2 FN2+ + H2O + Br-

[ONBr] = k2 [H2NO2+][Br-]/(k3[NH2])

r =k3k2 [H2NO2+][Br-] [FNH2]/(k3[NH2])

apply ss assumption to H2NO2+

H+ + HNO2 + FNH2 FN2+ + 2H2O

Rate Law: r = k [H+] [HNO2] [Br-]

r = k3 [ONBr][FNH2]

d[ONBr]/dt = 0

k2 [H2NO2+][Br-] = k3 [ONBr][NH2]

H+ + HNO2 H2NO2+

k-1

k2

H2NO2+ + Br- ONBr + H2O

r = k2k1 [Br-][H+][HNO2]

k-1 + k2[Br-]

k1[H+][HNO2]

k-1 + k2[Br-]

[H2NO2+] =

r =k2[H2NO2+][Br-]/[NH2]

apply ss assumption to H2NO2+

H+ + HNO2 + FNH2 FN2+ + 2H2O

Rate Law: r = k [H+] [HNO2] [Br-]

r = k1[H+][HNO2] = k-1[H2NO2+] + k2 [H2NO2+][Br-]

k-1 + k2[Br-]

H+ + HNO2 + FNH2 FN2+ + 2H2O

Rate Law: r = k [H+] [HNO2] [Br-]

SS

RDS r = k [H+][HNO2][Br-] k = (k2k1/k-1)

same as RDS mech. when ... k2[Br-] << k-1

2 (M + B P) RDS

Half Orders in Rate Law .....

D + 2B 2P

Reactant split in first step - prior to RDS

r = k2[M][B] & [M] = (k1/k-1 [D])1/2

r = k[B][D]1/2

Hg22+ Hg2+ + Hg fast

k-1

k2

Hg + Tl3+ Hg2+ + Tl+ slow

Term in denominator of rate law

Hg22+ + Tl3+ 2Hg2+ + Tl+

Look for P that is co-product prior to RDS

… or I that is P in first step & R after RDS

r = k2 [Hg][Tl3+]

[Hg] = k1[Hg22+]/(k-1[Hg2+])

r = k [Hg22+][Tl3+]/[Hg2+]

Still involve some type of collision.

A B (+ C)

A + M A* + M

A* B (+ C)

r = k[A]

k1

1. E + S ES

k2

k3

2. ES E + P

r = k3 [ES] k1[E][S] = (k2 + k3) [ES]

[ES] = k1/(k2 + k3) [E][S]

[ES] = [E][S]/KM(KM = (k2+k3)/k1

r = k3/KM [E][S]

If [S] >> KM then [ES] = [E]tot and …..

r = k3 [E]tot

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