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# Reaction Thermodynamics Review - PowerPoint PPT Presentation

Reaction Thermodynamics Review. D G rx – indicates direction of reaction given the current distribution of reactants and products. D G ˚ rx – indicates the free energy difference between reactants and products at their standard state concentrations).

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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 + bBcC + 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).

e.g.... A + C  I + D

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

1st Order Reactions

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]

2nd Order Reactions

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

Determining Reaction Order, n

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

Disadvantage: Requires reaction integrity over multiple half-lives

unless ‘fraction’ < ½ used.

Determining Partial Order (a)

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

AB*

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.

F2(g) + 2ClO2(g) 2FClO2(g)

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

H2 + I2 2HI

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

Ea = 160 kJ mol-1

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)

A +BC + D

C+ EB+ F

A + ED + 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.

H+ + HNO2 + FNH2 FN2+ + 2H2O

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.

k1

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)

k1

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]

k1

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-]

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

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

D  2M

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

k1

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+]

Unimolecular Reactions

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