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

Reaction Kineticsr – 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 reagent that influences the rate of a reaction.

Reaction Kinetics aA + bB cC + dD

Rate (r) = 1/ni dci/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/b/l = reaction orders with respect to A, B, L, respectively.

overall order, n = a + b + l.

A and B are reactants. L = catalyst, intermediate

Elementary Reactions

1. 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).

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

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.

e.g.... A + C I + D I + B F + C

stoichiometry: A + B D + FI is intermediate and C is catalyst r = k[A][C]

Experimental Goals1. Determine rate law – Find n and k e.g. r = k [A] [C], n = 2, and k = r/([A][C]).

2. Determine mechanism(s) consistent with rate expression

slowfast

3. Determine Activation energy by T dependence of k.

Most Common Reaction Orders

1st order:

2nd order:

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

rate = k [A][B]rate expression can include ......orders > 2, half-integral orders,inverse dependency – [X] in denominator

1st Order Reactionsr = -d[A]/dt = k[A]

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

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

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

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

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

half-life

if [A] = [Ao]/2 then t = t1/2 & .....

t½ is independent of [Ao]

t½ = 0.693/k

t½ = ln 2/k

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

ln 0.5 = -kt½

2nd Order Reactions half-lifer = - d[A]/dt = k[A]2

1/[A] – 1/[A0] = k t linear 1/[A] = kt + 1/[A0]

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

t dt

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

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

t½ = 1/(k[Ao])

1/[A] – 1/[Ao] = kt

as [Ao] t½

2/[Ao] - 1/[Ao] = kt½

1/[Ao] = kt½

0 Order Reactions (rare – some free radical reactions)

r = -d[A]/dt = k

∫AoA d[A] = k ∫0

t dt

[A] = kt + [Ao]

Plot [A] vs. t slope = k Yint = [Ao]

[A]

t

[Ao]

0 50 100 150 200 250 300 350 400 450 5000.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1st 2nd

115 (115) 230 (115) 345

167 (333) 500

1st order – t½ is constant regardless of [A]0.2nd order – t½ doubles as [A]0↓ by ½.plot ln t½ vs. ln [Ao] slope = 1-n

Determining Reaction Order, n

Half-life Method

Advantage: Single experimentDisadvantage: 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])

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

3. repeat for other reagents in rate expression

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

20.4 - p726

G A + B

C + D

AB*

Ea determines rate - DG° determines Equil.

Ea

DG

where N2 = # collisions leading to reaction & N1 = total # collisions

Boltzmann Distribution N2/N1 = exp(-Ea/RT)

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

Determining the Rate Law

F2 (g) + 2ClO2 (g) 2FClO2 (g) rate = k [F2]x[ClO2]y

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

Problem 13.70

X = 1

r2 = k•(2[F2])x•[ClO2]y = k•2x•[F2]x•[ClO2]y = 2x = 2r1 k•[F2]x•[ClO2]y k• [F2]x•[ClO2]y

y = 1

r2 = k•[F2]x•(4•[ClO2])y = k•[F2]x•4y•[ClO2]y = 4y = 4r1 k•[F2]x•[ClO2]y k•[F2]x• [ClO2]y

rate [A] [B] [L]1.00E-05 0.1 0.1 0.012.00E-05 0.2 0.1 0.011.00E-05 0.1 0.2 0.012.00E-05 0.1 0.1 0.02

k = A exp(-Ea/RT)

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

Arrhenius Activation Energy

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

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 ….

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

T (K) 103k 1/T ln k Fit

599 0.54 0.00167 -0.62 -0.63

629 2.5 0.00159 0.92 0.91

666 14 0.00150 2.64 2.63

683 25 0.00146 3.22 3.36

700 64 0.00143 4.16 4.05

H2 + I2 2HI

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

Ea = 160 kJ mol-1

REACTION MECHANISM

1. List of Elementary Steps2. Must add to overall stoichiometry3. Must be “consistent” with Rate Law

2. Steady State Method (SS)

1. Rate Determining Step Method (RDS)

C is an intermediateProduct in an early step, reactant in a later step. Doesn’t appear in stoichiometry. May appear in rate law

A + B C + DC + E B + FA + E D + F

B is a catalystReactant 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-]

H+ + HNO2 + FNH2 FN2+ + 2H2O

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.

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

H+ + HNO2 + FNH2 FN2+ + 2H2O

k1

k-1

H+ + HNO2 H2NO2+ fast

k2H2NO2

+ + Br- ONBr + H2O slow k3

ONBr + FNH2 FN2+ + H2O + Br- fast

r = k2 [H2NO2+] [Br-]

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

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

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

H+ + HNO2 + FNH2 FN2+ + 2H2O

k1

k-1

H+ + HNO2 H2NO2+

k2

H2NO2+ + Br- ONBr + H2O

k3ONBr + FNH2 FN2

+ + H2O + Br-

r = k3 [ONBr][FNH2] d[ONBr]/dt = 0

k2 [H2NO2+][Br-] = k3 [ONBr][FNH2]

r = k3 k2 [H2NO2+][Br-] [FNH2]/(k3[FNH2])

apply ss assumption to H2NO2+

[ONBr] = k2 [H2NO2+][Br-]/(k3[FNH2])

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

H+ + HNO2 + FNH2 FN2+ + 2H2O

k1

k-1

H+ + HNO2 H2NO2+

k2

H2NO2+ + Br- ONBr + H2O

r = k2 [H2NO2+][Br-]/[FNH2]

apply ss assumption to H2NO2+

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

+][Br-]

[H2NO2+] = k1[H+][HNO2]

k-1 + k2[Br-]r = k2k1 [Br-][H+][HNO2] k-1 + k2[Br-]

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

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

r = k2k1 [Br-][H+][HNO2] k-1 + k2[Br-]

SS

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

H+ + HNO2 + FNH2 FN2+ + 2H2O

Half Orders in Rate Law .....

Reactant split in first step - prior to RDS D + 2B 2P

D 2M2 (M + B P) RDS

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

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

Look for P that is co-product prior to RDS… or I that is P in first step & R after RDS

Term in denominator of rate lawHg2

2+ + Tl3+ 2Hg2+ + Tl+

Hg22+ Hg2+ + Hg fast

k1

k-1

Hg + Tl3+ Hg2+ + Tl+ slow k2

r = k2 [Hg][Tl3+]

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

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

A B (+ C)

Unimolecular ReactionsStill involve some type of collision.

A + M A* + M

A* B (+ C)

r = k[A]

Enzyme Kinetics Mechanism:

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