CHEM1612 - PharmacyWeek 11: Kinetics – Arrhenius Equation

Dr. Siegbert Schmid

School of Chemistry, Rm 223

Phone: 9351 4196

E-mail: siegbert.schmid@sydney.edu.au

Unless otherwise stated, all images in this file have been reproduced from:

Blackman, Bottle, Schmid, Mocerino and Wille,     Chemistry, John Wiley & Sons Australia, Ltd. 2008

     ISBN: 9 78047081 0866

Lecture 32 - 3

2 NO2 (g) + F2 (g) 2 NO2F (g) reaction (1)

A reaction mechanism is a series of elementary reactions (or steps) that add up to give a detailed description of a chemical reaction

Step 1 NO2 + F2 NO2F + F slow

Step 2 NO2 + F NO2F fast

2 NO2 +F2 2 NO2F Overall

An elementary reaction is not made up of simpler steps. For elementary reactions the stoichiometric coefficients are equal to

the exponents in the rate law:

e.g. for step 2 the rate law is rate = k2 [NO2][F].

Reaction Mechanism

reactionintermediatek1

k2

Lecture 32 - 4

Overall reaction is: 2 NO2 (g) + F2 (g) 2 NO2F (g)

A rate-determining (or rate-limiting) step is an elementary reaction that is the slowest step in the mechanism.

e.g. Step 1: NO2 + F2 NO2F + F (slow)

The exponents in the overall rate law (overall reaction) are the same as the stoichiometric coefficients of the species involved in the rate-limiting elementary process (if no intermediates are involved; otherwise their concentrations need to be expressed in terms of the reactants used), in this case :

For the overall reaction (1) Rate = k [NO2][F2]

Rate-Determining Step

Lecture 32 - 5

Reaction Mechanism - Example 2The following elementary steps constitute a proposed mechanism for a

reaction:

Step 1 A + B X fast k1[A][B] =k-1[X]

Step 2 X + C Y slow

Step 3 Y D fast Overall reaction?

A + B + C D What are the reaction intermediates? X and Y

Show the mechanism is consistent with the rate law: rate = k [A] [B][C]

Rate = k [X][C] = k1/k-1[A][B][C] = k [A] [B][C]

k1

k-1

k2

k-2

k3

k-3

Lecture 32 - 6

For the reaction: NO2 + CO NO + CO2

The experimentally determined rate equation is: rate = k[NO2]2

Show the rate expression is consistent with the mechanism:

Step 1 2 NO2 N2O4 fast equilibrium

Step 2 N2O4 NO + NO3 slow

Step 3 NO3 + CO NO2 + CO2 fast

Overall

Reaction Mechanism - Example 3

k1

k2

k3

k-1

Lecture 32 - 7

A proposed mechanism for the reaction:

2NO(g) + Br2(g) 2NOBr(g)

consists of two elementary reactions:

NO + Br2 NOBr2 fast equilibrium

NOBr2 + NO 2NOBr slow

Task: Confirm that this mechanism is consistent with the stoichiometry of the reaction and the observed rate law:

Rate = k[NO]2[Br2]

Reaction Mechanism – Example 4

Lecture 32 - 8

The Oscillating Iodine Reaction Three solutions are added together quickly: iodate IO3

- in acid,

malonic acid, and H2O2.

The resultant solution oscillates in colour several times until it finally stops.

The overall reaction is:

IO3- + 2 H2O2 +CH2(CO2H) + H+ → ICH(CO2H)2 + 2 O2 + 3 H2O

Lecture 32 - 9

Low temperature: slow reactions

Egg cooking time at 80oC ~ 30 min

Bacterial growth at 4oC = slow

High temperature: faster reactions

Egg cooking time at 100oC ~ 5 min

Bacterial growth at 30oC = fast

Temperature vs. Rate

Lecture 32 - 10

A + B → C

Reactants must collide to react

- Orientation and energy factors slow down the reaction

Not all collisions produce a reaction

- Need enough energy

Not all collisions are effective, they need a particular orientation

- Orientation does not depend on T

Collision Theory

Lecture 32 - 11

Consider:

A + B C + D

For reaction to occur:colliding moleculesmust have energy greater than Ea

Ea = activation energy

Increasing temperatureincreases no. of moleculeswith energy greater than Ea

Po

ten

tia

l En

erg

y

Reaction Progress

Po

ten

tia

l En

erg

y

Ea

Ea

A + B

A + B

C + D

C + D

Activated complex (TS)

Activated complex (TS)

Exo

Endo

Activation Energy

Lecture 32 - 12

Only a small fraction of the possible collision geometries can result in a reaction.

Orientation is independent of T.

Orientation and CollisionNO2Cl + Cl· → NO2 + Cl2

Lecture 32 - 13

The reaction constant k depends on several factors:

k = Z·p·f

Z·p = A (frequency factor)

f = exp(– Ea / RT )

where Ea = activation energy, R gas constant, T temperature (K)

Collision Theory

Z: collision frequencyp: orientation probability factor(the fraction of collisions with proper orientation)f: fraction of collisions with sufficient energy

Po

ten

tia

l En

erg

y

Ea

Reactants

Products

Lecture 32 - 14

Reaction Rate Depends on TemperatureChemiluminescent reaction

The Cyalume stick glows more brightly in the hot water beaker because the reaction is faster.

The glow lasts for longer in a cold beaker, because the reaction is slower.

Lecture 32 - 15

The Arrhenius Equation

RTEAk /- a e

k

Ea

k = A

k ~ 0

The Arrhenius equation describes the temperature

dependence of the rate constant, k:

Lecture 32 - 16

Consider the decomposition of H2O2:

2 H2O2(aq) 2 H2O(l) + O2(g)

k (s–1) T(K)

7.77 x 10–6 298

Rate increases exponentially with temperature. Temperature

Rat

e co

nst

ant,

k

Rate and Temperature

Lecture 32 - 17

To calculate Ea, rearrange the Arrhenius equation (k = A e – Ea / R T) as:

ln k = ln A – Ea / R T

If k1 and k2 are the rate constants at two temperatures T1 and T2 respectively, then we can also calculate Ea easily:

ln k1 = ln A – Ea / R T1 ln k2 = ln A – Ea / R T2

And subtracting the two, the term (ln A) disappears, so we get:

Using Arrhenius Equation

212

1 11ln

TTR

E

k

k a

Lecture 32 - 18

For the reaction:

2 NO2(g) 2NO(g) + O2(g)

The rate constant k = 1.00 · 10-10 s-1 at 300 K and the activation energy Ea = 111 kJ mol-1.

(a) What are the values for A and k at 273 K?

(b) What is the value of T when k = 1.00· 10-11 s-1?

Method: Make use of, and rearrange k = A e – Ea / R T

Arrhenius Equation – Example 1

Lecture 32 - 19

(a) Find the value of A (independent of T):

A = k eEa / R T

= 1.00· 10-10 s-1 · exp [111000 J mol-1 / (8.314 J mol-1 K –1· 300 K)] = 2.13 ·109 s-1 (three significant figures)

Calculate the value of k at 273 K:

k = 2.13 · 109 s-1 exp (– 111000 J mol-1) / (8.314 J mol-1 K –1·273 K) = 1.23 · 10-12 s-1 (three significant figures)

(b) Calculate the temperature when k = 1.00· 10-11s-1

T = Ea / [R· ln (A/k)] = 111000 J mol-1 / (8.314·46.8) J mol-1 K-1 = 285 K (three significant figures)

Arrhenius Equation – Example 1

Lecture 32 - 20

Increasing the temperature increases the number of collisions with sufficient energy to react, i.e. with energy > Ea.

Distribution of Collision Energy

Blackman Figure 14.13

Lecture 32 - 21

Energy Landscape in Chemical Reactions

A + B

C + D

Activatedstate

Ea (forw)

Ea

(rev)

Exothermic reaction Endothermic reaction

A + B

C + D

Ea (forw)

Ea

(rev)

A + B C + D

Forward reaction is faster than reverse Reverse reaction is faster than forward

Figure from

Silberberg, “C

hemistry”,

McG

raw H

ill, 2006.

Lecture 32 - 22

If a reaction has a rate constant k of 2.0 · 10-5 s-1 at 20.0C and 7.32·10-5 s-1 at 30.00C, what is the activation energy ?

Answer:

ln {(2.0 · 10-5)/ (7.32 · 10-5)} = - (Ea/8.314) (1/293 - 1/303.00)

Ea = 96 kJ mol-1 (TWO SIGNIFICANT FIGURES)

212

1 11ln

TTR

E

k

k a

Arrhenius Equation – Example 2

Lecture 32 - 23

Determining EaIdeally you would require many more than two values to determine Ea.

ln k = ln A – Ea / R T

Plot lnk versus 1/T and get a line with a slope of –Ea/R.

Lecture 32 - 24

The rate constant of a particular reaction triples when the temperature is increased from 25 C to 35 C. Calculate the activation energy, Ea, for this reaction.

ln (1/3) = - (Ea / 8.314)(1/298 - 1/308)

-1.099 = - Ea(1.310 x 10-5)

Ea = 8.4·104 J mol-1 or 84 kJ mol-1

212

1 11ln

TTR

E

k

k a

Arrhenius Equation – Example 3