Reaction order• The rate law can be written in a generalized form:

v = k [A]a[B]b…. where a is the order of the reaction with respect to the

species A, and b is the order of the reaction with respect to the reagent B.

• The reaction order is (a + b +… . ).

• The order of a chemical reaction needs not to be an integral. Certain reactions do not have an overall order !!!

Example 1: v = k [A]1/2[B]1

Example 2: v = k (zero order reaction, such as ……)

• How to determine the unit of k?

Determination of the rate law

• Isolation method:

v = k [A]a[B]b -----> v = k’[B]b

• Method of initial rates (often used in conjunction with the isolation method):

v = k [A]a

at the beginning of the reaction

v0 = k [A0]a

taking logarithms gives:

logv0 = log k + a log[A0]

therefore the plot of the logarithms of the initial rates against the logarithms of the initial concentrations of A should be a straight line with the slope a (the order of the reaction).

Self-test 22.3: The initial rate of a reaction depended on the concentration of a substance B as follows:

[B]0/(mmol L-1) 5.0 8.2 17 30

v0/(10-7 mol L-1s-1) 3.6 9.6 41 130

Determine the order of the reaction with respect to B and calculate the rate constant.

Solution:

Log([B]0) -2.30 -2.086 -1.770 -1.523

Log(v0) -6.444 -6.018 -5.387 -4.886

-7

-6

-5

-4

-3

-2

-1

0

-2.5 -2 -1.5 -1 -0.5 0

Series1

22.3 Integrated rate law

• First order reaction:

A Product

The solution of the above differential equation is:

or: [A] = [A]0e-kt

• In a first order reaction, the concentration of reactants decreases exponentially in time.

][][

Akdt

Ad

ktA

A

0][

][ln

Self-test 22.4: In a particular experiment, it was found that the concentration of N2O5 in liquid bromine varied with time as follows:

t/s 0 200 400 600 1000

[N2O5]/(mol L-1) 0.110 0.073 0.048 0.032 0.014

confirm that the reaction is first-order in N2O5 and determine the rate constant.

Solution: To confirm that a reaction is first order, plot ln([A]/[A]0) against time and expect a straight line:

t/s 0 200 400 600 1000

ln([A]/[A]0) 0 -0.410 -0.829 -1.23 -2.06

-2.5

-2

-1.5

-1

-0.5

0

0 200 400 600 800 1000 1200

Series1

Half-lives and time constant

• For the first order reaction, the half-live equals:

therefore, is independent of the initial concentration.

• Time constant, , the time required for the concentration of a reactant to fall to 1/e of its initial value.

for the first order reaction.

kt

221

ln/

k

1

Second order reactions

• Case 1: second-order rate law: (e.g. A → P)

• Can one use A + A → P to represent the above process?

• The integrated solution for the above function is:

or

• The plot of 1/[A] against t is a straight line with the slope k.

2][][

Akdt

Ad

ktAA

0

11

][][

0

0

1 ][

][][

Akt

AA

021

1

][/ Akt

• Case 2: The rate law (e.g. A + B → Product)

• The integrated solution (to be derived on chalk board) is :

]][[][

BAkdt

Ad

ktABAA

BB)][]([

]/[][

]/[][ln 00

0

0

22.4 Reactions approaching equilibrium

Case 1: First order reactions:A → B v = k [A]B → A v = k’ [B]

the net rate change for A is therefore

if [B]0 = 0, one has [A] + [B] = [A]0 at all time.

the integrated solution for the above equation is [A] =

As t → ∞, the concentrations reach their equilibrium values: [A]eq = [B]eq = [A]0 – [A]eq =

]['][][

BkAkdt

Ad

])[](['][][

AAkAkdt

Ad0 0]['])['( AkAkk

0]['

' )'(

Akk

kek tkk

'

]['

kk

Ak

0

'

][

kk

Ak

0

• The equilibrium constant can be calculated as K =

thus:

• In a simple way, at the equilibrium point there will be no net change and thus the forward reaction will be equal to the reverse reaction:

k[A]eq = k’ [B]eq

thus

the above equation bridges the thermodynamic quantities and reaction rates through equilibrium constant.

• For a general reaction scheme with multiple reversible steps:

eq

eq

A

B

][

][

'k

kK

'][

][

k

k

A

B

eq

eq

...''2

2

1

1

k

k

k

kK

Determining rate constants with relaxation method

• After applying a perturbation, the system (A ↔ B) may have a new equilibrium state. Assuming the distance between the current state and the new equilibrium state is x, one gets

[A] = [A]eq + x; [B] = [B]eq - x;

Because one gets dx/dt = - (ka + kb)x

therefore

is called the relaxation time

xkkxBkAxkdt

Adbaeqbeqa )()]([)][(

][

dt

dx

dt

Ad

][

/texx 0ba kk

1

Example 22.4: The H2O(l) ↔ H+(aq) + OH-(aq) equilibrium relaxes in 37 μs at 298 K and pKw = 14.0. Calculate the rate constants for the forward and backward reactions.

Solution: the net rate of ionization of H2O is

we write [H2O] = [H2O]eq + x; [H+] = [H+]eq – x; [OH-] = [OH-]eq – x

and obtain:

Because x is small, k2x2 can be ignored, so

Because k1[H2O]eq = k2[H+]eq[OH-]eq at equilibrium condition

= =

hence k2 = 1.4 x 1011 L mol-1 s-1

k1 = 2.4 x 10-5 s-1

]][[][][ OHHkOHk

dt

OHd221

2

2222121 xkOHHkOHkxOHHkk

dt

dxeqeqeqeqeq ][][][)][]([

)][]([ eqeq OHHkk 211

eq

eqeq

OH

OHH

k

k

][

][][

22

1

eq

w

OH

LmolK

][

)(

2

21

1

655Lmol

K w

.

)()][][( // 212122

1wweqeq KKKkOHHKk

• Self-test 22.5: Derive an expression for the relaxation time of a concentration when the reaction A + B ↔ C + D is second-order in both directions.

To be demonstrated on in class

22.5 The temperature dependence of reaction rates

• Arrhenius equation:

A is the pre-exponential factor; Ea is the activation energy. The two quantities, A and Ea, are called Arrhenius parameters.

• In an alternative expression

lnk = lnA -

one can see that the plot of lnk against 1/T gives a straight line.

RTEaeAk /

RT

Ea

Example: Determining the Arrhenius parameters from the following data:

T/K 300 350 400 450 500

k(L mol-1s-1) 7.9x106 3.0x107 7.9x107 1.7x108 3.2x108

Solution:

1/T (K-1) 0.00333 0.00286 0.0025 0.00222 0.002

lnk (L mol-1s-1) 15.88 17.22 18.19 18.95 19.58

The slope of the above plotted straight line is –Ea/R, so Ea = 23 kJ mol-1.

The intersection of the straight line with y-axis is lnA, so A = 8x1010 L mol-1s-1

0

5

10

15

20

25

0 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035

Series1

The interpretation of the Arrhenius parameters

• Reaction coordinate: the collection of motions such as changes in interatomic distance, bond angles, etc.

• Activated complex

• Transition state

• For bimolecular reactions, the activation energy is the minimum kinetic energy that reactants must have in order to form products.

Applications of the Arrhenius principle

Temperature jump-relaxation method:

consider a simple first order reaction:

A ↔ B

at equilibrium:

After the temperature jump the system has a new equilibrium state. Assuming the distance between the current state and the new equilibrium state is x, one gets [A] = [A]eq + x; [B] = [B]eq - x;

][]['][ ' BkAk

dt

Adba

0dt

Ad ][

''' ][][' eqbeqa BkAk

22.6 Elementary reactions

• Elementary reactions: reactions which involves only a small number of molecules or ions.

A typical example:

H + Br2 → HBr + Br

• Molecularity: the number of molecules coming together to react in an elementary reaction.

• Molecularity and the reaction order are different !!! Reaction order is an empirical quantity, and obtained from the experimental rate law; molecularity refers to an elementary reaction proposed as an individual step in a mechanism. It must be an integral.

• An elementary bimolecular reaction has a second-order rate law:

A + B → P

• If the reaction is an elementary bimolecular process, then it has second-order kinetics; However, if the kinetics are second-order, then the reaction might be complex.

]][[][

BAkdt

Ad

22.7 Consecutive elementary reactions

• An example:

239U → 239Np →239Pu

• Consecutive unimolecular reaction

A → B → C

The rate of decomposition of A is:

• The intermediate B is formed from A, but also decays to C. The net rate of formation of B is therefore:

• The reagent C is produced from the unimolecular decay of B:

][][

Akdt

Ad1

][][][

BkAkdt

Bd21

][][

Bkdt

Cd2

• Integrated solution for the first order reaction (A) is:

• Then one gets a new expression for the reactant B:

the integrated solution for the above equation is:

when assuming [B]0 = 0.

• Based on the conservation law [A] + [B] + [C] = [A]0

tkeAA 10

][][

][ e[A]][

1k-0 Bkk

dt

Bd t21

012

1 21 ])[(][ Aeekk

kB tktk

012

2112

1 ][][ Akk

ekekC

tktk

Example. In an industrial batch process a substance A produces the desired compound B that goes on to decay to a worthless product C, each step of the reaction being first-order. At what time will B be present in the greatest concentration?

Solution: At the maximum value of B

Using the equation 25.7.6 and taking derivatives with respect to t:

In order to satisfy

= 0

tmax =

The maximum concentration of B can be calculated by plugging the tmax into the equation.

0dt

Bd ][

12

210121

kk

ekekAk

dt

Bd tktk

)(][][

0dt

Bd ][

tktk ekek 2121

2

1

21

1

k

k

kkln