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# DIFFERENTIAL RATES OF REACTION - PowerPoint PPT Presentation

aA + bB cD + dD. DIFFERENTIAL RATES OF REACTION. Rate = d[B]/dt. Rate’ = -d[A]/dt. -d[A]/dt = (1/2)d[B]/dt. Net rate = forward rate – reverse rate. THE EMPIRICAL RATE LAW (EXPERIMENTAL). aA + bB ----> cC + dD.

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aA + bB cD + dD

DIFFERENTIAL RATES OF REACTION

Rate = d[B]/dt

Rate’ = -d[A]/dt

-d[A]/dt = (1/2)d[B]/dt

Net rate = forward rate – reverse rate

aA + bB ----> cC + dD

It is often (but not always) possible to express the rate of a reaction in the form

• k is the rate constant; independent of [A],[B],[C],[D], but dependent on T

• m,n are the reaction “orders” with respect to the reagents; usually integers

• m + n is the “overall order”

• In general, the orders bear no relationship to the stoichiometry of the reaction. That is, m, n are not related to a and b. The orders m, n must be experimentally determined.

A + B ----------> C

INITIAL RATE = Ri = C/t  k[A0]m[B0]n

where t is chosen sufficiently small near t = 0 so that

[A]  [A0] and [B]  [B0]

RATE = k [A][B]2

First order

A------> Products

(1)

k has units of time-1

First order reactions have some interesting and unique properties. For example, write the rate law in the form:

(2)

A constant fraction of A disappears per unit time = k

Equation (1) may be directly integrated between time 0 and t with the result

(3)

(4)

First order rates can be characterized by a "half-life": By definition, the half-life, t1/2, is the time at which [A] = [A]o /2.

Substituting this condition in the integrated rate law, we find

(5)

Thus the time it takes the reaction to proceed 50% is independent of the initial amount of material! 1Kg or 1mg will react 50% in the same time

A------> Products

(6)

k has units conc-1time-1

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

The rate constant depends on the units of concentration, and the probability of reaction per unit time is not a constant throughout the reaction, but depends on the amount of [A] present at any time

-d[A]/[A] 1/dt = k [A]

Equation (6) can be directly integrated from [A]o to [A] and t=0 to t to give

(7)

The half-life may be derived in the same way as for first order

(not so useful; depends on [A]o)

A + B ------> products (first order in each, second order overall)

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

If [A]o = [B]o integration gives the same result as above, ie, for either A or B, 1/[A] - 1/[A]o = kt

If [A]o ≠ [B]o, the result is

Where x is the amount of A or B reacted at any time.

4HBr + O2 2Br2 + 2H2O

N2O5

2NO2 + 1/2O2

Rate = -d[

N2O5]/dt = k’[

N2O5]

N2O5

NO2 + NO3

NO2 + NO3

NO + O2 + NO2

NO + NO3

2NO2

STOICHIOMETRY AND MECHANISMS OF CHEMICAL REACTIONS

Stoichiometric equations

Stoichiometric equation

Stoichiometric equations do not imply the mechanism of the reaction; the above equation does not imply that the reaction occurs by the simultaneous collision of 4 molecules of HBr and 1 of O2. That would be most improbable due to the low probability of a simultaneous collision of 5 molecules!

Reaction mechanisms

Consider the “simple” reaction

Stoichiometric equation

Empirical rate law

“elementary

steps”

Mechanism

N2O5]/dt = k[

N2O5]

PROPERTIES OF ELEMENTARY REACTIONS (STEPS)

1. Elementary reactions are classified according to “molecularity”, the number of molecules involved in the reaction

• unimolecular: one molecule of reactant

• bimolecular: two molecules of reactant

• termolecular: three molecules of reactant

2. The pathway of an elementary step is known; it occurs as written. Thus, the rate law can be derived theoretically

unimolecular

N2O5

NO2 + NO3

(forward rate only; reverse is bimolecular)

A unimolecular step is first order. However, a first order reaction is not necessarily unimolecular

NO2 + NO3

NO + O2 + NO2

Bimolecular

A direct collision between NO2 and NO3 is required

Rate = (# of collisions per second/cm3)x(probability of a reaction per collision)

# of collision per second/cm3 = a[NO2][NO3]

probability of a reaction per collision = b (a constant)

Rate = k[NO2][NO3]

A bimolecular elementary step is second order, first order in each reactant

Termolecular

Similarly, for a termolecular process

A + B + C products

Rate = k[A][B][C]

A termolecular process is third order, first order in each reactant (very rare).

Reversible unimolecular

Parallel unimolecular

Consecutive unimolecular

Consecutive reversible unimolecular

1. A connection between kinetics and thermodynamics

The rate law

(1)

Net rate = forward rate - reverse rate

At equilibrium

d[A]/dt = 0

k1[A]eqm = k-1[B]eqm

(2)

For any number of intermediates

2. The principle of microscopic reversibility (PMR)

Consider an overall reversible process

A possible mechanism

The PMR specifically disallows this mechanism

"Each and every elementary step in a reversible reaction must have equal forward and reverse rates at equilibrium"

Another possible mechanism

Uncatalyzed pathway

Catalyzed pathway (C is the catalyst)

From thermodynamics, it is clear that the catalyst cannot change the equilibrium constant

K = [B]eqm/[A]eqm

In the presence of the catalyst, the reaction can proceed by both pathways, and the overall rate is

At equilibrium, -d[A]/dt = 0

The equilibrium constant should not involve C; what’s wrong? We have stipulated only that the overall rate at equilibrium is = 0, but the PMR requires that the rate of each pathway be = 0 at equilibrium!

For each step having equal forward and reverse rates:

Uncatalyzed pathway

Catalyzed pathway

PARALLEL UNIMOLECULAR wrong? We have stipulated only that the overall rate at equilibrium is = 0, but the PMR requires that the rate of each pathway be = 0 at equilibrium!

1. The rate law

A disappears with first order kinetics, with a rate constant k = k1 + k2. The overall rate is determined by the fastest step.

At the completion of the reaction, the ratio of products is