Chemical kinetics is the study of the rates of chemical reactions,
how reaction rates change under
varying conditions, and what molecular events occur during the overall
reaction.
The rate of a chemical reaction is expressed in molarity/time.
When time is in seconds,
the units are: M/s = M.s-1
= mol/L.s.
The unit M/s (or M.s-1)
is read as molarity per second and mol/L.s
as moles per
liter per second.
The rate of a chemical reaction is dependent upon:
a) the concentration of the reactants.
b) the temperature of the reaction.
c) the nature of the reactants (solid, liquid, gas).
The greater the surface area per unit volume, the faster the reaction.
d) the presence of a catalyst.
REACTION RATE
REACTION RATE is the change in concentration of reactant or product with time. This is the increase in molar concentration of product of a reaction per unit time or the decrease in molar concentration of reactant per unit time.
Consider the following reaction: 2 A(g) ---> B(g)
The rate of reaction can be defined in either of two terms:
1. increase in concentration of PRODUCT. Rate = D[B]/Dt
[B] = molarity of B
D[B] = change in concentration = [B]final - [B]initial
Dt = a given time interval.
2. decrease in concentration of REACTANT.
Rate
= -D[A]/Dt
Stoichiometry must be considered when expressing rates of reactions. Because two moles of A decompose for every one mole of B being formed in a given amount of time the rates are not equal. The decomposition of A is twice the rate of formation of B. Or the rate of formation of B is 1/2 rate of decomposition of A.
Rate = D[B]/Dt = -1/2 D[A]/Dt
These two expressions known as the GENERAL RATE OF REACTION, have the same value, regardless of which reactant or product is being studied and are derived by dividing the rate of disappearance of a reactant or the rate of formation of a product by the stoichiometric coefficient of that reactant or product in the balanced equation.
Applied to a more general reaction: aA + bB ---> cC + dD
Rate = -1/a D[A]/Dt
= -1/b D[B]/Dt =
1/c D[C]/Dt = 1/d
D[D]/Dt
Time must also be considered and specified when quoting a rate because rate changes as the reaction proceeds. The general reaction rate gives the average rate over a time interval Dt.
The instantaneous rate is the rate at a particular time.
The smaller the time intervals, the closer the average rate will approach the instantaneous rate.
If the time interval is very short, the equation gives the instantaneous rate.
The instantaneous rate is the slope of the tangent to a concentration-versus-time curve.
EXAMPLE #1: CALCULATING AVERAGE REACTION RATE
Consider the following reaction.
I1-(aq) + ClO1-(aq) ---> IO1-(ag) + Cl1-(aq)
At a certain constant temperature, the iodide, I1- , concentration was analyzed and found as follows:
TIME
[I1-]
3.00 s 0.0231
M
7.00 s 0.0168
M
Calculate the average rate of reaction of I1- during the time interval.
-D[I1-]/Dt
= -(0.0168 M - 0.0231 M)/(7s - 3 s) = 1.575 x 10-3 M/s
EXAMPLE #2: CALCULATING AVERAGE REACTION RATE
The following redox reaction was carried out at 25.0oC.
2 FeCl3(aq) + SnCl2(aq) ---> 2 FeCl2(aq)
+ SnCl4(aq)
The net ionic equation is as follows. 2 Fe3+(aq) +
Sn2+(aq) ---> 2 Fe2+(aq) + Sn4+(aq)
The concentration of Fe3+ ion at the beginning of a an experiment
was determined and then again after 4.50 min.
The data is shown below. What is the average rate of reaction
of FeCl3?
TIME
[FeCl3]
0 min
0.04662 M
4.50 min 0.03429 M
-D[FeCl3]/Dt
= -(0.03429 M - 0.04662 M)/(4.50 min) = 2.74 x 10-3 M/min
EFFECT OF CONCENTRATION ON RATE OF REACTION
RATE LAW is an equation that relates the rate of a reaction to the concentration of reactants (and catalyst) raised to various powers.
GENERAL FORM of RATE LAW
C
aA + bB -----> dD + eE
[ C = catalyst ]
Rate = k[A]m[B]n[C]p
The exponents m, n, and p are frequently, but not always integers, and
must be determined experimentally.
| NOTE: Neither the concentration
of a given reactant nor the rate of reaction are constant.
However, the ratio of concentration to rate is a constant: Rate/[A] = k or Rate = k[A]m |
For a homogenous chemical reaction, it is usually possible to express the rate as an algebraic function (rate law) of the concentrations of the reactants. In most reactions, this mathematical relationship (rate law) is simply the product of the concentration of the reactants, with each concentration term being raised to an exponent power. The exponent is called the order of the reaction with respect to that particular reactant. The overall order of the reaction is the sum of the exponents from all the concentration terms. Moreover, the exponents on the concentration terms cannot by predicted by the balanced overall chemical reaction. The exponential values are determined only from experimental kinetic data.
The rate law relates the concentration dependence of the reactants to the rate of the chemical reaction. The rate law for the reaction the following reaction is shown below. 2 A(g) ---> B(g)
Rate = k [A]m [B]n
[] = molarity; m,n = order of each reactant; k = rate constant.
The numerical value of k depends on:
a. The particular reaction.
b. The temperature.
c. The presence of a catalyst.
The units of k depend on the form of the rate law.
REACTION ORDER
Reaction order the exponent of the concentration of that species in the rate law; as determined experimentally.
The overall order of a reaction equals the sum of the orders of the reactant species in the rate law.
| Cyclopropane ----> Propylene
Rate = k[Cyclopropane] First order in Cyclopropane. |
| 2NO(g) + 2 H2(g) ---> N2(g) + 2 H2O(g)
Rate = k[NO]2[H2] Second order in NO (nitric oxide). First order in hydrogen. Third order overall. |
|
H1+
CH3COCH3(aq) + I2(aq) ---------> CH3COCH2I(aq) + HI(aq) Acetone Rate = [Acetone][H1+] First order in acetone. Zero order in iodine; that is the rate law contains [I2]0 = 1 First order in H1+ Second order overall. |
Although reaction orders frequently have whole-number values (particularly
1 or 2), they can be fractional. Zero and negative orders are also possible.
| In general as m increase the rate increases
as [A] increases.
If m is zero the rate is unaffected by [A] increases. If m is negative, the rate decreases as [A] increases. |
MAKING SENSE OUT OF REACTION ORDER
Consider the following reaction and rate law: 2A ---> B
Rate = k[A]m
| m | RXN ORDER | Effect of doubling [A] on rate | Rate = k[A]m = X M/s |
| -1 | Decrease by 0.5 | Rate' = k[2A]-1 = 0.5X M/s | |
| 0 | Zero | None | Rate' = k[2A]0 = X M/s |
| 1/2 | One half-order | Increase by 1.4 | Rate' = k[2A]0.5 = 1.4X M/s |
| 1 | First-order | Increase by 2 | Rate' = k[2A]1 = 2X M/s |
| 2 | Second-order | Increase by 4 | Rate' = k[2]2 = 4X M/s |
| 3 | Third-order | Increase by 8 | Rate' = k[2]3 = 8X M/s |
DETERMINING THE RATE LAW
One way to determine the rate law is a procedure called the method of initial rates.
A series of experiments are set up in which the initial concentrations of some reactants are held constant and others are varied in convenient multiples.
EXAMPLE 1: DETERMINING THE RATE LAW
For the reaction, A + B ---> C, the rate law is:
Rate = k[A]m[B]n
Rate data for the reaction was collected as follows:
Run [A]
[B] Rate
1. 0.10
0.01 1.2 x 10-3
2. 0.10
0.04 4.8 x 10-3
3. 0.20
0.01 2.4 x 10-3
Because the rate in Run 2 increased fourfold when [B] was increased by a factor of 4 the value of n is 1.
The reaction is first-order in B.
Because the rate in run 3 increased by a factor of 2 when [A] was increased by a factor of 2, the value of m is 1.
The reaction is first-order in A.
Rate = k[A][B]
The rate constant can now be calculated using any set of data.
EXAMPLE 2: DETERMINING THE RATE LAW
Consider the following reaction: A + 3 B ---> C
The rate law is: Rate = k[A]m[B]n
In a set of 4 experiments the following data was obtained.
| Experiment | [A]initial | [B]initial | Initial rate of formation of C (M/s) |
| 1 | 0.02 | 0.02 | 0.025 |
| 2 | 0.04 | 0.02 | 0.100 |
| 3 | 0.02 | 0.04 | 0.050 |
| 4 | 0.04 | 0.04 | 0.200 |
(Initial rate)1 = k [0.02]m[0.02]n = 0.025 M/s
(Initial rate)2 = k [0.04]m[0.02]n = 0.100 M/s
Dividing Rate2 by Rate1 gives:
(Initial rate)2
k [0.04]m[0.02]n
0.100 M/s
=
=
= 4
(Initial rate)1
k [0.02]m[0.02]n
0.025 M/s
Canceling quantities which are the same gives: [0.04]m/ [0.02]m = [2]m = 4
Therefore, m = 2.
| NOTE:
ln Na = a ln N ln [2]m =(m)(ln 2)= ln 4 m = 2 |
Dividing Rate3 by Rate1 gives:
(Initial rate)3
k [0.02]m[0.04]n
0.050 M/s
=
=
= 2
(Initial rate)1
k [0.02]m[0.02]n
0.025 M/s
[0.04]n/[0.02]n = [2]n = 2
Therefore, n = 1.
The reaction is second-order with respect to A and first-order with respect to B and third-order overall.
Rate = k[A]2[B]
The rate constant, k can now be calculated by using any of the four sets of data and plugging in the given values.
Using the data from experiment 1: 0.050 M/s = k[0.02]2[0.04]1
Solving for k gives: k = 3125 M-2s-1
RELATION BETWEEN REACTANT CONCENTRATION AND TIME
By applying calculus the rate law can be changed into another useful mathematical form. The derived mathematical equation will depend on the order of the rate law.
These equations allow the concentration to be predicted at any given time during the progress of a reaction. The time required for a reaction to proceed to any percent completion can be calculated.
First-Order Rate Law
First order rate law can be changed into: ln([A]o)/[A]t)
= kt or log [A]t = -kt/2.303 + log [A]o
| EXAMPLE 1 FIRST-ORDER
Consider the reaction A ---> B + C which follows first-order kinetics. If the rate constant is 4.50 x 10-4 s-1, what is the concentration of A after 5 minutes if the initial concentration is 0.100 M? ln([A]o)/[A]t) = kt ln([0.100])/[A]t) = (4.50 x 10-4 s-1)(300 s) [0.100])/[A]t = 1.145 [A]t = 0.0874 M |
| FIRST-ORDER EXAMPLE
How long will it take for the concentration from Example 1 to go from 0.1 M to 0.03 M? ln([A]o)/[A]t) = kt ln([0.100])/[0.03]) = (4.50 x 10-4 s-1)(t) t = 2675 s = 44.6 min |
Second-Order Rate Law
Second order rate law can be changed into: 1/[A]t
- 1/[A]o = kt
| Second-Order Example
A second order reaction starts with an initial concentration of 0.020 M of the reactant. If the rate constant is 10 L/mol.sec, calculate the time required to decrease the concentration to 0.0050 M? 1/[A]t - 1/[A]o = kt 1/[0.0050] - 1[0.020] = (10 L/molsec)t t = 15.0 sec |
HALF-LIFE OF A REACTION
First-Order Half-Life
The half-life, t1/2, of a reaction is the time that it takes
for the reactant concentration to decrease to one-half its
initial value.
t1/2 = 0.693/k
| FIRST-ORDER HALF-LIFE EXAMPLE
The half-life for the first-order decomposition of N2O5 at 67oC is 30.3 seconds. If the initial concentration of N2O5 is 0.160 M, then determine the concentration of N2O5 after 90.8 seconds. A. Calculate the rate constant. t1/2 = 0.693/k k = 0.693/30.3 s = 2.287 x 10-2 s-1 B. Calculate the concentration. ln([A]o)/[A]t) = kt ln([0.160])/[N2O5]) = (2.287 x 10-2 s-1)(90.8 s) [N2O5] = 0.0201 M |
Radioactive decay is a first-order process and the half-life formula can be applied.
Second-Order Half-Life
t1/2 = 1/(k[A]o)
Second-order half-life depends on concentration and becomes larger as the reaction progresses. The value of the half-life for a second-order reaction is twice as long as the preceding one. The fact that the half-life changes with time is evidence that the reaction is not first-order.
STUDY PROBLEMS
1. The oxidation of ammonia occurs via
the reaction, 4NH3(g) + 3O2(g) ----> 2N2(g)
+ 6H2O(g)
If the rate
of formation of N2 is 2.0 mol/ltr.sec then calcuate the rate
of H2O formation.
(a) 6.0 mole/ltr.sec (b) 0.333 mole/ltr.sec (c) 12.0 mole/ltr.sec (d) 4.0 mole/ltr.sec
2. For the reaction and data below calculate
the rate constant.
NO2(g)
+ O3(g) ----> NO3(g) + O2(g)
[NO2]
[O3]
rate
5.0 x 10-5
1.0 x 10-5 0.022
5.0 x 10-5
2.0 x 10-5 0.044
2.5 x 10-5
2.0 x 10-5 0.022
(a) 8.8 x 1011 (b) 1.2 x 105 (c) 4.4 x 107 (d) 2.2 x 104
3. The reaction 2NO(g) +
1H2(g) ---> N2(g) + 2H2O(g)
follows the rate law, rate= k [NO]2 [CH2]
From the information,
we can predict that the initial rate of reaction will double if:
(a) the initial CNO is doubled,
while initial CH2 remains constant.
(b) the initial CH2 is
doubled, while initial CNO remains constant.
(c) the initial CNO and
CH2 are both doubled
(d) the temperature is doubled, without
changing initial concentrations.
(e) the initial reactants are mixed
in the ratio of 1 mole NO: 1 mole H2
4. The recombination of iodine atoms
to form molecular iodine follows second order kinetics. If the initial
concentration of I
was 0.086
M and after 2.0 minutes the concentration is 1.2 X 10-12 M,
find the rate constant.
2 I(g) ----> I2(g)
(a) 0.12 min-1
(b) 0.12 moles
liter-1 min-1
(c) 6.9 X
109 liter mole-1 sec-1
(d) 5.4 X
10-8 liter mole-1 hour-1
(e) 6.9 X
109 liter2 mole-2 sec-1
5. Nitric oxide reacts with hydrogen at 1000 K according to the reaction:
2 NO(g) + 2 H2(g) ----> N2(g) + 2 H2O(g)
The rate law
is rate = k[NO]2[H2]. When time is in minutes the
units of k are:
(a) moles/ltr.min
(b) moles/ltr.min2 (c) ltr.2/moles2min
(d) moles2/ltr.min3
6. BA decomposes at 600o
C with a rate constant of 4.5/min. How long will it take for the
concentration of BA to
decrease from
0.40 M to 0.050 M?
(a) 9.4 min (b) 0.46 min (c) 0.21 min (d) 6.5 min
7. The half-life for the first order
decomposition of N2O5 at 67o C is 30.3
seconds. If the initial concentration N2O5 is
0.160 M, then
determine the concentration of N2O5 after 90.8 seconds.
(a) 0.0200
(b) 0.0845 (c) 0.00341
(d) 0.104
8. A second order reaction starts with
an initial concentration of 0.020 M of the reactant. If the rate constant
is 10 ltr/mol
sec, calculate
the time required to decrease the concentration to 0.0050 M?
(a) 0.069 sec (b) 45.0 sec (c) 23.2 sec (d) 15.0 sec
ANSWERS
1. a, 2. c,
3. b, 4. c, 5. c,
6. b, 7. a, 8. d