Chemical kinetics
Kinetics is the study of the
rates at which reactions occur – how quickly reactants are converted to
products
·
Examples of fast
reactions: combustion of hydrogen gas, neutralization of a strong acid with a
strong base, etc.
·
Examples of slow
reactions: rusting of iron, ripening of fruit, etc.
Rates of reactions vary based
on reaction conditions.
·
Ex. Coal does not
burn (react with oxygen) at room temperature, but once heated up the reaction
proceeds very rapidly.
·
Since reactions
occur as the result of molecular collisions, factors which affect collisions
will also affect reaction rates
Reaction rates are expressed
in the units of mol/(L·s) …
different units of time can be used depending on how slow or fast the reaction
is.
·
Written as an
expression of how M conc. of a reactant or product changes over time …
D[A]/Dt where A is a reactant or
product.
·
In order to study
reaction kinetics, the conc. of a reactant or product must be able to be
measured over the course of the reaction
·
Spectrophotometers
can be used to make these measurements for some reactions
For the generic reaction aA + bB à cC + dD
·
a, b, c, and d are
coefficients
·
A and B are
reactants, C and D are products
·
As the reaction
proceeds in the direction indicated, concentrations of A and B decrease while
concentrations of C and D increase
·
D[A]/Dt and D[B]/Dt would have negative values while D[C]/Dt and D[D]/Dt would have positive values
·
All of these
changes are related to each other by the coefficients … for every a mol/L of
decrease in [A] there must be b mol/L decrease in [B], c mol/L increase in [C],
and d mol/L increase in [D]
·
Rate of reaction
refers to the number of moles of reaction that occur per unit time … for
every a mol of A used up, 1 mol of rxn occurs, etc.
·
Rate of reaction is
always a positive number, so rates of reactant decrease must be multiplied by –1
to express rate of reaction.
·
Rate of rxn = (-1)
1 mol rxn
D[A] =
-D[A]
a mol A
Dt
a(Dt)
·
=
-D[B] =
D[C] =
D[D]
b(Dt)
c(Dt)
d(Dt)
For the reaction N2
+ 3 H2 à 2 NH3
Rate of reaction =
-D[ N2] = -D[ H2] =
D[ NH3]
(Dt) 3(Dt) 2(Dt)
Write expressions for the rate of the following rxn:
H2 (g)
+ ICl (g)
à I2 (g) + 2 HCl
(g)
·
These expressions
refer to an average rate over the unit time being studied.
The nature of the reactants
affects the rate of the reaction.
·
Physical state:
gases tend to react more rapidly than liquids; aqueous solutions of solids react
more rapidly than dry solids; more thoroughly mixed liquids react faster than
layered or immiscible liquids, etc.
·
Degree of
subdivision can play an important role for liquids and
solids.
·
In all of these
cases increasing the amount of surface area of reactants exposed to each other
increases the rate of the reaction.
Concentration of reactants
affects reaction rate
·
The rate law
equation shows a relationship between concentration of reactants and the rate of
the reaction.
·
Rate =
k[A]x[B]y …
k =
the specific rate constant for the rxn at a particular
temperature
A,
B, etc. are reactants
x, y, etc.
describe the relationship to conc.
·
Note: values for k,
x, y, etc. must be determined experimentally and are not necessarily related to
any coefficients from the equation.
·
Significance of
exponent values:
Exponent = 1, rate is directly proportional to conc. If
conc doubles, rate doubles.
Exponent = 2, rate is directly proportional to the
square of conc. If conc doubles, rate quadruples.
Exponent = 0, rate is independent of conc. The reactant
is usually not shown in the rate law.
·
Order of reaction:
the value of x is said to be the order of the reaction with respect to reactant
A (if x=2 the rxn is second order with respect to A), etc.
·
The overall order
of the reaction is the sum of all the exponents in the rate
law.
·
H2O2 (aq) + 3 I- (aq) + 2 H+ (aq) à 2 H2O (l) +
I3- (aq) rate =
k[H2O2][I-]
·
This reaction is
1st order with respect to H2O2, 1st
order with respect to I-, and zero order with respect to
H+. Overall, the
reaction is 2nd order.
·
Note: the value of
k would have to be calculated from experimental data, units for k will depend on
the overall order of the reaction so that rate units = mol/L·time
The rate law can be deduced
from experimental data using the method of initial rates.
·
Use experimental
data to find the ratio of change in rate to change in
conc.
·
Solve for k using
one set of data.
·
Check value for k
using another set of data
See problems 21, 25, and 26 pp.
598-599
Integrated Rate
Equation
·
Relates
concentration and time
·
Can also be used to
calculate half-life – the time it takes for ½ of a reactant to be converted to
product
·
Differs based on
order of reaction.
For first order reactions in A
and first order overall:
ln
([A]o/[A]) = kt
[A]o = initial concentration
[A] = concentration at time t
Rearranging
t =
at the half-life [A] =
½[A]o
t½ =
t½ = ln(2)/k = 0.693/k
Also
rearranging,
ln [A]o –
ln [A] = kt
ln [A] = –
kt + ln
[A]o
·
This is the
equation of a line (y = mx +b).
·
A graph of ln [A] vs. t gives a linear relationship for first order
reactions.
·
The slope of the
line = -k
·
The y-intercept of
the line = ln [A]o
For second order reactions in
A and second order overall:
1/[A] – 1/[A]o = kt
t½ = 1/(k[A]o)
1/[A] = kt +
1/[A]o
A graph of 1/[A] vs. time is linear
For zero order
reactions:
[A] = [A]o - kt
t½ = [A]o/2k
A graph of [A] vs. time is
linear
1.
The rate law for
the reaction of sucrose in water, C12H22O11 +
H2O à 2 C6H12O6 is rate =
k[C12H22O11]. After 2.57 hours at 25°C, 6.00 g/L of
C12H22O11 has decreased to 5.40 g/L. Evaluate k for this reaction at
25°C.
2.
The rate constant
for the decomposition of nitrogen dioxide 2 NO2 à 2 NO + O2 with a laser beam is 1.70
M-1●min-1.
Find the time in seconds needed to decrease 2.00 mol/L of NO2
to 1.25 mol/L.
3.
The first order
rate constant for the radioactive decay of radium-223 is 0.0606
day-1. What is the half
life of radium-223?
4.
The following data
were obtained from a study of the decomposition of a sample of HI on the surface
of a gold wire. (a) Plot the data
to find the order of the reaction, rate constant, and the integrated rate
equation. (b) Calculate the HI
concentration in mmol/L at 600. s.
|
t
(seconds) |
[HI] mmol/L |
|
0. |
5.46 |
|
250. |
4.10 |
|
500. |
2.73 |
|
750. |
1.37 |
Collision theory of reaction rates
For a reaction to occur,
molecules must collide.
·
For a successful
collision (one which leads to reaction) the molecules
must:
1.
have sufficient
energy (required to rearrange electrons as bonds break and
form)
2.
have the
proper orientation.
Transition state theory
During a reaction, as reactants
are converted to products, bonds must be broken and
formed.
·
A short-lived,
high-energy intermediate is formed where bonds are partially broken and
formed. This is referred to as a
transition state.
·
Activation energy
is how much energy reactant molecules need to reach the transition
state.
·
If more of the
reactant molecules have sufficient energy to reach the transition state, the
reaction will proceed at a faster rate.
Reaction mechanisms
·
the step by step
pathway by which a reaction occurs, going through a series of
intermediates
·
some reactions only
require one step; others require two or more
·
the reaction order
for an individual step is equal to the sum of the reactant coefficients for that
step
·
the slow step (and
steps preceding the slow step) of a mechanism determines the overall rate of the
reaction
·
If rate = k[A]x[B]y, then the values of x and y
relate to the reactant coefficients in the slow step of the mechanism for this
reaction and preceding steps.
·
Terminology
describing steps:
Unimolecular
= one reactant in the step
Bimolecular = two reactants in the
step
Termolecular
= three reactants in the step
·
The step which is
described as the slow step has the highest activation energy. Consequently, it limits the rate of the
overall reaction.
·
Requirements for an
acceptable mechanism:
1.
Sum of all steps
should give the balanced equation – reaction intermediates are formed and then
consumed in subsequent steps
2.
Mechanism must
agree with overall determined rate law.
·
Note: for one step
mechanisms, the coefficients will equal the exponents in the rate
law.
·
For multi-step
mechanisms, the exponents will equal the coefficients of the reactants in the
slow step + the coefficients of reactants in preceding (fast equilibrium)
steps. Reaction intermediates are
not included.
Example: 2 NO2 (g) + F2
(g) à 2 NO2F (g)
Rate = k[NO2][F2]
Possible
mechanism:
NO2
+
F2
à NO2F
+
F
Slow
F
+
NO2
à NO2F
Fast
Is this an acceptable
mechanism?
1.
Sum of all steps
should give the balanced equation – it does
2.
Mechanism must
agree with overall determined rate law
Slow
step has coefficients of 1 for NO2 and 1 for F2. These agree with the exponents of the
rate law.
The
step which limits the rate is bimolecular.
·
We can never prove
that a mechanism is correct, only that it fits the rate law
data
The mechanism for the reaction
of nitrogen dioxide with carbon monoxide to form nitric oxide and carbon dioxide
is thought to be
NO2
+
NO2
à NO3
+
NO
Slow
NO3
+
CO
à NO2
+
CO2
Fast
Write the net chemical equation
and the rate law expected for this reaction.
The mechanism for the
decomposition of hydrogen peroxide is
H2O2
à 2
OH
H2O2
+
OH
à
H2O
+
HO2
HO2
+
OH
à
H2O
+
O2
And the rate law is rate =
k[H2O2].
Which step is the slow
step?
Write the net chemical equation
and the rate law for the following mechanism:
2 NO ßà
N2O2
Fast
N2O2
+
Br2
à 2
NOBr
Slow
Temperature, activation
energy, and the rate constant
·
As temperature
increases, the average kinetic energy of molecules in a sample
increases
·
More molecules will
have the required activation energy to successfully collide (react), so the
reaction proceeds at a faster rate
·
Rate =
k[A]x[B]y
·
With an increase in
temperature [A], [B], x and y do not change. The only change can be in the value of k
(increases with temperature).
·
The Arrhenius
equation shows the specific relationship between temperature (T), activation
energy (Ea), and the rate constant (k)
k = Ae–Ea/RT
or
ln k = ln A –
Ea/(RT)
·
A is a constant
with the same units as k. It
expresses the fraction of collisions which lead to reaction when all
concentrations are 1 M
·
R is the universal
gas constant, with the same energy units as are used for Ea (1
cal = 4.184 J = 4.129 x 10-2 L atm). R = 8.314 J/(mol K). mol is
interpreted as mol of reaction
·
e is the
constant 2.718 upon which natural logarithms are based.
·
Note that large
values of Ea would make k smaller; large values of T would make k
larger.
The Arrhenius equation allows
us to predict how changes in temperature would effect changes in rate of
reaction.
·
Consider a reaction
run at some temperature (T1) with a specific rate constant
k1. This reaction is
then run at a new temperature T2, and has a specific rate constant
k2.
·
Assuming that A and
Ea do not vary with temperature, the following relationship can be
established
ln (k2/k1) = Ea/R
(1/T1 – 1/T2)
·
For many chemical
reactions this implies that raising T 10 °C above room temperature approximately doubles the rate
of reaction
The rate constant of a reaction
is tripled when the temperature is increased from 298 K to 308 K. Find the activation energy for the
reaction.
Catalysts
·
Substances that
increase the rate of a reaction by providing alternative pathways with lower
activation energies
·
Arrhenius equation
would predict that if Ea decreases while conc. and T remain constant,
k will increase (reaction proceeds at a faster rate)
·
Within a reaction
mechanism catalysts are used in initial steps to get the reaction started, but
regenerated in subsequent steps – net reaction shows that the catalyst is not
consumed
·
Catalysts are often
written over the arrow in an equation to indicate that they are needed but not
consumed in the reaction.
·
Biological
catalysts are referred to as enzymes, with reactant molecules referred to as
substrates.