CHEMICAL KINETICS: THE RATES AND MECHANISMS OF CHEMICAL REACTIONS
Chemistry
CHEMICAL
KINETICS:
THE RATES AND MECHANISMS OF CHEMICAL REACTIONS
THE RATES AND MECHANISMS OF CHEMICAL REACTIONS
Chemical kinetics is the study of
the speed or rate of a reaction under various conditions. Spontaneity is also important AND a
spontaneous reaction does NOT imply a rapid reaction. The changing of diamond into graphite is
spontaneous but so slow that it is not detectable even in a lifetime.
A mechanism is a sequence of
events at the molecular level that controls the speed and outcome of the
reaction.
FACTORS THAT AFFECT REACTION RATES
The following conditions affect the speed of a
chemical process:
1. Nature
of the reactants--Some reactant molecules react in a hurry, others
react very slowly. Pointers: o Physical
state- gasoline (l) vs.
gasoline (g) ; K2SO4(s) + Ba(NO3)2(s)
→ no rxn.;
while both of these in the aqueous state react.
o Chemical
identity - What is reacting? Usually
ions of opposite charge react very rapidly.
Usually, the more bonds between reacting atoms in a molecule,
the slower the reaction rate.
Substances with strong bonds (larger bond
energies) will react much more slowly.
Examples: metallic sodium reacts
much faster with water than metallic calcium. Oxidation of methane can be
increased with an increase in temperature; photosynthesis is very slow and
changes very little with an increase in temperature.
2. Concentration of reactants--more
molecules, more collisions.
3. Temperature--heat >em up & speed
>em up; the faster they move, the more likely they are to collide.
o An
increase in temperature produces more successful collisions that are able to
overcome the needed activation energy, therefore, a general increase in
reaction rate with increasing temperature.
o In
fact, a general rule of thumb is that a 10°C
increase in temperature will double the reaction rate. o * This actually depends on the magnitude
of the Ea* and the temperature range.
4. Catalysts--accelerate chemical
reactions but are not themselves transformed.
o Biological
catalysts are proteins called enzymes.
o A
catalyst is a substance that changes the rate of reaction by altering the
reaction pathway. Most catalysts work by
lowering the activation energy needed for the reaction to proceed, therefore,
more collisions are successful and the reaction rate is increased.
o Remember! The catalyst is not part of the chemical
reaction and is not used up during the reaction.* (May be homogeneous or heterogeneous
catalysts.) Ex. H2O2 decomposes
relatively slowly into H2O and O2; however; exposure to
light accelerates this process AND with the help of MnO2, it goes
extremely FAST!! Note: A catalyst lowers the activation energy
barrier. Therefore, the forward and
reverse reactions are both accelerated to the same degree.
o *
(Some homogeneous catalysts actually appear in the rate law because their
concentration affects the reaction.
Ex. NO catalyzing O3 )
*AP is a registered trademark of
the College Board, which was not involved in the production of, and does not
endorse, this product. © 2008 by René McCormick. All rights reserved.
4. Surface area of reactants--exposed
surfaces affect speed.
o Except
for substances in the gaseous state or solution, reactions occur at the
boundary, or interface, between two phases.
o The
greater surface area exposed, the greater chance of collisions between
particles, hence, the reaction should proceed at a much faster rate. Ex. coal dust is very explosive as opposed to
a piece of charcoal. Solutions are
ultimate exposure!
5. Adding an inert gas has NO EFFECT on
the rate [or equilibrium] of the reaction.
THE COLLISION THEORY OF REACTION RATES
¾ Particles
must collide.
¾ Only
two particles may collide at one time.
¾ Proper
orientation of colliding molecules so that atoms in the can come in contact
with each other to become products.
¾ The collision must
occur with enough energy to overcome the electron/electron repulsion of the
valence shell electrons of the reacting species and must have enough energy to
transform translational energy into vibrational energy in order to penetrate
into each other so that the electrons can rearrange and form new bonds.
¾ This
new collision product is at the peak of the activation energy hump and is
called the activated complex or the transition state. At this point, the activated complex can
still either fall to reactants or to products.
¾ With
all of these criteria met, the reaction may proceed in the forward
direction. Amazing that we have
reactions occurring at all!
¾ in
the forward direction. Amazing that we
have reactions occurring at all!
CHEMICAL REACTION RATES
The speed of a reaction is
expressed in terms of its “rate”, some measurable quantity is changing with
time.
The rate of a chemical reaction is measured by the
decrease in concentration of a reactant or an increase in concentration of a
product in a unit of time.
Rate = change in
concentration of a species
time interval
When writing rate expressions, they can be written
in terms of reactant disappearance or product appearance. Rate is not constant, it changes with
time. Graphing the data of an experiment
will show an average rate of reaction.
You can find the instantaneous
rate by computing the slope of a straight line tangent to the curve at that
time. 
reaction rate--expressed as the
Δ in concentration of a reagent per unit time or Δ[A]/Δt focus either on the
disappearance of reactants or the appearance of products
-
rate of Δ of a reactant is always negative
-
rate of Δ of a product is always positive
Consider:
2 NO2(g) → O 2(g) + 2 NO(g)
Oxygen can appear only half as
rapidly as the nitrogen dioxide disappears while NO appears twice as fast as
oxygen appears.
Examine the data table and calculate the AVERAGE rate
at which [NO2] changes in the first
50.0 seconds: (Remember the square brackets are shorthand for molarity!)
RATE = −Δ [NO2] = −[.0079]−[0.0100]
Δt 50.0
s
= −[−4.2 × 10−5 mol/L • sec]
= 4.2 × 10−5 mol/L •sec or M s−1
Note that the rate is NOT constant but decreases with time. The rates given below are average rates.
−Δ [NO2] (× 10-5)
Δt
|
Time period (s)
|
4.2
|
0 → 50
|
2.8
|
50 → 100
|
2.0
|
100 → 150
|
1.4
|
150 → 200
|
1.0
|
200 → 250
|
To determine the value of the rate at a particular time,
the instantaneous
rate, compute the slope of a
line tangent to the curve at that
point. Why the negative on NO2?
Chemical
Kinetics: The Rates and Mechanisms of Chemical Reactions 3
RELATIVE RATES: We can consider
the appearance of products along with the disappearance of reactants. The reactant’s concentration is declining,
the products is increasing. Respect the
algebraic sign AND respect the stoichiometry. [Divide the rate of change in
concentration of each reactant by its stoichiometric coefficient in the
balanced chem. eqn. and this is foolproof and a breeze!]
Thus.....
Rate of
reaction = - 1Δ[NO2] = 1 Δ[NO] =
Δ
[ O2]
2 Δtime 2 Δtime Δtime
Of course you can change these once the ratio is set.
You might prefer -1 : +1 : +2
Relative Rates from the balanced equation:
Using the coefficients from the balanced equation,
students should be able to give relative rates.
For example: 4 PH3 (g) Æ P4(g) +
6 H2(g)
1
⎡Δ[PH3]⎤ ⎡ [P4]
⎤ 1 ⎡Δ[H2]⎤
Initial rate rxn. = −
4 ⎢ Δtime ⎥⎦ = +⎢⎣
Δtime⎦⎥ = + 6 ⎢⎣Δtime ⎥⎦
⎣
Exercise
What are the relative rates of change in
concentration of the products and reactant in the decomposition of nitrosyl
chloride, NOCl?
2
NOCl (g) →
2 NO(g) + Cl2(g)
1
⎡Δ[NOCl]⎤ 1 ⎡ [NO] ⎤ ⎡Δ[Cl ]⎤
   Initial rate
rxn. = −
⎢⎣ Δtime ⎥⎦ =
+ 2 ⎢⎣Δtime⎦⎥ =
+⎢⎣Δtime2 ⎥⎦
2
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Differential RATE LAW: AN INTRODUCTION
Reactions are reversible. So far, we’ve only considered the forward
reaction. The reverse is equally
important. When the rate of the forward
= the rate of the reverse we have EQUILIBRIUM!
To avoid this complication we will discuss reactions soon after mixing--initial
reactions rates, and not worry about the buildup of products and how that
starts up the reverse reaction.
initial reaction rates--begin
with pure reactants, mix thoroughly, then measure speed of rxn. over time
The presence of products can alter results dramatically and
lead to confusing results. We’ll be
talking initial reaction rates throughout our discussions!
Rate
= k[NO2]n = −⎡⎢⎣
Δ[NOΔt 2]⎤⎥⎦
Δ[NOΔt 2]⎤⎥⎦
The rate expression or rate law expression is the relation between
reaction rate and the concentrations of reactants given by a mathematical
equation.
CONCENTRATION AND REACTION RATE DATA
THE DIFFERENTIAL RATE LAW OR RATE EXPRESSION: Rates
generally depend on reactant
concentrations. To find
the exact relation between rate and concentration, we must conduct experiments
and collect information.
C
aA
+ bB xX
→
where
C is a catalysts, the rate expression will always
have the form:
Initial rxn rate = k[A]m[B]n[C]p
k = rate
constant
[A] =
concentration of reactant A
[B] =
concentration of reactant B
[C] =
concentration of the catalyst--won=t see this too often in AP m = order of reaction for reactant A n = order of reaction for reactant B p = order of reaction for the catalyst C
Exponents can be zero, whole numbers or fractions AND MUST BE DETERMINED BY EXPERIMENTATION!!
THE RATE CONSTANT, k
-
temperature dependent & must be evaluated by
experiment.
-
Example:
rate = k[A]
-
and k
is 0.090/hr, therefore when [A] = 0.018 mol/L
- rate = (.0090/hr)(0.018 mol/L) = 0.00016
mol/(L• hr)
ORDER OF A REACTION
-
order with respect to a certain reactant is the exponent on its concentration term in
the rate expression
-
order of the reaction is the sum of all the
exponents on all the concentration terms in the expression
-
DETERMINATION OF THE RATE EXPRESSION
aA + bB → xX
-
initial rate = k[A]om[B]on
-
the little subscript “o” means “original” or at
“time zero”
1. Zero order: The change in concentration of reactant has
no effect on the rate. These are not very common. General form of rate
equation: Rate = k
2. First order: Rate is directly proportional to the
reactants concentration; doubling [rxt], doubles rate. These are very common! Nuclear decay reactions usually fit into this
category. General form of rate equation:
Rate = k [A]1 = k[A]
3. Second order: Rate is quadrupled when [rxt] is doubled and
increases by a factor of 9 when [rxt] is tripled etc. These are common, particularly in gas-phase
reactions. General form of rate equation:
Rate = k [A]2 or Rate = k[A]1[B]1
which has an overall order of two
(second order).
4. Fractional
orders are rare, but do exist!
Recall our general rate expression: Rate = k[A]m[B]n
If
m = 0 ; reaction is zero order with
respect to A
If
m = 1 ; reaction is 1st order with
respect to A
If
m = 2 ; reaction is 2nd order with
respect to A
If
n = 0 ; reaction is zero order with
respect to B
If
n = 1 ; reaction is 1st order with
respect to B
If
n = 2 ; reaction is 2nd order with
respect to B
Adding the orders of each reactant gives the overall order of the reaction.
EXAMPLE
Experiment Number
|
Initial Rate mol/(L• hr)
|
Initial concentration
[A]o
|
Initial concentration
[B]o
|
1
|
0.50 × 10−2
|
0.50
|
0.20
|
2
|
0.50 × 10−2
|
0.75
|
0.20
|
3
|
0.50 × 10−2
|
1.00
|
0.20
|
4
|
1.00 × 10−2
|
0.50
|
0.40
|
5
|
1.50 × 10−2
|
0.50
|
0.60
|
APPLY “TABLE LOGIC”
1. Look
for two trials where the concentration of a reactant was held constant.
2. Next,
focus on the other reactant. Ask yourself how it’s concentration changed
for the same two trials. Was it doubled?
Was it tripled? Was it halved?
3. Once
you have determined the factor by which the concentration of the other reactant was changed, determine
how that affected the rate for those same two trials. Expect easy math!
Did changing the concentration have zero effect on
the rate? If so, then it is zero
order. Did the rate double when the
concentration of the other reactant
was doubled? If so, it is first order.
Did the rate quadruple as a result of the reactant’s concentration
doubling? If so, it is second order. Did
the rate increase by a factor of eight?
If so, it is third order.
THINK of the concentration doubling
as the number “two”. So, rate = k[reactant] becomes, rate = k[2]m and you
are trying to determine the value of m. If the rate
doubled think (2 rate) = k[2]m and more simply 2 = [2]m, so m = 1. If the rate
quadrupled, then think simply, 4 = [2]m and m must equal 2 to make that a true statement, and so on…
4. Finally,
examine the data table again. This time
look for trials where the concentration of the reactant you just determined the
order for is held constant and repeat steps 1-3 above.
For our example data table, the rate stays the same
regardless of the concentration of [A], therefore, it is zero order with respect to A. However, the rate doubles with a doubling of
[B] and triples with a tripling of [B].
This indicates the rate is first order with respect to [B].
Summary: Initial reaction rate = k[A]oo[B]o1 = k[B]o1 = k[B]
The overall reaction rate order = 1 + 0 = 1st order overall.
Now. . .
Use a set of the data to calculate k: 0.0050 mol/(L•hr) = k[0.20 mol/L]1 k = 2.5 × 10−2 /hr or 2.5 × 10−2 hr−1
You should get the same value with any line of data!
Ugly algebraic method is sometimes necessary:
rate 1 = k[reactant]m [reactant]n
rate
2 k[reactant]m [reactant]n
Select a trial where one reactant concentration is
held constant SO THAT IT CANCELS; the k’s will also cancel
Using trails 1 & 4:
0.50 × 10-2 = k [0.50]m [0.20]n
so…. ½ = [ ½ ]n
and ∴ n must be ONE to make that true! It
1.00 × 10-2 k [0.50]m [0.40]n
It’s just the long-hand version of table logic.
If you have a case where one reactant is never held
constant, then you can either add an “expect” column to the table since you
usually can determine the order of at least one reactant and can predict the
“expected” change in the rate. Then
proceed as usual and compare your “expected rate” to the actual rate for
doubling the concentration of the other
reactant. OR just plug into ugly
algebra, just use the method that makes you both quick and accurate!
Exercise
In the following reaction,
a Co-Cl bond is replaced by a Co-OH2 bond.
[Co(NH3)5Cl]+2
+ H2O →
[Co(NH3)5H2O]+3 + Cl
Initial rate = k{[Co(NH3)5Cl]+2}m
Using the data
below, find the value of m in the
rate expression and calculate the value of k. Exp.
Initial Concentration
Initial rate
of [Co(NH3)5Cl]+2 mol/(L• min) (mol/L)
1
1.0 × 10-3 1.3 × 10-7
2
2.0 × 10-3 2.6 × 10-7
3
3.0 × 10-3 3.9 × 10-7
4 1.0 × 10-3 1.3 × 10-7
m
= 1; k = 7700 min−1
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Exercise
The reaction between bromate ions and bromide ions
in acidic aqueous solution is given by the equation
BrO3-
(aq) +
5 Br – (aq) +
6 H+ (aq) → 3 Br2 (l) + 3 H2O (l)
The table below gives the results of four experiments. Using these data, determine the orders for
all three reactants, the overall reaction order, and the value of the rate
constant. What is the value of k?
What are the units of k?
Experiment Initial [BrO3-]
Initial [Br –] Initial [H+] Measured initial
rate (mol/L•s)
1
0.10 0.10
0.10 8.0 × 10-4
2
0.20 0.10
0.10 1.6 × 10-3
3
0.20 0.20
0.10 3.2 × 10-3
4
0.10 0.10
0.20 3.2 × 10-3
rate = [BrO3-] [Br –] [H+]2;
overall order = 4; k = 8.0 L3/mol3
is
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