Download - Week 10.1 chemical kinetics
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Prepared by:
Mrs Faraziehan Senusi
PA-A11-7C
Collision Model
Catalysis
Chapter 5
Chemical Kinetics
Reaction Rates
Reference: Chemistry: the Molecular Nature of Matter and Change,
6th ed, 2011, Martin S. Silberberg, McGraw-Hill
Rate Laws
Reaction mechanism
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Kinetics
• Chemical kinetics is the study of reaction rates,
the changes in concentrations of reactants (or
products) as a function of time.
• Chemical kinetics is the study of rates of
chemical reactions, the factors that affect
reaction rates, and the mechanisms (the series
of steps) by which reactions occur.
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FACTORS THAT INFLUENCE
REACTION RATE
1. Physical State of the Reactants
– In order to react, molecules must come in
contact with each other.
– The more homogeneous the mixture of
reactants, the faster the molecules can
react.
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2. Concentration of Reactants
– A reaction can occur only when the
reactant molecules collide.
– As the concentration of reactants increases,
the more molecules present in the
container, the more frequently they
collide, and the more often a reaction
between them occurs.
– Thus, reaction rate is proportional to the
concentration of reactants:
Rate α collision frequency α concentration
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3. Temperature
– At higher temperatures, reactant
molecules have more kinetic energy,
move faster, and collide more often and
with greater energy.
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• Presence of a Catalyst
– Catalysts speed up reactions by changing
the mechanism of the reaction.
– Catalysts are not consumed during the
course of the reaction.
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Reaction Rates
• A rate is a change in some variable per unit of time.
• Rates of reactions can be determined by monitoring
the change in concentration of either reactants or
products as a function of time.
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• Consider a general reaction, A B .
• We quickly measure the starting reactant
concentration (conc A1) at t1, allow the reaction to
proceed, and then quickly measure the reactant
concentration again (conc A2) at t2.
• The change in concentration divided by the change
in time gives the average rate:
• We use the minus sign simply to convert the
negative change in reactant concentration to a
positive value for the rate.
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• If instead we measure the product to determine
the reaction rate, we find its concentration
increasing over time.
• That is, conc B2 is always higher than conc B1.
• Thus, the change in product concentration, Δ[B],
is positive, and the reaction rate for A B
expressed in terms of B is
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In this reaction, the
concentration of butyl
chloride, C4H9Cl was
measured at various times.
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
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The average rate of the
reaction over each
interval is the change in
concentration divided
by the change in time:
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
t
ClHC
94 rate average
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• Note that the average
rate decreases as the
reaction proceeds.
• This is because as the
reaction goes forward,
there are fewer
collisions between
reactant molecules.
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
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• A plot of concentration vs.
time for this reaction
yields a curve like this.
• The slope of a line tangent
to the curve at any point is
the instantaneous rate at
that time.
• All reactions slow down
over time.
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
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Reaction Rates and Stoichiometry
• In this reaction, the ratio of C4H9Cl to C4H9OH is
1:1.
• Thus, the rate of disappearance of C4H9Cl is the
same as the rate of appearance of C4H9OH.
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
t
OHHC
t
ClHC
9494 rate
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• What if the ratio is not 1:1?
2 HI(g) H2(g) + I2(g)
• Therefore,
t
I
t
HI
2
2
1 rate
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• To generalize, then, for the reaction
aA + bB cC + dD
t
D
dt
C
ct
B
bt
A
a
1111 rate
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Concentration and Rate
One can gain information about the rate
of a reaction by seeing how the rate
changes with changes in concentration.
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Comparing Experiments 1 and 2,
when [NH4+] doubles, the initial rate doubles.
NH4+
(aq) + NO2−
(aq) N2(g) + 2 H2O(l)
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Likewise, comparing Experiments 5 and 6,
when [NO2−] doubles, the initial rate doubles.
NH4+
(aq) + NO2−
(aq) N2(g) + 2 H2O(l)
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• This means
Rate [NH4+]
Rate [NO2−]
Rate [NH+] [NO2−]
or
Rate = k [NH4+] [NO2
−]
• This equation is called the rate law, and k is the
rate constant.
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Rate Laws
• A rate law shows the relationship between the
reaction rate and the concentrations of reactants.
• For a general reaction,
• the rate law has the form
• The exponents m and n, called the reaction orders,
define how the rate is affected by reactant
concentration.
• The overall reaction order can be found by adding
the exponents on the reactants in the rate law.
aA + bB + … cC + dD + …
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Rate = k [NH4+] [NO2
−]
• This reaction is
First-order in [NH4+]
First-order in [NO2−]
• This reaction is second-order overall.
NH4+
(aq) + NO2−
(aq) N2(g) + 2 H2O(l)
Examples:
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• If the rate doubles when [A] doubles, the rate depends on
[A] raised to the first power, [A]1, so m = 1.
• Similarly, if the rate quadruples when [B] doubles, the
rate depends on [B] raised to the second power, [B]2, so n
= 2.
• In another reaction, the rate may not change at all when
[A] doubles; in that case, the rate does not depend on [A],
or, to put it another way, the rate depends on [A] raised to
the zero power, [A]0, so m = 0.
• Keep in mind that the coefficients a and b in the general
balanced equation are not necessarily related in any way
to these reaction orders m and n.
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A key point to remember is that the components of the rate law:
rate, reaction orders, and rate constant must be found by experiment;
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Determining Reaction Orders
Experimentally
Example:
Consider the reaction between oxygen and nitrogen monoxide, a key
step in the formation of acid rain and in the industrial production of
nitric acid:
O2(g) + 2NO(g) 2NO2(g)
The rate law, expressed in general form, is Rate = k[O2]m[NO]n
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• Table shows experiments that change one reactant concentration
while keeping the other constant.
• If we compare experiments 1 and 2, we see the effect of doubling
[O2] on the rate.
• First, take the ratio of their rate laws:
• Because k is a constant and [NO] does not change between these
two experiments, these quantities cancel.
• Substituting the values from Table, we obtain
• Dividing, we obtain and rounding the figure:
• The reaction is first order in O2 .
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• To find the order with respect to NO, we compare experiments 3
and 1 , in which [O2] is held constant and [NO] is doubled:
• As before, k is constant, and in this pair of experiments [O2]
does not change, so these quantities cancel:
=
• we obtain and rounding the figure
• The reaction is second order in NO.
• Thus, the rate law is
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Determining the Rate Constant
• The rate constant is specific for a particular reaction at a particular
temperature.
• The experiments with the reaction of O2 and NO were run at the
same temperature, so we can use data from any to solve for k.
From experiment 1,
The average value of k for the five experiments in Table is 1.72 x 103 L2/mo12·s
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• Units for the rate constant :
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Integrated Rate Laws
• The equation that relates concentration and time
is the integrated rate equation.
• We can also use it to calculate the half-life, t1/2,
of a reactant-the time it takes for half of that
reactant to be converted into product.
• The integrated rate equation and the half-life are
different for reactions of different order.
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First-Order Processes
a represents the coefficient of reactant A
in the balanced overall equation.
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• Using calculus to integrate the rate law for a first-order
process gives us
ln [A]t
[A]0 = −akt
Where
[A]0 is the initial concentration of A.
[A]t is the concentration of A at some time, t,
during the course of the reaction.
• Manipulating this equation produces…
ln [A]t − ln [A]0 = − akt
ln [A]t = − ak t + ln [A]0 …which is in the form
y = m x + b
Therefore, if a reaction is first-order, a plot of ln[A] vs. t
will yield a straight line, and the slope of the line will be –ak.
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• Using ideal gas equation,
PV=nRT
0
0
00
lnln
ln
ln/
/ln
][
][ln
][
PktP
ktP
P
P
P
RTP
RTP
A
A
RT
PA
V
n
o
A
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ln [A]t = -akt + ln [A]0
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Consider the process in which methyl
isonitrile is converted to acetonitrile.
CH3NC CH3CN
Example:
This data was collected for this
reaction at 198.9°C.
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• When ln P is plotted as a function of time,
a straight line results.
• Therefore,
The process is first-order.
k is the negative slope, k = 5.1 10 −5 s-1.
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Second-Order Processes
For a simple second order reaction,
aktAA
dtaAd
adt
Ad
a
dt
Ad
t
Arate
a
0
2
2
2
][
1
][
1
k [A]
][
k[A] ][
k[A] rate
law, rate From
][][
productA
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The decomposition of NO2 at 300°C is described
by the equation
NO2 (g) NO (g) + 1/2 O2 (g)
and yields data comparable to this:
Time (s) [NO2], M
0.0 0.01000
50.0 0.00787
100.0 0.00649
200.0 0.00481
300.0 0.00380
Example:
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• Graphing ln [NO2] vs. t yields:
Time (s) [NO2], M ln [NO2]
0.0 0.01000 −4.610
50.0 0.00787 −4.845
100.0 0.00649 −5.038
200.0 0.00481 −5.337
300.0 0.00380 −5.573
• The plot is not a straight line,
so the process is not first-
order in [A].
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• Graphing 1/[NO2] vs. t,
gives this plot.
Time (s) [NO2], M 1/[NO2]
0.0 0.01000 100
50.0 0.00787 127
100.0 0.00649 154
200.0 0.00481 208
300.0 0.00380 263
• Because this is a straight
line, the process is
second-order in [A].
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Example
A reaction, A B + C is carried out at a particular temperature.
As the reaction proceeds, we measure the molarity of the reactant,
[A] at various time, as tabulated below:
a) Determine the order of the reaction.
b) State the rate-law expression for the
reaction.
c) Calculate the value of the rate
constant, k at this temperature.
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Solution:
a) Determine the order of the reaction.
From graphical plots, the reaction is deduced to be 1st order
since plots of ln[A] vs. t gives straight line (linear).
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b) State the rate-law expression for the reaction.
Rate law expression: r = k[A]
c) Calculate the value of the rate constant, k at this temperature.
For 1st order reaction, slope, m = -ak
rate constant, k = - (m/a) = 0.29 min-1
29.05.15.8
)2.0()83.1(
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Zero -Order Processes
ta
adt
Ad
a
dt
Ad
t
Arate
a
k [A][A]
k ][
k[A] rate
law, rate From
][][
productA
0
0
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• Half-life is defined as
the time required for
one-half of a reactant to
react.
• Because [A] at t1/2 is
one-half of the original
[A],
[A]t = 0.5 [A]0.
Half-Life
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For a first-order process, this becomes
0.5 [A]0
[A]0 ln = −kt1/2
ln 0.5 = −kt1/2
−0.693 = −kt1/2
= t1/2 0.693
k NOTE: For a first-order
process, the half-life
does not depend on [A]0.
First-Order Processes
ln [A]t
[A]0 = −kt
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For a second-order process,
1
0.5 [A]0 = kt1/2 +
1
[A]0
2
[A]0 = kt1/2 +
1
[A]0
2 − 1
[A]0 = kt1/2
1
[A]0 =
= t1/2 1
k[A]0
Second-Order Processes
ktAA
0][
1
][
1
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2/10
2/10
2/100
0
2k
[A]
k0.5[A]
k [A]0.5[A]
k [A][A]
t
t
t
t
Zero -Order Processes
For a zero-order process,
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Example
The decomposition of ethane (C2H6) to methyl radicals is
a first order reaction with a rate constant of 5.36 x 10-4 s-1
at 700°C.
C2H6 (g) 2CH3 (g)
Calculate the half-life of the reaction in minutes.
= t1/2
0.693
k
Answer:
t1/2 = 21.5 min
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Example
The gas-phase decomposition of NOBr is second order in [NOBr],
with k = 0.810 M-1.s-1 at 10°C. We start with 4.00 x10-3 M NOBr in a
flask at 10°C. How many seconds does it take to use up 1.50 x 10-3 M
of this NOBr?