1 spontaneity – inherent tendency for a reaction to occur; does not mean speed reactions have...
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Spontaneity – inherent tendency for a reaction to occur; does not mean speed
Reactions have different rates: 2 H2 + O2 2 H2O SLOW Explosions FAST
To be useful, reactions must occur at a reasonable rate.
Chapter 12:Chemical Kinetics
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Chemical Kinetics
The area of chemistry concerned with the speed (rate) of a reaction and reaction mechanisms.
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Main Goal of Kinetics Understand the steps by which a
reaction takes place What factors determine how fast food
spoils? Design of fast setting material for
dental fillings What affects the rate that steel rusts? What affects the rate the fuel burns in
cars?
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Reaction Rate FactorsReaction Rate Factors Physical state of reactants
Concentrations of reactants
Temperature
Presence of catalyst
Force of collisions
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Reaction Rate
Change in concentration (conc) of a reactant or product per unit time.
(where A is a reactant or product) (note that rate is always a + value)
Unit typically M/s
t
A Rate
tt
at t Aat t ARate
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1 2
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Progress of a hypothetical rxn
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Sample Exercise 1 Using the data given in slide 6, calculate the
average rate at which A disappears over the time interval 20 s to 40 s.
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2 NO2(g) 2 NO(g) + O2(g)
Time(s)
[NO2] [NO] [O2]
0 0.0100 0 0
50 0.0079 0.0021 0.0011
100 0.0065 0.0035 0.0018
150 0.0055 0.0045 0.0023
200 0.0048 0.0052 0.0026
250 0.0043 0.0057 0.0029
300 0.0038 0.0062 0.0031
350 0.0034 0.0066 0.0033
400 0.0031 0.0069 0.0035
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Find the avg rate the first 50 s of NO2 change.
We want to work with positive numbers so the equation will be:
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Concentration of nitrogen dioxide, nitric oxide, and oxygen versus time.
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What do we discover about the rate of this reaction by looking at the graph on slide 10?
Instantaneous rate – value of the rate at particular time; found by calculating the slope of a line tangent to the curve at that point
At 100 s:
M/s104.2s 110
M 0026.0
t
NO- Rate 52
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Reaction Rates
All reactions slow down over time.
Therefore, the best indicator of the rate of a reaction is the instantaneous rate near the beginning of the reaction. (called initial rate)
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
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Using the graph on slide 12, calculate the instantaneous rate of disappearance of reactant at t = 0.
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Rate Laws
Rate = k[A]m [B]n
k = rate constant, dependent on temperature
m & n = reaction orders
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The rate law for any reaction must be determined experimentally, it cannot be predicted by merely looking at the chemical equation.
*usually involves only the concentration of reactants
Rate Laws
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Types of Rate Laws
Differential Rate Law: expresses how rate depends on concentration. (Often just called “rate law”)
Integrated Rate Law: expresses how concentration depends on time.
They are inter-related, so once you experimentally determine one, the other can be found.
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WHY???!!!!
We can work backward from the rate law to infer the steps by which the reaction occurs which helps us to find the slowest step – then we can figure out how to speed it up!
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Method of Initial Rates
Initial Rate: the “instantaneous rate” just after the reaction begins (just after t = 0)
The initial rate is determined in several experiments using different initial concentrations.
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Method of Initial Rates: Sample 1 NH4
+(aq) + NO2−(aq) N2(g) + 2 H2O(l)
Rate = k [NH4+]m[NO2
−]n
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Determine the values of n and m.
nm
nm
7-
7
mol/L) (0.200 mol/L) k(0.0100
mol/L) (0.200 mol/L) k(0.0200
10 x 5.4
108.10
1 Rate
2 Rate
mols/L.s
mols.L.s
x
m)0.2( mol/L.s) (0.01o0
mol/L.s) (0.0200
m
m
m1 Rate2 Rate )0.2(00.2
The value of m is 1
Method of Initial Rates: Sample 1
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A similar method is used to find n.
n
n
7-
7
M) 0202.0(M)200.0(k
M) 0.0404k(0.200M)(
mol/L.s 108.10
mol/L.s 106.21
4 Rate
5 Rate
x
x
n)00.2()( 2.00 n100..200
The value of n is also 1.
Method of Initial Rates: Sample 1
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The values of n and m are both 1 and the rate law is:
]NO][k[NH Rate -24
This rate law is first order in both NO2- and NH4
+ .
The overall reaction order is the sum of n and m.
The reaction is second order overall.
Method of Initial Rates: Sample 1
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• For the reaction
NH4+(aq) + NO2
-(aq) N2(g) + 2H2O(l)
we can observe that – as [NH4
+] doubles with [NO2-] constant the rate doubles,
– as [NO2-] doubles with [NH4
+] constant, the rate doubles,
– We conclude rate [NH4+][NO2
-].
• Rate law:
• The constant k is the rate constant.
]NO][NH[Rate 24k
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]NO][k[NH Rate -24
(0.200M)k(0.0100M) M/s 10 x 4.5 -7
Then
l/Ms 107.2 k 4M) M)(0.200 (0.0100
M/s 10 x 5.4 -7 x
Calculate the rate constant k
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Using Initial Rates to Determine Rate Laws
A reaction is zero order in a reactant if the change in concentration of that reactant produces no effect.
A reaction is first order if doubling the concentration causes the rate to double.
A reaction is nth order if doubling the concentration causes an 2n increase in rate.
Note that the rate constant does not depend on concentration.
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Overall Reaction Order
Sum of the order of each component in the rate law.
rate = k[H2SeO3][H+]2[I]3
The overall reaction order is 1 + 2 + 3 = 6.
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Kinetic Sample Problem 12 H2 + 2 NO 2H2O + N2 at 800 K
Experiment [H2] [NO] Initial rate in M/min
1 .001 .006 .025
2 .002 .006 .050
3 .003 .006 .075
4 .009 .001 .0063
5 .009 .002 .025
6 .009 .003 .056
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Complete Sample Problems 2-4
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3
24-
2
2
22
1/minM 106 k
BAk Rate .4
1/minM 10 3.0 k
CAk Rate 3.
min1/atm 2.2
HNOk Rate 2.
k
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Integrated Rate Laws
Using calculus to integrate the rate law for a first-order process gives us
ln[A]t
[A]0
= −ktWhere[A]0 is the initial concentration of A, and
[A]t is the concentration of A at some time, t, during the course of the reaction.
At
A Rate k
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Integrated Rate Laws
Manipulating this equation produces…
ln[A]t
[A]0
= −kt
ln [A]t − ln [A]0 = − kt
ln [A]t = − kt + ln [A]0
…which is in the form
y = mx + b
On purple equation sheet
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First-Order Processes
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 -k.
ln [A]t = -kt + ln [A]0
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First-Order Processes
Consider the process in which methyl isonitrile is converted to acetonitrile.
CH3NC CH3CN
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First-Order Processes
This data was collected for this reaction at 198.9 °C.
CH3NC CH3CN
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First-Order Processes
When ln P is plotted as a function of time, a straight line results.
Therefore, The process is first-order. k is the negative of the slope: 5.1 10-5 s−1.
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If you know the initial concentration and k, the concentration at any time can be calculated.
Example: The decomposition of a certain insecticide in water follows first-order kinetics with a rate constant of 1.45 yr-1. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0 x 10-7 g/cm3.
(a) What is the concentration of the insecticide on June 1 of the following year?(b) How long will it take for the concentration of the insecticide to decrease to 3.0 x 10-7 g/cm3?
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Complete sample problems on wkst.
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Second-Order Processes
Similarly, integrating the rate law for a process that is second-order in reactant A, we get
1[A]t
= kt +1
[A]0
also in the form y = mx + b
tA
1
A
1
0
kt
On purple equation sheet
Rearranged to give
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Second-Order Processes
So if a process is second-order in A, a plot of vs. t will yield a straight line, and the slope of that line is k.
1[A]t
= kt +1
[A]0
1[A]
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Second-Order ProcessesThe decomposition of NO2 at 300°C is described by the equation
NO2 (g) NO (g) + 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
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Second-Order Processes
• Plotting ln [NO2] vs. t yields the graph below.
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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Second-Order Processes
• Graphing vs. t, however, 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].
1[NO2]
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Half-Life 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.
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Half-LifeFor 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
0.693kNOTE: For a first-order
process, then, the half-life does not depend on [A]0.
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Half-LifeFor a second-order process,
10.5 [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
1k[A]0
NOTE: For a second order reaction, The half-life is dependent upon the initial concentration.
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Plot [A] vs t.
Plot ln[A] vs t.
Plot 1/[A] vs t.
Testing for a Rate LawTesting for a Rate Law
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Summary of the Kinetics for Reactions that are Zero, First, and Second Order in [A] Order
Zero First Second
Rate law: Rate = k Rate = k[A] Rate = k[A]2
Integrated rate law:
[A] = -kt + [A]0
ln[A] = -kt + ln[A]0
1/[A] = -kt + 1/[A]0
Plot needed to give a straight line:
[A] vs. t ln[A] vs. t 1/[A] vs. t
Relationship of rate constant to the slope of straight line:
Slope = -k Slope = -k Slope = k
Half-life:t1/2 = [A]0/2k
t1/2 = 0.693/k t1/2 = 1/k[A]0
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Temperature and Temperature and RateRate
Most reactions speed up as temperature increases. (E.g. food spoils when not refrigerated.)
Two light sticks are placed in water, one at room temp. and one in ice
the one at room temp. is brighter than the one in ice. The chemical reaction, responsible for
chemiluminescence is dependent on temp. The higher the temp., the faster the
reaction and the brighter the light. As temperature increases, the rate
increases.
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Collision Model
Key Idea: Molecules must collide to react.
However, only a small fraction of collisions produces a reaction. Why?
Arrhenius: An activation energy must be overcome.
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The Collision Model
The more molecules present, the greater the probability of collision and the faster the rate.
The higher the temperature, the more energy avail. to the molecules and the faster the rate.
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The Collision Model
Complication: not all collisions lead to products. In fact, only a small fraction of collisions lead to product.
In order for reaction to occur the reactant molec. must collide in the correct orientation and with enough energy to form products.
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Arrhenius Equation
Arrhenius: molecules must possess a minimum amount of energy to react. Why? In order to form products, bonds must be broken
in the reactants. Bond breakage requires energy.
Activation energy, Ea, is the minimum energy required to initiate a chemical reaction.
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Arrhenius Equation
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Arrhenius Equation Consider the rearrangement of
acetonitrile:
In H3C-NC, the C-NC bond bends until the C-N bond breaks and the NC portion is perpendicular to the H3C portion. This structure is called the activated complex or transition state.
The energy required for the above twist and break is the activation energy, Ea.
Once the C-N bond is broken, the NC portion can continue to rotate forming a C-CN bond.
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Arrhenius Equation
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Arrhenius Equation (continued)
k = rate constant A = frequency factor (a measure of the probability
of a favorable collision) Ea = activation energy T = temperature R = gas constant (8.31 J/molK)
Both A and Ea are specific to a given reaction
k = Ae-Ea/RT
ln k = + ln AR
-Ea
T
1
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Catalysis
Catalyst: A substance that speeds up a reaction without being consumed
Enzyme: A large molecule (usually a protein) that catalyzes biological reactions.
Homogeneous catalyst: Present in the same phase as the reacting molecules.
Heterogeneous catalyst: Present in a different phase than the reacting molecules.
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Catalysis
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•Enzymes are biological catalysts.Enzymes are biological catalysts.•Most enzymes are protein molecules with Most enzymes are protein molecules with large molecular masses (10,000 to 106 amu).large molecular masses (10,000 to 106 amu).•Enzymes have very specific shapes.Enzymes have very specific shapes.•Most enzymes catalyze very specific Most enzymes catalyze very specific reactions.reactions.•Substrates undergo reaction at the active site Substrates undergo reaction at the active site of an enzyme.of an enzyme.
EnzymesEnzymes