Chemical Kinetics Two Types of Rate Laws 1.Differential- Data table contains RATE AND CONCENTRATION DATA. Uses “table logic” or algebra to find the order

ChemicalKinetics

Two Types of Rate Laws

1. Differential- Data table contains RATE AND CONCENTRATION DATA. Uses “table logic” or algebra to find the order of reaction and rate law

2. Integrated- Data table contains TIME AND CONCENTRATION DATA. Uses graphical methods to determine the order of the given reactant. K=slope of best fit line found through linear regressions

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Integrated Rate Law

• Can be used when we want to know how long a reaction has to proceed to reach a predetermined concentration of of some reagent

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Graphing Integrated Rate Law

• Time is always on x axis

• Plot concentration on y axis of 1st graph

• Plot ln [A] on the y axis of the second graph

• Plot 1/[A] on the y axis of third graph

• Your are in search of a linear graph

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Results of linear graph• Zero order: time vs concentration= line

y= mx+ b

[A]= -kt + [A0 ]

A- reactant A,

A0 - initial concentration of A at t=0

l slope l= k, since k cannot be negative, and k will have a negative slope

Rate law will be rate=k[A]0

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Results of linear graph• First order: time vs ln [ ]= line

y= mx+ b

ln [A]= -kt + ln [A0 ]

A- reactant A,


l slope l= k, since k cannot be negative, and k will have a negative slope


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Results of linear graph• second order: time vs 1/ [ ]= line

y= mx+ b

1/[A]= kt + 1/ [A0 ]

A- reactant A,


k=slope


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First-Order Processes

Consider the process in which methyl isonitrile is converted to acetonitrile.

CH3NC CH3CN

© 2012 Pearson Education, Inc.

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This data were collected for this reaction at 198.9 C.

CH3NC CH3CN


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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 of the slope: 5.1 105 s1.


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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 table:

Time (s) [NO2], M

0.0 0.01000

50.0 0.00787

100.0 0.00649

200.0 0.00481

300.0 0.00380

12


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Using the graphing calculator• L1=time

• L2=concentration-- if straight line, zero order

• L3=ln concentration-- if straight line- 1st order

• L4= 1/concentration– if straight line—2nd order

• Perform 3 linear regressions

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Second-Order Processes

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 a straight line, so the process is second-order in [A].


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Second-Order Processes• Graphing vs.

t, however, gives this plot Fig. 14.9(b).

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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• Determine the rate law and calculate k

• What is the concentration of N2 O5 at 600s?

• At what time is the concentration equal to 0.00150 M?

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The decomposition of N2 O5 was studied at constant temp2 N2 O5 (g) 4 NO2(g) + O2(g)

[N2 O5 ] Time (s)

0.1000 0

0.0707 50

0.0500 100

0.0250 200

0.0125 300

0.00625 400

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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-Life

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

0.693k

Note: For a first-order process, then, the half-life DOES NOT DEPEND ON CONCENTRATION!!!!!!!!

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Half-Life

First order decay is what is seen in radioactive decay

= t1/2

0.693k

This is the equation used to calculate the half-life of a radioactive isotope

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Problem

• Exercise• A certain first-order reaction has a half-

life of 20.0 minutes.• a. Calculate the rate constant for this

reaction.• b. How much time is required for this

reaction to be 75% complete?• 3.47 × 10−2 min−1; 40 minutes

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Half-Life

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

1k[A]0

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Half life zero order

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Half life first order

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Half life second order

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Summary table

order

zero First second

Rate Law Rate=k rate-=k[ A] rate= k [A]2

Integrated rate law in form of y=mx+b

[A]t = -kt + [Ao ] ln[A]t = -kt + ln[Ao ]

Rate law in data packet on AP exam

Does not appear ln[A]t -ln[Ao ] = -kt 1/[A]t -1/[Ao ] = kt

slope Slope = -k Slope= -k slope=k

Half-life [A0 ]/2k 0.693/k 1/kA0

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Reaction Mechanisms

The molecularity of a process tells how many molecules are involved in the process.


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Reaction Mechanisms

• Chemical reactions proceed via a sequence of distinct stages.

• The sequence is known as the mechanism and each part of the mechanism is known as a “step”.

• The rate of the reaction is only dependant upon the slowest step, also known as the rate determining step or RDS.

• only reactants that appear in the rate determining step appear in the rate equation and vice-versa


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Multistep Mechanisms• In a multistep process, one of the steps will

be slower than all others.• The overall reaction cannot occur faster than

this slowest, rate-determining step.


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• Example #1The reaction below

W + Y Z

Has the following mechanism

W R slow

R + Y Q fast

Q Z fast

Here the rate only depends on the concentration of _______________

and therefore the rate equation only contains this reactant

The rate equation is, ______________ in the rate determining step is 1.

Note: R and Q are not reactants or products, but are rather they are called ________________________, produced in one step, but then are used up in a subsequent step.


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Note:

1. In all valid mechanisms the sum of the individual fast and slow steps must be the same as the overall chemical equation.

2.The stoichiometric coeffiecient of a substance that appears in the slow step is the power that the

concentration of that substance is raised to in the rate equation.

3. If a substance is present at the beginning of a reaction AND present in the same form at the

end of the reaction, it can be identified as a catalyst


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Slow Initial Step• A proposed mechanism for this reaction is

Step 1: NO2 + NO2 NO3 + NO (slow)

Step 2: NO3 + CO NO2 + CO2 (fast)

• The NO3 intermediate is consumed in the second step.

• As CO is not involved in the slow, rate-determining

step, it does not appear in the rate law.


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The reaction below

A + B C + D

Has the following mechanism

A Q fast equilibrium

Q + B C + D slow

Here the slow step contains Q and B, and _____is an intermediate

Can intermediates be featured the rate equation? _____________

Since the formation of Q is dependent on A, Q can be replaced by A in the rate equation. Therefore the rate equation is given as_________________ The orders w.r.t A and B are________________________________

since the stoichiometric coefficient of B in the rate determining step is 1, and the stoichiometric coefficient of A (which Q depends upon) is also 1.


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Temperature and Rate

• Generally, as temperature increases, so does the reaction rate.

• This is because k is temperature-dependent.


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Reaction Coordinate Diagrams• The diagram shows the

energy of the reactants and products (and, therefore, E).

• The high point on the diagram is the transition state.

• The species present at the transition state is called the activated complex.

• The energy gap between the reactants and the activated complex is the activation-energy barrier.


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Maxwell–Boltzmann Distributions

• Temperature is defined as a measure of the average kinetic energy of the molecules in a sample.

• At any temperature there is a wide distribution of kinetic energies.


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• As the temperature increases, the curve flattens and broadens.

• Thus, at higher temperatures, a larger population of molecules has higher energy.


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• If the dotted line represents the activation energy, then as the temperature increases, so does the fraction of molecules that can overcome the activation-energy barrier.

• As a result, the reaction rate increases.


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Maxwell–Boltzmann DistributionsThis fraction of molecules can be found through the expression

where R is the gas constant and T is the Kelvin temperature.

f = e−Ea/RT


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Arrhenius Equation

Svante Arrhenius developed a mathematical relationship between k and Ea:

k = Ae

where A is the frequency factor, a number that represents the likelihood that collisions would occur with the proper orientation for reaction.

−Ea/RT


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Arrhenius Equation

Taking the natural logarithm of both sides, the equation becomes

ln k = ( ) + ln A1T

y = mx + b

Therefore, if k is determined experimentally at several temperatures, Ea can be calculated from the slope of a plot of ln k vs. .

Ea

R

1T


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Catalysts• Catalysts increase the rate of a reaction by

decreasing the activation energy of the reaction.

• Catalysts change the mechanism by which the process occurs.


Documents

Chemical Kinetics Two Types of Rate Laws 1.Differential- Data table contains RATE AND CONCENTRATION DATA. Uses “table logic” or algebra to find the order