Chemical Kinetics

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Chemical Kinetics

• Area of chemistry that deals with rates or speeds
  at which reactions occur.
• Examples of why the study of kinetics is
  important
• --allows determination of expiration dates on
  food.
• --allows determination of how fast a medicine
  will work.
• --increases efficiency of chemical processes
  more product, less waste.

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Some Reactions Occur at Very Fast Rate
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Some Reactions are Very Slow
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Collision Theory In order for molecules to
react, they must be properly oriented. More
collisions provides a better chance of correct
orientations. Higher rate of properly oriented
collisions equals faster reaction.
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Proper Orientation
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Notice the Orientation
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Not Properly Oriented
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Not Properly Oriented
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Properly Oriented
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REACTION !
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Several factors influence the rate of
collisionsand therefore, the rate of a reaction.
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Factors Affecting Rates of Reaction

1. Concentration of reactants
2. Temperature at which the reaction occurs
3. The presence of a catalyst
4. Physical State / Surface area of reactants.

a brief overview.
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Increase Temperature

• Temperature is a measure of the kinetic energy of
  molecules.
• Kinetic Energy ½ mv2
• Temperature related to the SPEED of the molecule

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Increased temperature increased collision rate
increase rate of reaction.
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Concentration
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High Concentration of Reactants Much interaction
Low Concentration of Reactants Little interaction
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Oxygen in Atmosphere
Pure Oxygen
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Physical States and Surface Area
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Gas
Gas
Gas
Gas
Different States Less Interaction
Different States Less Interaction Possible
Liquid
Liquid
Liquid
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Gas
Same States More Interaction Possible
Gas
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Mg metal reacts with high temperature steam, but
not water.
Mg (s) H2O (l) ? No Reaction Mg (s) H2O (g) ?
MgO H2
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When solids and gases react, the reaction is
limited to the surface of the solid.
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Grain Elevator Fire
Grinding chemicals with a mortar and pestle
increases reaction rate.
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Catalysts
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Reactants
Catalyst (Enzyme)
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Enzyme reactants Complex
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Enzyme products Complex
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Products
Unchanged Enzyme
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Reactants
Catalyst Enzyme
Enzyme reactants Complex
Enzyme products Complex
Products
Unchanged Enzyme
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Activation Energy
Activation Energy
Activation Energy-- energy need to get a reaction
started
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Activation Energy
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Activation Energy
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Activation Energy
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Activation Energy
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Activation Energy
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Activation Energy
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Activation Energy
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Activation Energy
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Activation Energy
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Activation Energy
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Activation Energy
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As a review the different factors that help
increase the probability that reactants will
align in the correct way and with enough energy
for a reaction to occur are...

• Increase temperature
• Increase concentration
• Physical State and Surface Area
• The presence of a catalyst

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Now for the details and calculations!
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The Rate of a Chemical Reaction is the rate of
change in the concentration of reactants and/or
products per unit time.
Reaction rates are determined by measuring the
concentration of one or more chemicals involved
at different times during the chemical reaction.
Reaction rates are expressed as the number of
moles per liter that react each second during the
reaction.
The units of a reaction rate aremol L-1 s-1 M
s-1.
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Considering Reaction A ? B
The reaction rate is a measure of how quickly A
is consumed or B is produced.

• Average rate of reaction can be written
• (remember ? final initial)

This is a measure of the average rate of
appearance of B.
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• Average rate can also be written in terms of A

This is the rate of disappearance of A (negative
value makes final rate positive).
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Data for reaction A?B
Notice the decrease in rate as A is used up.
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? Moles A ? Moles B
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Start with one mole of A at time zero, measure
amounts of A and B at given time intervals.
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Data for Reaction A?B
Time A B
0 seconds 1.0 mol 0.0 mol
20 seconds 0.54 mol 0.46 mol
40 seconds 0.30 mol 0.70 mol
Calculate the average rate of appearance of B
over the time interval from 0 to 40 seconds.
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You Try One!
Time A B
0 seconds 1.0 mol 0.0 mol
20 seconds 0.54 mol 0.46 mol
40 seconds 0.30 mol 0.70 mol
Calculate the average rate of disappearance of A
over the time interval from 0s to 40s.
Same Value Notice negative sign makes value
positive.
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Average Reaction Ratereaction rate over a given
amount of time. Example Change in
concentration during the time period of 20s-40s
during the reaction (as in previous examples).
Instantaneous Reaction Ratereaction rate at a
given moment in time. Notes Initial Rate where
time 0 is an instantaneous rate.
Use kinetic curve to calculate this.
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Sometimes the unit on the y axis is moles (Liters
are assumed).
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Remember the equation for finding the slope (math
class).
We use the same equation to determine the
instantaneous (for a specific time) rate of a
reaction from a kinetic curve.
(
)
Reaction rates have values - sign makes rate
of disappearance .
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Rates in Terms of Concentration

• The volume in the reaction vessel remains
  constant.
• Therefore, in analyzing the reaction A?B we can
  actually consider the rate in terms of Molarity.
• This gives us the units M/s (molarity per second).

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Consider the reaction between butyl chloride and
water
C4H9Cl(aq) H2O(l)? C4H9OH(aq) HCl(aq)
Brackets indicate concentration.
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• Using the curve created from the data, we can
  determine the instantaneous rate for any given
  point on the curve. (slope of tangent line
  tangent line touches curve at point of interest).

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Calculate the rate of disappearance of C4H9Cl at
t 0.
0.06 M 0.1M
Rate -
200 s 0.0 s
2 x 10-4 M/s
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Calculate the rate of disappearance of C4H9Cl at
t 300.
0.02 M 0.055 M
Rate -
600 s 300 s
1 x 10-4 M/s
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Rates and Stoichiometry

• When mole ratios of equations are not 11.
• aA bB?cC dD

This is how your book explains this. ? Too
Complicated!?
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The lengthy explanation provided by your text
basically boils down to A PICKET FENCE
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• If the rate of decomposition of N2O5 in a
  reaction vessel is 4.2 x 10-7 M/s, what is the
  rate of appearance of NO2?
• 2N2O5(g)? 4NO2(g) O2(g)

Given 4.2 x 10-7 mol L-1 s-1 N2O5
4.2 x 10-7 mol L-1 s-1 N2O5
4 mol NO2

2 mol N2O5
8.4 x 10-7 mol L-1 s-1 NO2
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• If the rate of decomposition of N2O5 in a
  reaction vessel is 4.2 x 10-7 M/s, what is the
  rate of appearance of O2?
• 2N2O5(g)? 4NO2(g) O2(g)

Given 4.2 x 10-7 mol L-1 s-1 N2O5
4.2 x 10-7 mol L-1 s-1 N2O5
1 mol O2

2 mol N2O5
2.1 x 10-7 mol L-1 s-1 O2
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What are the 4 factors that affect the rate of a
chemical reaction?

• Concentration of reactants
• Temperature at which the reaction occurs
• The presence of a catalyst
• State and Surface area of solid or liquid
  reactants.

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Concentration and Reaction Rates

• Increasing concentration of reactants increases
  reaction rate. Examples Add more wood to fire
  step on gas in automobile.
• How do you think the reaction rate would change
  over time?
• Reaction rates usually decrease over time.
• Due to decreasing concentration of reactants.

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Compare reaction rates with varying
concentrations of reactants
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The effect of concentration on a chemical
reaction is expressed in a rate law.
k rate constantbrackets concentration of
reactantsThe exponents m and n are called
reaction orders.
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Reaction Order

• General form for rate laws

Reactions show how changing a concentration on a
given reaction influences the rate of a
reaction. Sum of m and n is the overall reaction
order.
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• Reaction orders are determined experimentally and
  do not relate to coefficients of balanced
  equation (We will see how to determine these in a
  bit)
• .
• In most rate laws reaction orders are 0, 1, or 2.
• Can be fractional or negative at times.

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• In the above equation NH4 NO2- are raised to
  the first power.
• The reaction is 1st order for NH4
• The reaction is 1st order for NO2-
• The reaction is 2nd order overall

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Units for Rate Constant (k)

• The units for the rate constant depend on the
  order of the rate law.

Z overall order of reaction
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• The rate law of the following reaction was
  experimentally determined to be
• H2(g) I2(g)? 2 HI(g)
• Rate k?H2??I2?
• What is the order of reaction for H2? I2?
• 1st order.
• What is the overall reaction order?
• 2nd order
• What are the units for the rate constant of the
  reaction above?

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• What are the units for the rate constant of the
  reaction below?
• 2N2O5(g) ? 4NO2(g) O2(g)
• Rate k N2O5

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Determining Rate laws

• Rate law for any reaction must be determined
  experimentally.
• Zero order for a reactant means concentration
  changes have no effect on reaction rate (straight
  line on graph).

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Example of data that would give us a 0 order of
reaction.
C
As we double B and keep A constant, rate doesnt
change.
Reaction order of B 0.
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• 1st order means concentration changes give
  proportional changes in reaction rate doubling
  concentration doubles rate (parabola on graph
  because rate decreases as reactant is used up).

• 2nd Order rate law, increasing concentration
  results in a rate increase equal to the
  concentration increased to the second power.
  (parabola is steeper in beginning ramp at end).
• Double con. 22 4 (rate increase)
• Triple con. 32 9 (rate increase)

1st
2nd
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What is the reaction order of A?
C
As we double A and keep B constant, rate
increases by factor of 4.
Reaction order of A 2 because 22 4.
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Once we know the reaction orders of the reactants
involved, we can write the rate law. Since the
reaction order of A was 2 and the reaction order
of B was 0, the rate law is Rate kA2 We
dont include B in the rate law because any
number raised to the 0 power 1.
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• A particular reaction was found to depend on the
  concentration of the hydrogen ion. The hydrogen
  concentration was changed in each of the 3
  reactions and the following rates of reactions
  were found in each experiment.
• H (M) 0.0500 0.100 0.200
• Initial rate(M/s) 1.6x10-7 3.2x10-7
  6.4x10-7
• What is the order of the reaction in H?1st
  order
• Predict the concentration of H when initial
  rate 0.80 x 10-7 M/s.0.250 M

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What is the reaction order of NH4?
1
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What is the reaction order of NO2-?
1
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What is the rate law of the reaction?
Rate kNH4NO2-
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What is the overall order of the reaction?
2
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Once we determine the rate law we can apply the
experimental data to the rate law to find the
value of k.
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Consider the following data for the upcoming
slides.
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Calculate the value of the rate constant in the
following scenario.

• NH4 0.0200M
• NO2- 0.200M
• Rate 10.8 x 10-7 M-1s-1

Rate k NH4 NO2-
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• When we know the rate law and the rate constant
  for a reaction, you can calculate the rate of
  reaction for any given concentrations of
  reactants.

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• What is the rate of reaction if the following
  concentrations are present.
• NH4 0.100M
• NO2- 0.100M
• k 2.7 x 10-4 M-1s-1

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NH4 0.100 M NO2- 0.100 M k 2.7 x 10-4
M-1s-1
Rate k NH4 NO2-
Rate 2.7 x 10-4 M-1 s-1 (.100 M) (.100 M)
Rate 2.7 x 10-6 M / s
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• Typical Problem
• Determine rate law from data.
• Determine value of k including units.
• Determine rate given concentrations not in data
  table.
• Common question Does k change as concentration
  changes? Answer No, k only changes with
  temperature.

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Note The previous slides related to reactants
that were both 1st order. This is not always the
case. Handout Explanation of Determining Rate
Law(Read in Class) Practice Problem Handout
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Change in Concentration with time

• Rate law tells how rate changes as concentration
  changes at a particular temperature.
• Using calculus, we can convert the rate laws into
  equations that can give us the concentrations of
  reactants or products at any time during a
  reaction.
• Dont worry. No calculus! The derived equations
  are are given on the AP exam. You must know how
  to use the equations.

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Equations Listed on AP Exam
Used for First Order Reactions
Used for Second Order Reactions
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First-Order Reactions

• Equation to relate beginning concentration (A0)
  to concentration at any time (At).
• Given any of the 3 quantities, we can solve for
  the fourth (t, k, At, A0).
• Make sure units of At and A0 are the same.

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• The logarithm of a number is the power to which
  10 must be raised to equal that .
• Example logarithm of 1000 3, because 10
  must be raised to 3 to equal 1000.
• ln is the natural logarithm. Natural logarithms
  are the power to which e (2.718) must be raised
  to equal that number.
• Example The ln of 7.39 is equal to 2.
  Therefore, e2.00 7.39.

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A certain first order reaction has a rate
constant of 4.5 x 10-3 s-1. What is the
concentration of a 0.050 M sample after it reacts
for 75.0 s?
Because we need the inverse of the natural
logarithm (antilogarithm), type into calculator
ex -3.334 At 0.356 M
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• m is the slope (-k)
• b is the y-intercept of the line (lnA0)

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2N2O5(g) ? 4NO2(g) O2(g)

• The decomposition of dinitrogen pentoxide is a
  1st order reaction with a rate constant of 5.1 x
  10-4s-1 at 45ºC.
• a.) If initial conc. is 0.25M, what is the
  concentration after 3.2 min.?
• b.) How long will it take for the concentration
  of N2O5 to decrease from 0.25M to 0.15M?

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A0 0.25 k 5.1 x 10-4 s-1
t 192 s
Type into calculator ex -1.485 At 0.2265 M
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How long will it take for the concentration of
N2O5 to decrease from 0.25M to 0.15M?
A0 0.25 At 0.15 k 5.1 x 10-4
s-1 t ?
T 1.0 x 103 s
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Pressure can be used as unit of concentration.
(remember pressure is directly related to n/V in
ideal gas equation (PV nRT).
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(CH3)2O?CH4(g) H2(g) CO(g)

• The decomposition of dimethyl ether (CH3)2O, at
  510ºC is a first order process with a rate
  constant of 6.80 x 10-4s-1
• If the initial pressure of (CH3)2O is 135 torr,
  what is its partial pressure after 1420 s?

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A0 .1776 atm k 6.80 x 10-4 s-1
t 1420 s
Type into calculator ex -2.694 At 0.0676 atm
51 torr
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Second-Order Reactions

• Rate of reaction depends on the reactant
  concentration raised to the second power or
  depends on concentrations of two different
  reactants, each to the 1st power.

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Here is the equation given on the AP Exam.
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• Plot of 1/At versus T gives a straight line.
• Slope k
• Y-intercept 1/A0

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The decomposition of NO2 is second order. The
rate constant for the reaction is k 0.543 M-1
s-1. If the initial concentration of NO2 in a
closed vessel is 0.0500 M, what is the
concentration after 0.500 hr?
Answer 1.00 x 10-3 M
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Using Lab Data to Predict Order First order
reactions give straight lines when we graph
lnAt vs. t. Second order reactions give
straight lines when we graph 1/At vs. t. Once
we graph the line, the slope can be determined,
which is k. See Sample exercise 14.8 on page 540.
Homework 14.37-14.38 Should use graph paper.
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Half-life (t1/2)

• Half life is the time it takes the initial
  concentration of reactant to drop to one-half its
  value.
• Consider a reactant that is 0.120 M. At some
  point in the reaction it will be 0.060 M. This
  would occur after one half-life. After a second
  half-life passes, the concentration would be
  0.030 M.
• If the half life of the reaction above is 30
  seconds, what will the concentration of the
  reactant be after 2 minutes?
• Answer 0.0075 M

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One way to deal with half-lives is to consider
the 2 in ½ to the power of the number of half
lives. Example A reaction has a half-life of
10 seconds. If its initial concentration is
10.0 M, what is the concentration after 30
seconds? This represents 3 half lives. 23 8.
1/8 .125 .125 x 10.0 M 1.25
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• In 1st order reactions the concentration of the
  reactant decreases by 1/2 in a series of regular
  time intervals (t1/2). The decreasing rate
  depends only on k and doesnt depend on A0.

Not given on exam. Can derive from equation
given earlier (see page 541 if interested.
Easier to just memorize).
Half-life is used to describe radioactive decay
and elimination of medications from the body.
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CH2CH2 CH2 ? CH3CH CH2

• Conversion of cyclopropane to propene in the gas
  phase is a first-order reaction with a rate
  constant of 6.7 x 10-4s-1 at 500ºC. Calculate the
  half-life of the reaction.

Answer 1034 s
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Half-Life of 2nd Order reactions

• Unlike 1st order reactions, the half-life of
  second order reactions is dependent on the
  initial concentration of reactant.

Just know the information above. We wont be
doing any calculations with these to save time.
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Temperature Rate

• Most reactions increase in rate with increasing
  temperature.
• This is due to an increase in the rate constant
  with increasing temperature.

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Collision Model

• Collision model for chemical reactions is based
  on Kinetic Molecular Theory.
• Number of collisions increase with increasing
  concentration.
• Collisions increases with increasing molecular
  speed (temp.)

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Activation Energy
Ea

• Minimum amount of energy required to initiate a
  chemical reaction.
• Varies from reaction to reaction
• According to the Collision Model, this energy is
  the KE of colliding molecules.

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Activation energy must be enough to overcome
initial resistance for the reaction to take place.
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• The difference in energy between the starting
  point and the highest energy along the pathway is
  the activation energy (Ea).
• The arrangement of the atoms at the highest
  energy point is called the activated complex or
  transition state.

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• Kinetic Molecular theory states that molecules in
  a sample have a wide variety of energy levels.
• Increasing the temperature of a sample places a
  larger fraction of the molecules with energy of
  Ea.
• The lower the activation energy the faster the
  reaction.
• See figure 14.16 page 546 and sample exercise
  14.10 on 547

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Which energy diagram represents a highly
exothermic reaction that has a small activation
energy? Which reaction would be the slowest? Is
this reaction endothermic or exothermic?
Endothermic
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Catalyst

• Catalyst substance that changes the rate of
  reaction without undergoing permanent change
  during the process.
• Homogenous catalyst catalyst that is present in
  the same phase as the reactant molecules.

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• As a general rule catalysts change the rate of
  reaction by the lowering of Ea (activation
  energy).
• Usually this is done by giving a completely
  different mechanism for the reaction.
• This lowers the overall Ea.

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Catalyzed pathway is two steps, both with lower
Ea than the original reaction pathway.
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Heterogeneous Catalysis

• Catalyst exists in a different phase from the
  reactant molecules.
• Usually a solid in contact with gas or liquid
• Often times metal or metal oxides.

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• Adsorption usually the initial step of
  heterogeneous catalysis.
• Uptake of molecules into the interior of another
  substance.
• Increasing surface area of catalyst increases
  effectiveness.
• Active sites places on the surface of a
  molecule where adsorption takes place.

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Enzymes

• Enzymes are biological catalyst of reactions that
  occur in living systems.
• Most enzymes are large protein molecules.
• They are selective in the reactions that they
  catalyze.

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• Active sites
• Lock-and-key model

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Energy Diagram Handout (Please read when I am
finished going over the rest of the slides). We
wont be practicing any of the next few slides
because of the lack of time. I will simply go
through them quickly so you are aware of the
equations and what they are related to.
Hopefully, questions related to these wont show
up on the exam, but if they do, at least seeing
the equations might help.
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The fraction of the molecules that possess Ea is
given by the following equation (not given on
exam equation sheet, will likely be given in
question).

• R gas constant (8.314 J/mol)
• T Absolute temperature
• Ea the activation energy/mol
• f fraction of molecules that possess Ea

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Calculate the fraction of atoms in a sample of
argon gas at 400K that have an energy of 10.0 KJ
or greater.
Because we need units to cancel and R uses
Joules, we must make sure activation energy is in
joules.
10.0 KJ 1.00 x 104 J/mol T400K
fe-3.0070 4.94 x 10-2
1/ 4.94 x 10-2 20 Roughly 1/20 atoms has
this energy.
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Orientation Factor

• Collisions can occur between molecules with
  sufficient Ea, and still no reaction may occur.
• The collisions must occur with proper orientation
  of the molecules that are reacting.

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Observations of Arrhenius

• Increase in rate in relationship to temperature
  increase was non-linear.
• Reaction rate obeyed an equation based on 3
  factors
• f
• of collisions per second
• Fraction of collisions with proper orientation.

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Arrhenius Equation

• k the rate constant
• R gas constant (8.314 J/mol)
• T Absolute temperature
• Ea the activation energy
• A frequency factor (constant at varied
  temperatures related to frequency of collisions
  and probability of proper orientation).

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Taking the natural log of both sides gives a
formula in straight line form
Given on Exam

• Graph of ln k versus 1/T will be a straight line
  with a slope of Ea/R and a y-intercept of ln A.

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Simplifying the equation to calculate a rate
constant (k1) at a temperature (T1) when the Ea
and rate constant (k2) at a temperature (T2) are
given.
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• The rate constant of a first order reaction is
  3.46 x 10-2 s-1 at 298 K. What is the rate
  constant at 350 K if the activation energy is
  50.2 kJ/mol?

143

• k1 3.46 x 10-2 s-1 k2 ?
• T1 298 K T2350 K

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k2 0.702 s-1
145

• The first-order rate constant for the reaction of
  methyl chloride with water to produce methanol
  and hydrochloric acid is 3.32 x 10-10 s-1 at
  25ºC. Calculate the rate constant at 40ºC if the
  activation energy is 116 kJ/mol.

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Reaction Mechanisms

• The process by which a reaction occurs.
• Chemical equations only show beginning and ending
  substances
• Can show in detail bond breaking and forming and
  structural changes that occur during a reaction.

148
Elementary steps

• Processes that occur in a single event or
  step, are elementary processes.
• Particles collide with sufficient energy and
  proper orientation for reaction to occur.

149

• The number of molecules that participate as
  reactants define the molecularity of the step.
• One molecule Unimolecular
• 2 reactant molecules bimolecular

150
Multistep Mechanisms of Rx

• Often times chemical reactions are a result of
  multiple steps not show by the overall equation.

This Rx below 225ºC occurs as 2 elementary steps.
151

• 1st step
• 2 NO2 molecules collide
• The NO3 then collides with CO and transfers an O.
• The elementary steps must add to result in the
  overall chemical equation.

152

• The proposed mechanism for the above reaction is

a.)Is the proposed mechanism consistent with the
overall rx? b.)Identify any intermediates.
a.) yes b.) Mo(CO)5
153
Rate Laws of Elementary Steps

• Every reaction is made up of a series of
  elementary steps
• Rate laws reflect the relative speeds of these
  steps.
• Rate laws must be determined experimentally

154
Elementary Step Reactions

• The rate law of any elementary step is based
  directly on its molecularity.
• Unimolecular 1st order (A?product)
• Rate kA
• Bimolecular 2nd order (AB ?prod)
• Rate kAB

155
Rate Laws for All elementary steps
156

• Consider the reaction above.
• Write the rate law for the reaction, assuming it
  has a single elementary step.
• Is a single step mechanism likely for this
  reaction? Why or why not?

1. Rate kNO2Br2
2. No, termolecular rxs are rare

157
Rate Laws of Multistep Mechanisms

• Most reactions involve multiple steps.
• Often one step is much slower than the other.
• The Rate determining step is the slowest step
  in the reaction.

158

• The slowest step of a multi-step reaction
  determines the overall rate of reaction.

slow
fast
159
Step 1 is the rate determining step, because it
is the slow step.Therefore the rate of the
overall reaction is the rate of step 1.

• Rate k1NO22
• Step one is a bimolecular process
• This rate law does in fact follow the observed
  rate of reaction.

160
Initial fast step mechanisms

• The first step in a multi-step mechanism of this
  type is no longer rate limiting.
• It is more difficult to derive the rate under
  these circumstances.

161

• The experimentally determined rate law for the
  reaction is
• Rate kNO2Br2
• Must find a reaction mechanism that is consistent
  with the rate law.

?
162

• An alternative that does not involve termolecular
  steps
• Step 1 NO(g) Br2(g) NOBr2(g) fast
• Step 2 NOBr2(g) NO 2NOBr(g) slow

k1
k-1
k2
163
Step 2 NOBr2(g) NO 2NOBr(g)
k2

• Step 2 is the Rate Determining step.
• The overall reaction is governed by the rate law
  for this step.

Intermediates are unstable and concentrations are
unknown.
164
The rate of the forward and reverse rx. are equal
in dynamic equilibrium.

• k1NOBr2 k-1NOBr2
• Solve for concentration NOBr2

Substitute this relationship into rate law
165
2NO2 F2(g)?2NO2F(g)

• The rate law for the above reaction is Rate
  kNO2F2
• Suggested mechanism is
• NO2 F2 NO2F F slow
• F NO2 NO2F fast

k1
k2
Is this mechanism acceptable? Does it satisfy the
2 requirements?
166

• 1st requirement is that the sum of the steps
  gives the balanced equation
• NO2 F2 NO2F F
• F NO2 NO2F
• 2NO2 F2 F 2NO2F F

k1
k2
Overall 2NO2 F2(g)?2NO2F(g)
167

• 2nd requirement is that the mechanism agree with
  the experimentally determined rate law
• Mechanism states that 1st step is rate limiting.
• 1st step is bimolecular
• Therefore Rate kNO2F2

168

• The first step of a mechanism involving the
  reaction of bromine is
• Br2(g) 2Br(g) (fast/equilibrium)
• What is the expression relating concentration of
  Br to concentration of Br2?

k1
k-1