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Chemical Kinetics
Rates of chemical reactions and how they can be measured
experimentally and described mathematically
So far, we have worked with reactions that occur almost
instantaneously
• Precipitations– Ba2+ (aq) + SO4
2- (aq) BaSO4 (s)
• Acid-base reactions– HCl(aq) + NaOH (aq) NaCl (aq)+ H2O
(l)
Lots of reactions are much slower…
• Rusting– 4Fe (s) + 3O2 (g) 2Fe2O3 (s)
• Formation of ammonia– N2 (g) + 3H2 (g) 2NH3 (g)
• Formation of diamond– C (graphite) C (diamond)
In this chapter we will…
• Explore the factors that affect rates of reactions
• Quantify the influence of the above factors on the rates of reactions
• Determine what happens at the molecular level in reactions
REACTION RATES• The change in molar concentration
of a reactant or a product per unit time
• Units are Moles/L s
In an experiment, the decomposition of N2O5 in 100 mL of a 2.00 M CCl4 solution produced 0.500 L of oxygen at STP after 200 minutes. Calculate the concentration of unreacted N2O5 at this time. 2N2O5 (in CCl4) 4NO2 (in CCl4) + O2 (g)
• ? Moles oxygen produced• ? Moles N2O5 reacted
• ? Moles N2O5 initially
• ? Moles N2O5 left over• ? Moles/L
• 0.500 L x 1mole = 0.02232 mol O2
22.4 L
• 0.02232 mol O2 x 2N2O5 = 0.04464 mol N2O5
1 O2
• Initially… (0.100L) (2.00M) = 0.200 mol N2O5
• 0.200 mol – 0.04464 mol = 0.155 mol left over
• M = 0.155 mol/ 0.100 L = 1.55 M
Following this procedure we can calculate the concentration of unreacted N2O5 at any stage of the reaction and plot the data as shown
Mol/L
Time
To find the rate at any instant, draw a tangent to the curve and determine its slope
This gives the average rate over time, D[N2O5]/Dt
If you know the rate of one substance in a reaction, you can determine the rate of any other substance in the reaction by using a mole ratio2N2O5 (in CCl4) 4NO2 (in CCl4) + O2
(g)N2O5 decomposes at a rate of
0.23M/s,what is the rate of formation of NO2 ?0.23M N2O5 x 4 NO2 = 0.46 M/s NO2
s 2 N2O5
Rate Laws• Have the general form: rate =
k[A]n
• [A] = concentration of reactants and catalyst
• k = rate constant• n = order of reactant
(an integer or fraction)
Initial Rate Method• To determine the form of the rate
law, one MUST use experimental data
• Do a series of experiments in which the initial concentrations of the reactants are varied one at a time and record the initial rates of reaction
(OH-)
I- (aq) + OCl- (aq) OI- (aq) + Cl- (aq)
initial M Rate (mol/Ls)
I- OCl- OH-1) 0.010.01 0.01 6.1 x 10-4
2) 0.020.01 0.01 12.2 x 10-4
3) 0.010.02 0.01 12.3 x 10-4
4) 0.010.01 0.02 3.0 x 10-4
Rate = k[I-]x [OCl-]y [OH-]z
1: 6.1 x 10-4 = k[0.01]x[0.01]y[0.01]z
2: 12.2 x 10-4 = k[0.02]x[0.01]y[0.01]z
Solve for x…
0.5 = 0.5x
X = 1
1: 6.1 x 10-4 = k[0.01]x[0.01]y[0.01]z
3: 12.3 x 10-4 = k[0.01]x[0.02]y[0.01]z
Solve for y
0.5 = 0.5y
Y = 1
1: 6.1 x 10-4 = k[0.01]x[0.01]y[0.01]z
4: 3.0 x 10-4 = k[0.01]x[0.01]y[0.02]z
Solve for z
2 = 0.5z (ohhh…this is tricky!)Log 2 = log 0.5z
Z log 0.5 = log 2Z = log 2/log 0.5
Z = -1
Rate = k[I-][OCl-]/[OH-]
The reaction is first order with respect to
I- and OCl-, and inverse first order with
respect to [OH-].What is the overall reaction order?(sum of exponents)What are the units of k for this
reaction?(plug units into the rate law and
solve)
Integrated Rate Law Method
First Order Rate Lawsrate = k [A]1 = - D [ A ] / D t
orrate = -D[A] = k D t
[A]
Integrating both sides gives…
-ln ( [A]t / [A]0 ) = kt
ln ( [A]0 / [A]t ) = kt
ln [A]t = -kt + ln [A]0time
ln[A]t
slope = - k
Half-life• time it takes for the reactant
concentration to reach one half of its initial value
• symbol t ½ • for first order:
t ½ = (ln 2) / k
Integrated Rate Law Method
Second Order Rate Lawsrate = k [A]2 = - D
[ A ] /D tintegrating gives
1 / [A]t = kt + 1/[A]0
t1/2 = 1 / k[A]0
Integrated Rate Law Method
Zero Order Rate Lawsrate = k [A]0 = k
integrating gives[A]t = -kt + [A]0
t1/2 = [A]0 / 2k
Collision Theory
The rate of a reaction depends on the
1. concentration of reactants2. temperature3. presence/absence of a
catalyst
The value of the rate constant is
dependent on the temperature
How can we explain the effect of
temperature on the rate of areaction?
NO(g) + Cl2(g) --> NOCl(g) + Cl- (g)
At 25oC: k= 4.9 x 10-6 L/mol sAt 35oC: k= 1.5 x 10-5 L/mol s
k has increased by a factor of THREE!
WHY?
The Collision Theory states that in order to react, molecules have to collide….
• with the proper orientation
• with an energy at least equal to Ea
activation energy (Ea ) =
required minimum energy for areaction to occur
k for a reaction depends on 3 things:
• Z = collision frequency (# collisions/second)higher temperature means more collisions
• f = fraction of collisions that occur with E > Ea
*this factor changes rapidly with Tf = e –(Ea/RT)
• p = fraction of collisions that occur with the proper orientation (independent of T)
Overall: k = pfz = A e -(Ea/RT)
• where A = pz• Arrhenius Equation: ln k = -Ea 1 +
ln AR T
• ln k2 = Ea 1 1 k1 R T1 T2
YOU TRY!
Calculate the activation energy for the reaction: 2HI (g) --> H2 (g) + I2 (g)
GIVEN: k at 650. K = 2.15 x 10-8 L/mol s and k at 700. K = 2.39 x 10-7 L/mol s
R = 8.3145 J / mol K1.82 x 105 J
Transition State Theory
Two reactants come together to form an
Activated Complex, or TRANSITION STATE which then separates to form theproducts.
In the activated complex, the N—Cl
bond has partially formed, while the
Cl—Cl bond has partially broken.
• Breaking bonds requires an input of energy while forming bonds
RELEASES energy.
• If E reactants > E products then the reaction is EXOTHERMIC
• If E reactants < E products then the reaction is ENDOTHERMIC
• A catalyst increases the rate of reaction by…LOWERING THE ACTIVATION ENERGY
• Homogeneous Catalysis- catalyst is in the same phase as the reactants
• Heterogeneous Catalysis- catalyst is in a different phase than the reactants
Reaction Mechanisms
The overall balanced equation usually
represents the SUM of a series of simple
reactions called ELEMENTARY STEPSbecause they represent the progress
ofthe reaction at the molecular level.
The sequence of elementary steps iscalled the REACTION MECHANISM
NO2 (g) + CO (g) NO (g) + CO2 (g)Rate = k [NO2]2
The above reaction actually takes place in two steps:
1. NO2 + NO2 NO3 + NO SLOW
2. NO3 + CO NO2 + CO2 FAST
The reaction mechanism must satisfy two requirements:
1) Sum of elementary steps must be the overall reaction
2) The rate law indicated by the mechanism must match the experimentally determined rate law
2 H2O2(aq) O2(g) + 2 H2O(l)by experiment Rate = k[H2O2] [I-]
H2O2(aq) + I-(aq) IO-(aq) + H2O(l) SLOW
H2O2(aq) + IO-(aq) I-(aq) + H2O(l) + O2(g)
The overall rate of the reaction is controlled by the slow step also known as the…..
RATE DETERMINING STEP
H2O2(aq) + I-(aq) IO-(aq) + H2O(l) SLOW
H2O2(aq) + IO-(aq) I-(aq) + H2O(l) + O2(g)
When the slow step is used to determine
the rate law, we get:
What is the intermediate in this reaction?
What is the catalyst?
2N2O5 (g) ----------> 4NO2 (g) + O2 (g)
Rate = k[N2O5]
Why can’t this be a one-step reaction?
If it was a one-step, the overall reaction
would be that one step, and the rate law
would be rate = k[N2O5]2
A general example:
E + S MfastM E + P slow
Work out the rate law in terms of
reactants and catalyst:
Is this an acceptable mechanism? Why or why not?
Overall Reaction = 2NO2 (g) + F2 (g) 2NO2F (g) Rate = k[NO2] [F2]
Proposed MechanismStep 1: NO2 + F2 NO2F + F slow
Step 2: F + NO2 NO2F fast