# Kinetics

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Kinetics

 Studies the driving force behind reactions
 Explores the rate at which chemical reactions occur
 Covers time b/w start of reaction and point equilibrium is reached or reaction stops
 Used in many ways like setting expiration dates on consumer products
Reaction Rate:         rate of change in concentration per unit time
 # of moles per liter that react each second (units- mol/L s)
 can only be determined experimentally
 All data will be given to you on AP exam!
 Usu. associated with rate at which reactant is being used up (rate of disappearance) or
rate at which product is being created (rate of appearance)
 Data collected should track the appearance or disappearance of a substance over a
time interval
 Data usu. graphed to show progression of reaction in a graph called Kinetic Curve or
Concentration versus Time Curve
 See example graphs in textbook, Pages 558, 564, & 571
 Using Curve:
The rate is obtained by determining the slope of the curve at a desired pt. in time
3 step process
1. select desired pt on curve, and draw tangent to it
2. select 2 pts on tangent, and determine the concentrations, c, and times, t,
corresponding to these pts.
3. calculate the slope using equation:     Rate = _ C2-C1 = _ ∆C
t2-t1    ∆t
(equation look familiar math whizzes? Slope? Rise over run? y/x or conc/time)
 Example: aA + bB → cC + dD
-could measure conc. of one of the substances over time and calc other concs
stoichiometrically
-could also collect data about one of the substances and determine how conc changes
over various time intervals; therefore determining the rate!
- only 2 measurements are used, so the rate is considered an average and the graph
should be a curve instead of a straight line, therefore: average rate is much more
accurate for smaller time intervals
Average rate = ∆[A] = ∆C = [A]final – [A]initial
∆t     ∆t     tfinal - tinitial

   Since reactants disappear during reactions, rates calc by measuring a reactant will be
negative
   If product is measured, calculated rate will be positive
   Chemists always use a positive val. regardless, therefore;
o Rate = ∆Cproducts            OR              Rate = -∆Creactants
∆t                                       ∆t
   The sign is implied by calling the rate either rate for the appearance (implied positive
sign) of products or rate of disappearance (implied negative sign) of reactants
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Stoichiometry
   Rate expression for equation aA + bB → cC + dD
Rate = -(1/a)(∆[A]/∆t) = -(1/b)(∆[B]/∆t) = (1/c)(∆[C]/∆t) = (1/d)(∆[D]/∆t
   Example: For the reaction 2N2O5(g) → 4NO2(g) + O2(g) write the rate expression
for each substance.
   Answer: Based on the balanced equation, 2 moles of N2O5 decomposes for every 1
mole of O2 produced, so the rate of disappearance will be ½ the rate of appearance
Rate = -(1/2)(∆[ N2O5]/∆t) = (¼)(∆[NO2]/∆t) = (∆[O2]/∆t
   Now you try it!
Write the rate expression for each substance in the following reaction:
2N2O(g) → 2N2(g) + O2(g)

  Now we’ll use the following reaction in our next example:
2C2H6 + 7O2 → 4CO2 + 6H2O
Assume the rate was determined measuring CO2 produced and was found to be
2.50mol CO2 L-1s-1
Set up problem in form of stoichiometric question: “What will the rate of
consumption of C2H6 be if the rate of production of CO2 is 2.50 mol/L s?
?mol C2H6 = 2.50 mol CO2
Ls              Ls

Since denominators are the same, we simply need to convert mol CO2 into mol C2H6!

mol C2H6 = 2.50 mol CO2           X       2 mol C2H6
Ls            Ls                         4 mol CO2

= 1.25 mol C2H6
Ls
Rates of O2 and H2O may be calculated in same manner
 Reaction rates must specify which reactant or product it applies to

Factors that Effect Reaction Rates
   Conc of reactants
   Temp
↑ temp, ↑ reaction rates
↓ temp, ↓ reaction rates
   Catalysts: participate in reaction, but not included in balanced equation
   Other factors will have no effect on the rate if they do not change the conc, temp, or
catalyst

Rate Laws
   Concs expressed in rate law
   All rate laws start w/ same form: Rate = k[A}x[B]y[C]z
K=rate constant            []= conc of reactants
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Exponents determined from experimentation and have no relationship to coefficients
of balanced chemical equation!
   Determined by running several experiments in which reaction concs are changed, to
determine whether the change effects the rate
o If change occurs, the exponent for the change may be calculated
o No theoretical way to predict the exponents
   Example: Kinetic data for the peroxydisulfate reaction at 20.0ْ C
Experiment [S2O82-] (mol L-1) [I-] (mol L-1)         Initial Rate of Reaction
#                                                     (mol L-1s-1)
1              0.200             0.200                 2.2 X 10-3
2              0.400             0.200                 4.4 X 10-3
3              0.400             0.400                 8.8 X 10-3

Rate Law → Rate = k[S2O82-]x[I-]y           solve for exponents x & y
Focus your attention on the rate change from exp 1 to exp 2 as conc of S2O82- is
changed and write the rate law for each
Rate 1 = 2.2 X 10-3 = k(0.200)x(0.200)y
Rate2 = 4.4 X 10-3 = k(0.400)x(0.200)y
Then write ratios of the 2 equations w/larger # on top for simpler calculations.
Rate2 = 4.4 X 10-3 = k(0.400)x(0.200)y
Rate 1 2.2 X 10-3 k(0.200)x(0.200)y

Cancel the 2 k’s and (0.200)y factors out:   Rate2 = 4.4 X 10-3 = (0.400/0.200)x
Rate 1 2.2 X 10-3                   x=1
Now determine the value of y using exp 2 & 3. After all calculations are done, y=1 also.
Rewrite the rate law: Rate = k[S2O82-]1[I-]1

Now the rate constant may be calculated using the rate law expression from either exp.
Experiment 1: k = 2.2 X 10-3 mol L-1s-1             = 0.055L mol-1s-1
(0.200 mol L-1)(0.200mol L-1)

Experiment 2: k =       4.4 X 10-3 mol L-1s-1      = 0.055L mol-1s-1
-1          -1
(0.400 mol L )(0.200mol L )

Experiment 3: k =       8.8 X 10-3 mol L-1s-1          = 0.055L mol-1s-1
(0.400 mol L-1)(0.400mol L-1)
Keep track of units on constant b/c if is not set! It may change w/ the order of the
reaction.
If the rate constants are not equal or close to equal, something is wrong with the rate
law!

   The order of reaction is used to indicate the exponents in the rate law
   The example problem was 1st order w/respect to peroxydisulfate, 1st order w/respect
to iodide ions, and 2nd order overall
o Individual reactants have the same orders as their exponents
o Order of overall reaction is sum of all exponents
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REMEMBER THE DIFFERENCE
REACTION RATE
(or simply rate) varies with the conc of reactants and time. Rate always
has units of mol L-1s-1
RATE CONSTANT
This is a constant value at a fixed temperature for a given reaction. The
units of a rate constant depend on the order of the reaction.

Determine the rate law and value of the rate constant, with its units, for the following
data.

Experiment        [A] (mol L-1)      [B] (mol L-1)     Initial Rate of Reaction
#                                                         (mol L-1s-1)
1                 0.100              0.200                 1.1 X 10-6
2                 0.100              0.600                 9.9 X 10-6
3                 0.400              0.600                 9.9 X 10-6

Answer: Rate = k[A]0[B]2 = k[B]2
Substitute the rate and concs into the rate law to calculate k
1.1 X 10-6 mol L-1s-1 = k(0.200 mol B L-1)2        k = 2.75 X 10-5 L mol-1s-1

Practice Problems
What is the overall order of each of the following rate laws, and what are the units of the
rate constant, k, in each of these rate laws?
(a) Rate = k[A][B][C]
(b) Rate = k[X]2[Y]3
(c) Rate = k[M]2[N]
(d) Rate = k
(e) Rate = k[R]
Solution
(a) Order = 3. Units are L2 mol-2s-1
(b) Order = 5. Units are L4 mol-4s-1
(c) Order = 3. Units are L2 mol-2s-1
(d) Order = 0. Units are mol L-1s-1
(e) Order = 1. Units are s-1

Effect of Temp on Reaction Rates
   Svante Arrhenius developed an equation for the relationship b/w rate constant and
temp
   Arrhenius Equation: k = Ae-Ea/RT
K = rate constant        Ea = activation energy        R = univ. gas constant
T = Kelvin temp          A = proportionality constant         8.314 J mol-1K-1
E = base of natural logs

   When natural log taken, equation becomes:
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ln k =       -Ea + ln A
RT
 May be used to compare reaction rates at 2 different temps
ln (k1/k2) = (-Ea/R) (1/T1 – 1/T2)
 May also be used in place of rate constant
ln (rate1/rate2) = (-Ea/R) (1/T1 – 1/T2)
 When working these problems, keep in mind
(a) Larger rate constant (or rate) will always be associated w/ higher temps
(b) Activation energy always has a + sign
Practice Problems
1. At 200 K athe rate constant for a reaction is 3.5 X 10-3s-1, and at 250 K the rate
constant is 4.0 X 10-3s-1. What is the activation energy?
2. A common rule of thumb is that, near room temp, the rate of a reaction will
double with each 10ْ C rise in temperature, Estimate the activation energy needed
for this rule to hold true.
3. What is the rate of reaction at 450 C if the reaction rate is 6.75 X 10-6 mol L-1s-1 at
25 C? The activation energy was previously determined to be 35.5kJ mol-1
1. Ea = 1.11 X 103 J mol-1 = 1.11kJ mol-1
2. Ea = 5.01 X 104 J mol-1 = 5.01kJ mol-1
3. rate1 = 3.05 X 10-2mol L-1s-1 As expected, the reaction rate ↑ at higher temp
Applications of Rate Laws
   zero order has rate law in which exponents of reactants are all zero
o any # raised to zero power = 1
o therefore rate = k
o does not depend on concs
o rate constant will have units mol L-1s-1 like the rate
o catalytic reactions are often zero order
o conc vs time plot is a straight line
    st
1 order reaction
o Variety of ways a reaction may be 1st order
o Most common case- conc of 1 reactant has wxponent of 1 → Rate = k[A]
o As conc of reactant ↓, rate ↓
 When conc is half, rate will also be half the original
 Called half-life and works like nuclear decaying
o Plot is a curve when plotting conc vs time
o 1st order rate equation allows us to calc concs at any time after reaction has
started
Ln ([A]0/[A]t) = kt or ln[A]0 – ln[A]t = kt            [A]0: initial conc
[A]t:conc@time, t
 Plotting natural log vs time is a straight line w/ slope -k
   Practice Problem
A certain 1st order reaction has a rate constant of 4.5 X 10 -3s-1. how much of a 50.0
millimolar (mM) sample will have reacted after 75.0s?
Plug values into: ln[A]0 – ln[A]t = kt and solving for [A]t = 0.0356 M
50.0mM – 35.6mM = 14.4mM
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 Half life: after a half life [A]t = 0.5[A]0
o Substituting into previous equation gives:
o Ln ([A]0/[A]t) = -kt1/2         t1/2 = 0.693/k
 Practice Problem
What percentage of a sample has reacted after six half lives?
o Solution: after each ½ life, half of the sample present at the start of that ½ life will
be reacted. After one ½ life, ½ of the sample is left; after two ½ lives, ½ X ½ or ¼
will remain; after 3 half-lives, ½ X ½ X ½ or 1/8 remains, etc. So after 6 ½ lives
1/64 of the original is left, meaning 63/64 has reacted. The percentage that has
reacted is calculated as 63/64 X 100 = 98.4% reacted

   Second-Order Reactions
o Rate = k[A]2 or k[A][B]
o 1/[A]t – 1/[A]0 = kt
o ½ life equation: 1/(0.5[A]t ) – 1/[A]0 = kt1/2      t1/2 = 1/(k[A]0
o ½ life depends on starting concs of reactants
o Plotting 1/[A] vs time is a straight line w/ slope k

Theory of Reaction Rates
   Collision Theroy: reaction rate is = to frequency of effective collisions b/w reactants
o Effective collisions are those during which molecules collide w/sufficient
energy and in proper orientation
o Minimum energy required is activation energy, Ea
o As molecules collide, KE is converted to PE
 If PE increase is greater than Ea, reaction occurs
o # of effective collisions increase w/increase in temp since average KE
increases w/temp
o Example of effective orientation: HI + Cl → HCl + I

   When Cl collides w/ I, no reaction
   When Cl collides w/H, reaction occurs and Cl replaces I!
   Reaction Rate = Nfefo
   N = # of collisions per second (depends on temp and conc)
   fe = fraction of collisions w/min. energy (↑as temp ↑)
   fo = fraction of collisions w/min. orientation (remains constant)
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   Transition-State Theory
o Attempts to describe molecular configurations and energies as a collision of
reactants occurs
o As molecules approach, they don’t act like billiard balls and just bounce off
each other
o As they get closer, orbitals interact and distort each other
o These distortions weaken bonds so that they are broken upon contact and new
bonds are formed
o Refer to previous HI + Cl example w/effective collisions
 As Cl and HI approach, very electronegative Cl atteacts electrons that
H shares with I
 Therefore, H-I bond is weakened and H-Cl bond begins to form
 At moment of collision, H-I bond approx. ½ broken and H-Cl bond
approx. ½ formed (activation complex)
 When activation complex breaks apart:
1. effective collisions: new products
2. ineffective collisions: original reactants

o Reaction Profile: describes energy of reactants
 Plots the ↑ in PE as they approach, w/a max at moment of collision,
then the ↓ in PE as products recoil
 PE ↑ as molecules approach, interact, and become distorted; therefore,
KE is ↓ equally
 Ea often referred to as an energy barrier b/w reactants and products
 Based on same considerations as collision theory: minimum energy
and orientation at point of collision

   Difference in Theories
Collision theory: reactions are collisions b.w hard spheres
Transition-State Theory: collisions are interactions b/w sponge balls that are
deformed in the process
   Interpretation of Reaction Profiles
o Provide source of info about rates and are a graphical view of conversion of
reactants into products
o Rate constant (rate of reaction) inversely related to height of energy barrier, Ea
 When Ea low, large proportion of collisions are effective
 When Ea high, few collisions are effective

o Can be used to determine if endo or exothermic
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   PE difference b/w products and reactants = to heat of reaction, ∆H
   ∆H = PEprods - PEreactants
   Endo: PE of prods > PE of reactants (∆H+)
   Exo: PE of prods < PE of reactants (∆H-)

o Also shows forward and reverse reactions
o For reverse, start on products side
 Has different Ea and heat of reaction (opposite sign for ∆H)
o Also gives explanation for action of catalysts
 Speeds up reaction by providing alternative reaction pathway w/lower
energy barrier
 Decreases the Ea of both forward and reverse reactions by same amt so
equil is reached faster
 Equil position and heat of reaction not affected since PEs con’t change
 Does not increase amt of product formed
 Enzymes are natural catalysts found in living organisms for very
special reactions
 Most have optimum temp of approx. 36 C
 Usu. enzyme, E, and reactant(s), S, react to form an exzyme-
substrate, ES, complex
 ES then decomposes into prods, P, and orig. enzyme
 E + S ↔ ES ↔ P + E
 Specific for certain reactions because of lock and key model of
enzyme action
 Rate of reaction given by Michaelis-Menton Equation

Rate = K[E][S]           K: rate constant for formation of ES
C + [S]               complex
C: combo of other rate constants in
Process
   When substrate conc very large:
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Rate = k[E]

   When substrate conc very low:

Rate = (k/c)[E][S]

Reaction Mechanisms

        Chemical equations are written as a single step
        Most occur in a series of steps called elem. Reactions
        All elem. Reactions must add up to overall balanced equation
        Complete sequence of steps called reaction mechanism
        Elem. Reactions

   Mechanism usually involves collision of only 2 reactant
molecules
   Extremely rare to have more than 2
   Coefficients are exponents in rate law
   One step will always be slower than others (rate-determining
step or rate-limiting step) determines overall rate of reaction
   Example: H2 + 2ICl → I2 + 2HCl
Rate = k[H2 ] [ICl]

2 Step Mechanism: H2 + ICl → HI + HCl
HI + ICl → I2 + HCl

Rate law describes 1st step which is assumed to be slow step.
HI can be reacted with ICl to show that it is a much faster
reaction. Chemical species that are part of mechanism, but not
balanced equation called intermediate.

HI is intermediate in example.

Another possible mechanism is 3-step process

H2 + ICl → HI + HCl
H2 + ICl → HI + HCl
2HI → H2 + I2

By experimentation, it was determining that 3rd step is slowest.
This means rate determining state give a completely different
rate law. This is not the correct mechanism.

Determining Rate Laws for Elem. Reactions
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   To compare exper. rate law with rate laws of elem. reactions, it is
necessary to convert rate law with intermediate into one that has
only reactants

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