Kinetics & Collision Theory
BASIC IDEA - Before a rxn can occur the particles must hit each other
AND
this collision must have enough energy to cause a change.
To summarize reaction rates, they depend on TWO factors:
OR
Reaction
rate
=
number of x
collisions
percent of collisions with
enough energy to react
FACTORS EFFECTING REACTION RATES:
1.) NATURE OF REACTANTS
a) Fe2+(aq) + Ce4+(aq) -----> Fe3+(aq) + Ce3+(aq)
b) CH4 (g) + 2 O2 (g) ---> CO2 (g) + 2 H2O (g)
c) Mg (s) + 2 H+(aq) ---> Mg2+ (aq) + H2 (g)
2.) CONCENTRATION
3.) TEMPERATURE
4.) PRESENCE OF A CATALYST
1
Ch. 12 - Rates of Reactions
• The rate of a reaction is a positive quantity that tells us how the
concentration of a reactant or product changes with time.
2 NH3 (g) -----> N2 (g) + 3 H2 (g)
Unit for rate is:
Express the rate of reaction in terms of reactants/products
(-)∆ conc. NH3
time
or
∆ conc. N2
time
or ∆ conc. H2
time
∆ [NH3] ≠ ∆ [N2] ≠ ∆ [H2]
If rate of formation of N2 is 0.040 M/min, what is the rate in terms of
a. ∆ conc. NH3
b. ∆ conc. H2
• Rate may be average rate OR instantaneous (see graph)
• For instantaneous find slope of tangent on rate curve for that time.
-------------------------------------------------------------------------------------------2 N2O5 (g) -----> 4 NO2 (g) + O2 (g)
Time (min)
0
conc. N2O5 (M)
0.160
Inst. rate (M/min) 0.056
1
0.113
0.039
2
0.080
0.028
3
0.056
0.020
4
0.040
0.014
Q - What is the average rate at:
a. 3 min
b. 4 min
Differential rate law Integrated rate law –
2
2NH3 (g) ----> N2 (g) + 3 H2 (g)
Rate Expression:
If Zeroth Order n = 0
Rate = k [ NH3]n
Rate = k [NH3]0
k = Rate
[NH3]0
If First Order
n=1
Rate = k [NH3]1
k = Rate
[NH3]1
If Second Order n= 2
Rate = k [NH3]2
k = Rate
[NH3]2
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• Sometimes rates depend on more than once substance.
X + 2 Y + Z ----> Q + 2 R
Rate = k [X]1 [Y]0 [Z]2
Rxn is 1st order with respect to X
Zero order with respect to Y
2nd order with respect to Z
Reaction is 3rd order overall
How will rate change if:
A) [X] is doubled
B) [Y] is tripled
C) [Z] is tripled
D) If ALL of the above occur
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To Determine Order: Compare rate and concentration changes.
Concen. Δ
[A] triples
[B] doubles
[C] doubles
Effect on Rate
Rate triples
Rate unchanged
Rate quadruples
Order
Rate Law:
NOTE:
1) The concentrations of the products do not appear in the rate law because the reaction
rate is being studied under conditions where the reverse reaction does not contribute
to the overall rate.
2) The value of the exponent n must be determined by experiment (i.e. the reaction
mechanism must be determined); it cannot be written from the net equation.
3
Find Rate Order from the Following Data
Ex. 1
CH3CHO (g) --> CH4 (g) + CO (g)
Conc. CH3CHO (M) 0.10
Inst. rate (M/min)
0.085
0.20
0.34
0.30
0.76
0.40
1.4
n
Rate 2 =
Rate 1
Conc. 2
Conc. 1
log
Rate 2 = n log Conc. 2
Rate 1
Conc. 1
n =
Ex. 2
Conc.
0.010 M
0.015 M
Rate
0.0082 M/min
0.0277 M/min
Another Method:
NH4+ (aq) + NO2- (aq)  N2 (g) + 2 H2O (l)
Experiment
Initial [NH4+]
Initial [NO2-]
1
2
3
0.100 M
0.100 M
0.200 M
0.0050 M
0.010 M
0.010 M
Initial Rate
(mol/L•s)
1.35 x 10-7
2.70 x 10-7
5.40 x 10-7
Rate 2 = 2.70 x 10-7 M/s = k (0.100 M)n (0.010 M)m
Rate 1 1.35 x 10-7 M/s = k (0.100 M)n (0.0050 M)m
Rate 2 =
Rate 1
Rate 3 = 5.40 x 10-7 M/s = k (0.200 M)n (0.010 M)m
Rate 2 2.70 x 10-7 M/s = k (0.100 M)n (0.010 M)m
Rate 3 =
Rate 2
Rate Law:
4
Integrated rate laws – another method to find rate order
• 1st order rate law:
Rate = k [A]1
from calculus ln Ao = k t
where Ao = original conc.
A
A = Conc. @ time (t)
Rearrange and solve for ln A:
Zero
First
Second
Rate Law
Integrated Rate Law
Ao - A = k t
ln (Ao/A) = k t
1/A - 1/Ao = k t
Solve for y = mx + b
Plot needed to
give straight line
Slope of line
Solve for Half-life (T1/2)
(A = 1/2 Ao) OR (2A = Ao)
5
Examples - Use Table 12.6
1) For a 1st order reaction it is found that 20% of the sample (X) has
decomposed in 100 seconds.
(A) Find the value for the rate constant. (units?)
(B) What is the half-life of this reaction?
2) Redo assuming a 2nd order reaction and an initial concentration of 0.200 M.
(A) Rate constant?
(B) Half-life?
What is the order for radioactive decay?
Activation Energy and Potential Energy Diagrams
• Activation Energy - the energy required to form the activated complex
• The greater the activation energy the _____________ the reaction.
POTENTIAL ENERGY DIAGRAMS:
• Three key points:
(1) Start
(2) Activated Complex
(3) End
∆ H rxn =
Ea =
PE
PE
CO + NO2
CO + NO2
CO2 + NO
Reaction Coordinate
CO2 + NO
Reaction Coordinate
6
CATALYSTS - Change rate of reaction without being consumed.
•They do this by forming a different
______________ which has a _______ activation
energy.
• a catalyst is often a solid on whose surface a
reactant particle can be adsorbed, or held.
• Enzymes are biological catalysts.
Homogeneous Catalyst – all reacting species are in the same phase
Heterogeneous Catalyst –reacting species are in different phases
1)
2)
3)
4)
Adsorption and activation of the reactants
Migration of the adsorbed reactants on the surface
Reaction of the adsorbed substances
Escape, or desorption, or the products
Rates and Temperature
Kinetic Energy Distribution plot
# molec.
OR
# collisions
Kinetic Energy
KEY IDEAS:
1) At a given temp. (T1) not all molecules have the same kinetic energy.
Only a few have high energy.
2) When the temperature is increased (T2) the number of high energy molecules
increases greatly, even though the average increases only slightly.
3) A threshold energy line (E) is used to relate to chemical reaction. Only the
molecules beyond the threshold energy can react.
7
Summary of Diagrams
and applications
POTENTIAL ENERGY
DIAGRAMS:
KINETIC ENERGY DISTRIBUTIONS:
A) Draw a kinetic energy distribution.
A) Draw a PE diagram.
B) In the same diagram can you
draw the effect of adding a
catalyst. If so, do so.
C) In the same diagram, can you
show the effect of increasing
temperature? If so, do so.
========================================================
PE
Reaction Coordinate
Reaction Coordinate
# particles
PE
1) Which of the two rxns below is faster?
2) Show how the graphs below would change if a catalyst
were added.
3) Which graph below would have a larger increase in rate
if…
i) temp increased?
ii) a catalyst were added?
# particles
1) Which of the two reactions
below would be faster?
KE
KE
8
Rate and Temperature Cont.
Consider Collision Theory • Arrhenius proposed the idea of threshold (activation) energy. Function
of Molecules with energy needed to react.
• Rate and # collisions both increase exponentially with temperature.
• Reaction rate is much smaller than the calculated collision frequency.
• Arrhenius postulated that the # collisions having an energy greater than
or equal to the Ea is given by:
# collisions =
w/ Ea
tot. #
collisions
(-Ea/RT)
e
R = gas law constant (8.31 J/mol.K)
T = Temperature (K)
e = base for natural log
Ea = Activation Energy (J)
• Experiments show that the observed reaction rate is considerably
smaller than the rate of collisions w/ enough energy to surmount the
barrier.
k = (z p) e-Ea/RT
k = A e-Ea/RT (Arrhenius Equation)
a more useful form is:
ln k = - Ea 1 + ln (A)
R T
Thus slope (m) =
or Ea =
Another useful form ~ “Two Point Form”
- Use T2 as higher temp to
avoid negative logs
Examples • Find Ea if rate doubles when T changes 25 °C --> 35 °C.
• Given k = 0.0251 s-1 @ 0 °C, Ea = 116 kJ.
Find a) Ratio of rate increase at 100 °C
b) value of k at 100 °C
9
Reaction Mechanism
• Most reactions take place in a series of steps, each step usually
involving two particles.
• To find the rate expression for a multi-step mechanism:
1) Focus on the slow step (rate determining step - R.D.S.)
2) Coefficients for each reactant in R.D.S. are the order of reaction
with respect to each reactant
3) Eliminate any unstable intermediates from rate expression.
a. Write the equilibrium expression for the previous reaction.
b. Rearrange the equation and solve for the intermediate.
c. Substitute this equation in for the [intermediate] in the original
rate expression.
(1)
HBr + O2
Find Rate Expression
 If the slow step is
a) Step 1?
<==> HOOBr
(2) HOOBr + HBr <==> 2 HOBr
(3) HOBr +
HBr <==> H2O + Br2
b) Step 2?
(4) HOBr + HBr <==> H2O + Br2
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NET:
Another Example:
1) X2 (g)
<==> 2 X (g)
FAST
2) X (g) + A2 (g) ---> AX + A (g)
SLOW
3) A (g) + X2 (g) ---> AX + X (g)
FAST
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NET:
Find Rate Expression:
10