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Chapter 14 – Chemical Kinetics

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									Chapter 14 – Chemical Kinetics
• How fast a chemical A. Factors that affect reaction occurs rate • Only need to consider 1. Concentration of the forward reaction reactants 2. Temperature 3. Catalysts 4. Surface Area

[conc] R t

B. Reaction Rates 1. Rate is determined in the lab by experiment 2. Rate determined by measuring (-) disappearance of reactants (+) appearance of products

Example 1 – Rates of…
A+B C +D
RA   [ Af  Ai ] (t f  ti )
RC  [C f  Ci ] (t f  ti )

Rate of disappearance of A Note the NEGATIVE sign!!

Rate of appearance of C Note the POSITIVE sign!!

C4H9Cl + H2O  C4H9OH + HCl
Data Table 14.2 – Disappearance of C4H9Cl

Time (sec) 0.0 50.0

[C4H9Cl] (M) 0.100 0.0905 0.0820 0.0741
0.0671

[0.0741 M  0.0905 M ] R (150 s  50 s )

100.0 150.0

= 1.64 x 10-4 M/s

200.0

Example 2 – Rates with Coefficients aA + bB  cC + dD
1 [ A] RA   a t
1 [C ] RC  c t

Note that the coefficient becomes a reciprocal value for rate comparison

Example 3 – Rate Comparison 2 N2O5  4 NO2 + O2
RN 2O5 1 [ N 2O5 ]  2 t
Given:

RN O
2

5

= 4.2 x 10-7 M/s

RNO2

1 [ NO2 ]  4 t
RNO2 4 1   RN 2O5 1 2
RNO2  2(4.2x107 M / s)  8.4x107 M / s

Calculate the rate of appearance of NO2

1 1 RNO2  RN 2O5 4 2

Example 3 – Rate Comparison 2 N2O5  4 NO2 + O2
RN 2O5 1 [ N 2O5 ]  2 t
Given:

RN O
2

5

= 4.2 x 10-7 M/s

[O2 ] RO2  t
The rate of O2 appearance is ½ the rate of N2O5 disappearance

1 RO2  (4.2 x107 M / s)  2.1x107 M / s 2

Rate Law Expression
R = k [reactant]m
• R = rate law expression • k = rate constant units are M-1s-1 Note: k depends upon temperature and nature of reaction • m = order of reaction
– m=0  rate is independent of [ ]0 – m=1  rate is directly related to [ ]1 – m=2  rate is directly related to [ ]2

aA + bB  products
• R = k [A]m[B]n
m = order with respect to A n = order with respect to B Overall order of reaction is = m + n Note: order of reaction must be determined experimentally in the lab and cannot be simply concluded from the equation coefficients!!!!

• 2 N2O5  4 NO2 + O2 R = k[N2O5] • CHCl3 + Cl2  CCl4 + HCl R = k [CHCl3][Cl2] • H2 + I2  2 HI R = k [H2] [I2]

Method of Initial Rates
A +BC
EXP

[A]

[B]

Initial Rate (M/s)

1 2 3

0.100 M 0.100 M 0.100 M 0.200 M 0.200 M 0.100 M

4.0 x 10-5 4.0 x 10-5 16.0 x 10-5

First Order Reactions
[ A] RA    k[ A] t
• Using Calculus… ln [A]t – ln [A]0 = -kt or ln [A]t /[A]0 = -kt [A]0=original conc [A]t=conc @ time, t k = rate constant t = time

RA  k[A]

Graphing First Order Reactions
ln [A]t = -k t + ln [A]0

[A]

y

= mx +b

ln [A]

t

This is NOT a linear plot…. Scientists like linear plots

t

Example – 1st Order
• The decomposition of an insecticide in H2O is first order with a rate constant of 1.45 yr -1. On June 1st, a quantity of 5.0x10-7 g/cm3 washed into a lake. insect  product R = k [insect]
What is the concentration on June 1st next year? ans. [insect]t=1yr = 1.17x10-7 g/cm3 How long will it take for the [insect] to drop to 3.0x10-7 g/cm3? ans. t = 0.35 years = 4 months

a)
b)

1st Order Reactions, Half-Life
1 [ A]0 ln 2  kt1 2 [ A]0

The time that it takes for Original concentration to Drop to ½ of its original concentration.

1 ln  kt1 2 2
1 ln 2 t 1 2 k

0.693  t1 2 k

Second Order Reactions

RA  k[ A]

2

y

= mx +b
slope=k

• Using Calculus…

1 1 kt [ A]t [ A]0
[A]0=original conc [A]t=conc @ time, t k = rate constant t = time

1
[A]

t

2nd Order Reactions, Half-Life
t1/2= 1 k[A]0

To Determine Order You Must Graph the Data
y = mx +b y = mx +b
slope=k

1st Order
1
ln [A] [A]

2nd Order

t

t

Activation Energy, Ea
1. Molecules must collide to react 2. Not all collisions result in a reaction 3. The higher the collision frequency, the faster the reaction rate a. increase temperature b. increase pressure or decrease volume (for gas only) c. catalyst d. increase [conc]

Activation Energy, Ea
4. Activation energy, Ea – the minimum energy needed to start a reaction 5. Activated complex – intermediate product forming before the reaction is completed

Activated Complex

A* Ea E B
The bigger Ea, the slower the rate

Energy

A

Reaction progress
For A  B For B  A exothermic E (-) endothermic E (+) + Ea

Arrhenius Equation – Rate and Temperature

k  Ae

 Ea / RT

Ea ln k    ln A RT

k=rate const A=frequency Ea=Activation energy R=gas const 8.31 J/mol K T=Temperature (Kelvin)

Solving Arrhenius for Two Temperatures
Ea ln k1    ln A RT1
Ea ln k 2    ln A RT2

 Ea   Ea  ln k1  ln k2     ln A      ln A   RT   RT  1 2    

k1 Ea  1 1     ln  k2 R  T2 T1   

Graphing Arrhenius
Yintercept= ln A

ln k
Slope = - Ea R
Note: to obtain Ea, you must multiply slope by the gas constant
1/ T

k1 Ea  1 1     ln  T T  k2 R 2 1


								
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