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The Collision Model

Date post: 07-Feb-2016
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The Collision Model. The reaction rate depends on: collision frequency a probability or orientation factor activation energy (E a ) The reaction rate increases as the number of collisions between reacting species increase. Concentration temperature. Cl. Cl. Cl. Br. Br. Br. H. H. H. - PowerPoint PPT Presentation
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The Collision Model The reaction rate depends on: collision frequency a probability or orientation factor activation energy (E a ) The reaction rate increases as the number of collisions between reacting species increase. Concentration temperature
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Page 1: The Collision Model

The Collision Model

The reaction rate depends on: collision frequency a probability or orientation factor activation energy (Ea)

The reaction rate increases as the number of collisions between reacting species increase. Concentration temperature

Page 2: The Collision Model

The Collision Model

Collisions must occur in a particular orientation for reactions to occur.

For the reaction: Cl . + H - Br H - Cl + Br .

Cl . Br HDesired rxn cannot occur.

Cl . BrH

Cl . Br

H

Desired rxn cannot occur.

Desired rxn can occur.

Page 3: The Collision Model

The Collision Model

Collisions must occur with a specific minimum amount of energy in order for a reaction to take place.

Activation energy (Ea) the minimum energy the reactants must

have for a reaction to occur

the energy difference between the reactants and the transition state

Page 4: The Collision Model

The Collision Model

Transition state: a particular arrangement of atoms of the

reacting species in which bonds are partially broken and partially formed

the state of highest energy between reactants and products

a relative maximum on the reaction-energy diagram.

Page 5: The Collision Model

Reaction Energy Diagrams

Reaction energy diagram: a plot of potential energy changes that

occur as reactants are converted to products

Page 6: The Collision Model

Reaction Energy Diagrams

Given a reaction energy diagram for a chemical reaction, you should be able to identify the reactants, products, transition state, activation energy, the heat of reaction, and whether the reaction is endothermic or exothermic.

Page 7: The Collision Model

Reaction Energy Diagrams

Example: For each reaction energy diagram below, mark the location of the reactants, products and transition state. Identify the magnitude of Ea and Hrxn. Is each reaction endothermic or exothermic?

Page 8: The Collision Model

Arrhenius Equation

Reaction rate increases with temperature because: molecules have more kinetic energy more collisions occur greater number of collisions occur with

enough energy to “get over the hill” i.e. with energy greater than or equal to Ea

Page 9: The Collision Model

Arrhenius Equation

The Arrhenius Equation relates the value of the rate constant to Ea and the temperature:

k = Aewhere k = rate constant

Ea = activation energyR = gas constant (8.314 J/mol.K)T = temperature in KelvinA = frequency factor (a constant)

A is related to the frequency of collisions and the probability that the collisions are oriented favorably for reaction.

-Ea/RT

Page 10: The Collision Model

Arrhenius Equation

The Arrhenius Equation can be converted to another useful (experimentally useful, that is) form by taking the natural log of both sides:

ln k = - Ea + ln ART

A graph of ln k vs. 1/T (in K) gives a straight line with a slope of - Ea/R. Ea = - slope x R

Page 11: The Collision Model

Arrhenius Equation

The activation energy for a reaction can also be found in a non-graphical way if the rate constant at two or more temperatures is known:

ln k1 = Ea 1 - 1k2 R T2 T1

Page 12: The Collision Model

Arrhenius Equation

Example: Calculate the activation energy for the rearrangement of methyl isonitrile to acetonitrile using the following data.

You can solve this in one of several ways:

Graph ln k vs. 1/T and determine the slope;

Find ln k and 1/T and determine the slope using two well spaced points using (ln k)/(1/T);

Use the non-graphical method

Page 13: The Collision Model

Arrhenius Equation

Personally, I prefer the non-graphical method:

ln k1 = Ea 1 - 1k2 R T2 T1

Use two points that are well separated and convert T to Kelvin (K = oC + 273.15)

Temp (K) 1/T (K-1) k

462.9 2.160 x 10-3 2.52 x 10-5

524.4 1.907 x 10-3 3.16 x 10-3

Page 14: The Collision Model

Arrhenius Equation

Plug the values of k and T into the equation. Be careful to put k1 and T1 in the appropriate order:

ln k1 = Ea 1 - 1k2 R T2 T1

ln 2.52 x 10-5 = Ea 1 - __1__3.16 x 10-3 (8.314 J/mol K) 524.4 K 462.9 K

Solve for Ea.

Page 15: The Collision Model

Arrhenius Equation

ln 2.52 x 10-5 = Ea 1 - __1__3.16 x 10-3 (8.314 J/mol K) 524.4 K 462.9 K

Solve for Ea.

- 4.8315 = (- 3.047 x 10-5 mol/J) Ea

Ea = 1.59 x 105 J/mol

Page 16: The Collision Model

Arrhenius Equation

Alternately, you can solve it graphically.

First, pick two well-separated points and calculate 1/T (in Kelvin) and ln k.

Temp (K) 1/T (K-1) k ln k

462.9 2.160 x 10-3 2.52 x 10-5 -10.589

524.4 1.907 x 10-3 3.16 x 10-3 -5.757

Page 17: The Collision Model

Arrhenius Equation

Find the slope of the line:

y = (ln k)x (1/T)

Slope = -5.757 - (-10.589) = - 1.910 x 104 K

1.907 x 10-3 - 2.160 x 10-3

Page 18: The Collision Model

Arrhenius Equation

Since slope = - Ea/R

- Ea = - 1.910 x 104 K R

Ea = - (-1.910 x 104K) x 8.314 J/mol.K

Ea = 1.588 x 105 J/mol

Page 19: The Collision Model

Arrhenius Equation

Once you find the value for Ea, you can use it to find the frequency factor (A) for the reaction.

k = A e

A = k

e

-Ea/RT

-Ea/RT

Page 20: The Collision Model

Arrhenius Equation

To find A for the previous example, pick one of the data points given and plug in the values of k, Ea, and T.

A = 2.52 x 10-5 s-1

e

A = 2.096 x 1013 s-1

- 1.588 x 105J/ (8.314 J/molK x 462.9K)

Page 21: The Collision Model

Arrhenius Equation

Once you have the value for A and Ea for the reaction, you can calculate the value of the rate constant at any temperature using the Arrhenius equation:

For example, the value of k at 25oC (298 K) is:

k = (2.096 x 1013 s-1) e

k = 2.82 x 10-15 s-1

- (1.59 x 105 J/mol)/(8.314 J/mol K x 298 K)


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