Collision Model, Energy Diagrams & Arrhenius

EquationSection 7

Chemical KineticsChapter 12

aA + bB products Rate = k [A]n [B]m

How does temperature affect rate?

Collision Model

Molecules must collide to react

Increase temp.; increases frequency of collision

Only a small fraction of the collisions produces a reaction.

Why?

Threshold energy = activation energy = the energy that must be over come to produce a chemical reaction

Collision Model

Ea

2BrNO NO + Br2 Rate = k [A]n [B]m

2Br-N bonds must be broken, the energy comes from the kinetic energy of the molecules.

E has no effect on rate – rate depends on the size of the activation energy

Energy Diagram

At any temperature only a fraction of collisions have enough energy to be effective

Energy and Temperature

Possible orientations for a collision between two BrNO molecules

Collision Model

The fraction of effective collisions increases exponentially with temp.

# of collisions Ea = (total collisions) e-Ea/RT

e-Ea/RT = fraction of collisions with E Ea at T

Collision Model

Runiversal gas constant

8.3145 J/K• mol

                       

               

          

                   

Rate also depends on molecular orientation

Successful collisions

1. Energy Ea

2. Correct orientation

k = z p e-Ea/RT

z = collision frequency

p = steric factor (always less than one) reflects the fraction of collisions with effective orientation

A = frequency factor, it replaces zp. A=zp

k = A e-Ea/RT

Collision Model

k = A e-Ea/RT

Take the natural log of each side

For a reaction that obeys the Arrhenius equation, the plot of ln (k) vs 1/T give a straight line.

slope = -Ea /R

y-intercept = ln (A)

Most rate constants obey the Arrhenius equation which indicates that the collisions model is reasonable.

Arrhenius Equation

y = m x + b

ln (k) = R

Ea 1

T+ ln (A)x

slopey-intercept

Arrhenius Equation

Take ln(k2) – ln(k1)

Use algebra as on page 557.

ln (k) = R

Ea 1

T+ ln (A)x

ln (k) = R

Ea 1

T+ ln (A)x

11

22

k2 R

Ea 1

T1

_k1

1

T2

ln

=

The values of k1 and k2 measured at T1 and T2 can be used to calculate Ea

The reaction (CH3)3CBr + OH- (CH3)3COH + Br-

in certain solvent is first order with respect to (CH3)3CBr and zero order with respect to OH-. In several experiments, the rate constant k was determined at different temperatures. A plot of ln(k) vs. 1/T was constructed resulting in a straight line with a slope of -1.10 x 104 K and y-intercept of 33.5. Assume k has units of s-1.

a. Determine the activation energy for this reaction.

b. Determine the value of the frequency factor A

c. Calculate the value of k at 25 C.

End of Chapter Exercises #58

here

ANSWER

The activation energy for the decomposition of HI(g) to H2 (g) and I2 g) is 186 kJ/mol. The rate constant at 555 K is 3.52 x 10-7 L/mol•sec. What is the rate constant at 645 K?

End of Chapter Exercises #59

9.5 x 10-5 L/mol•sec

ANSWER

A certain reaction has an activation energy of 54.0 kJ/mol. As the temperature is increased form 22C to a higher temperature , the rate constant increases by a factor of 7.00. Calculate the higher temperature.

End of Chapter Exercises #61

51C

ANSWER