Collision Model, Energy Diagrams & Arrhenius Equation

Section 7

Chemical Kinetics Chapter 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

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

Natural Logrithms

R

universal gas constant

8.3145 J/K• mol

e y = x

The e constant or Euler's number is:

e ≈ 2.71828183 (irrational number)

ln(x ∙ y) = ln(x) + ln(y)

ln(x / y) = ln(x) - ln(y)

ln(x y) = y ∙ ln(x) Invented by John Napier

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

slope y-intercept

The reaction 2N2O5 4NO2 + O2 was studied at several temperatures and the following values of k were obtained. Calculate Ea.

Arrhenius Equation Sample Exercise page 555

T (C) T (K) 1/T (K) k (sec-1) ln (k)

20 2.0 x 10-5

30 7.3 x 10-5

40 2.7 x 10-4

50 9.1 x 10-4

60 2.9 x 10-5

293

303

313

323

333

3.41 x 10-3

3.30 x 10-3

3.19 x 10-3

3.10 x 10-3

3.30 x 10-3

-10.82

-9.53

-8.22

-7.00

-5.84

Slope = ln (k)

(1/T) =

-Ea

R

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

1 1

2 2

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 The rate constant for the gas phase decomposition of N2O5,

2N2O5 4NO2 + O2 , has the following temperature dependence:

End of Chapter problem #57

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

338 4.9 x 10-3

318 5.0 x 10-4

298 3.5 x 10-5

2.96 x 10-3

3.14 x 10-3

3.36 x 10-3

-5.32

-7.60

-10.26

Slope = ln (k)

(1/T) =

-Ea

R

Slope = -1.2 x 104 K; Ea = 1.0 x 102 kJ/mol

ANSWER

Make the appropriate graph using these data, and determine the activation energy

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

53, 55, 57, 58, 59, 61, 63

Exercises page 571-572