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 E a

2BrNO  NO + Br 2 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  E a = (total collisions) e -Ea/RT e -Ea/RT = fraction of collisions with E  E a at T Collision Model R universal gas constant J/K mol

Rate also depends on molecular orientation Successful collisions 1.Energy  E a 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 = -E a /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 EaEa 1 T +ln (A)x slope y-intercept

Arrhenius Equation Take ln(k 2 ) – ln(k 1 ) Use algebra as on page 557. ln (k) = R EaEa 1 T +ln (A)xln (k) = R EaEa 1 T +ln (A)x k2k2 R EaEa 1 T1T1 _ k1k1 1 T2T2 ln = The values of k 1 and k 2 measured at T 1 and T 2 can be used to calculate E a

The reaction 2N 2 O 5  4NO 2 + O 2 was studied at several temperatures and the following values of k were obtained. Calculate E a. Arrhenius Equation Sample Exercise page 555 T (  C) T (K)1/T (K)k (sec -1 )ln (k) x x x x x x x x x x Slope =  ln (k)  (1/T) = -E a R

57, 58, 59, 61 Exercises page

The reaction The rate constant for the gas phase decomposition of N 2 O 5, 2N 2 O 5  4NO 2 + O 2, has the following temperature dependence: End of Chapter problem #57 T (K)1/T (K)k (sec -1 )ln (k) x x x x x x Slope =  ln (k)  (1/T) = -E a R Slope = -1.2 x 10 4 K; E a = 1.0 x 10 2 kJ/mol ANSWER Make the appropriate graph using these data, and determine the activation energy

The reaction (CH 3 ) 3 CBr + OH -  (CH 3 ) 3 COH + Br - in certain solvent is first order with respect to (CH 3 ) 3 CBr 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 x 10 4 K and y-intercept of 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 H 2 (g) and I 2 g) is 186 kJ/mol. The rate constant at 555 K is 3.52 x L/molsec. What is the rate constant at 645 K? End of Chapter Exercises # x L/molsec ANSWER

A certain reaction has an activation energy of 54.0 kJ/mol. As the temperature is increased form 22  C to a higher temperature, the rate constant increases by a factor of Calculate the higher temperature. End of Chapter Exercises #61 51  C ANSWER