526 12 Chemical Reaction Mechanisms I: Rate Laws and Mechanisms
whered 12 is the collision diameter for substance 1 and substance 2, and whereμ 12
is the reduced mass of molecules of massm 1 andm 2 , defined in Eq. (9.8-27) and in
Appendix B:μ 12 m 1 m 2
m 1 +m 2or1
μ 121
m 1+
1
m 2(12.1-7)
We can now summarize our results for both types of bimolecular processes in gaseous
reactions:A gaseous bimolecular elementary process is second order overall, and for
a two-substance reaction it is first order in each substance.Exercise 12.2
Show that Eq. (12.1-6) is correct.EXAMPLE12.1
For the gas-phase reactionH 2 +I 2 −→2HIat 373.15 K, the rate constant is equal to 8. 74 × 10 −^15 L mol−^1 s−^1. Assume that the reaction
is elementary and estimate the fraction of collisions that leads to reaction.
Solution
From Table A.15,d 12
1
2
(2. 72 × 10 −^10 m+ 6. 53 × 10 −^10 m) 4. 54 × 10 −^10 mμ
(0.002016 kg mol−^1 )(0.2538 kg mol−^1 )
(0.002016 kg mol−^1 + 0 .2538 kg mol−^1 )(6. 022 × 1023 mol−^1 ) 3. 321 × 10 −^27 kgf
kπd 122(
8 kBT
πμ 12) 1 / 2
NAvf
(8. 74 × 10 −^15 L mol−^1 s−^1 )(0.001 m^3 L−^1 )π(4. 54 × 10 −^10 m)^2(
8(1. 3807 × 10 −^23 JK−^1 )(373.15 K)
π(3. 321 × 10 −^27 kg)) 1 / 2
(6. 022 × 1023 mol−^1 ) 1. 1 × 10 −^26Exercise 12.3
At 473.15 K the rate constant for the reaction of the previous example is equal to
9. 53 × 10 −^10 L mol−^1 s−^1. Estimate the fraction of collisions that leads to reaction at this
temperature.