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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 645
Answers to integrated activities
I18.2 Typical orders ofmagnitudes are qTm/NA ≈ 107, qR ≈ 10 per rotational degree of
freedom, qV ≈ 1 per vibrational degree of freedom, and qE ≈ 1. Vibrational and
electronic contributions will therefore be ignored from now on. According to
transition-state theory the rate constant is given by ([18C.10–794] and [18C.9–
794])
kr = κ
kT
h
RT
p−○
NAq−○C‡
q−○Cq−○B
e−∆E0/RT
At 298 K the factors in from the the ratio of partition functions evaluate to
1.5 × 1011 m3 mol−1 s−1, assuming κ = 1.
For a reaction between structureless particles A, B, and C‡ all have contribu-
tions from translation; in addition, C‡ has two rotational degrees of freedom,
therefore
kr = (1.5 × 1011 m3 mol−1 s−1) ×
NA × qTm × (qR)2
qTm × qTm
e−∆E0/RT
= (1.5 × 1011 m3 mol−1 s−1) ×
(qR)2
qTm/NA
e−∆E0/RT
= (1.5 × 1011 m3 mol−1 s−1) × (10)2
107
= (1.5 × 106 m3 mol−1 s−1) × e−∆E0/RT
In collision theory the rate constant is given by [18A.9–783]
kr = σNA (8kT
πµ
)
1/2
e−Ea/RT
For a typical value σ = 0.4 nm2 and a mass of 2 × 10−26 kg, at 298 K
kr = (1.7 × 108 m3 mol−1 s−1) × e−Ea/RT
Assuming that Ea ≈ ∆E0, collision theory gives a rate constant greater by about
a factor of 100, implying a steric factor of about 0.01; this is a plausible result.
If A and B are non-linear triatomics, then A, B and C‡ all have three rotational
degrees of freedom
kr = (1.5 × 1011 m3 mol−1 s−1) ×
NA × qTm × (qR)3
qTm × (qR)3 × qTm × (qR)3
e−∆E0/RT
= (1.5 × 1011 m3 mol−1 s−1) × 1
(qR)3 × qTm/NA
e−∆E0/RT
= (1.5 × 1011 m3 mol−1 s−1) × 1
(10)3 × 107
= (15 m3 mol−1 s−1) × e−∆E0/RT
�e steric factor is now 9 × 10−8 – very much smaller than for the reaction
between structureless particles.