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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 661
�e expression for θB then followswith some rearrangement and simpli�cation
θB =
αBpBαApA − (αApA + 1)αBpB
αBpBαApA − (αApA + 1)(αBpB + 1)
= −αBpB
αBpBαApA − αApAαBpB − αApA − αBpB − 1
= αBpB
αApA + αBpB + 1
which is the required expression.�at for θB is simply found by swapping the
indices A and B: the equations all remain valid under such a change.
Solutions to exercises
E19C.1(b) �e amount in moles of CO gas is found using the perfect gas law.
n = pV
RT
= (1 bar)×(105 Pa)
1 bar
× 3.75 × 10−6 m3
(8.3145 JK−1mol−1) × (273.15 K)
= 1.65... × 10−4 mol
which corresponds to NAn = (6.0221 × 1023mol−1) × (1.65 × 10−4 mol) =
9.94... × 1019 molecules.
A rough calculation of the surface area notes that the collision cross section
is σ = πd2, where d is the diameter of the colliding spheres. �erefore d =
(σ/π)1/2, and hence r = 1
2 (σ/π)1/2.�e area of onemolecule is πr2 = π 14 σ/π =
1
4 σ . In the tables, no value for the collision cross section of CO is given, so the
value for N2 is used.�e surface area is therefore (9.94... × 1019)× 1
4 × (0.43×
10−18 m2) = 11 m2 .
In fact circles do not cover a plane completely, and it can be shown that the
highest coverage which can be achieved is one in which the circles cover 0.91
of the area of the plane. �e estimate of the area therefore needs to be scaled
up by a factor of 1/0.91 ≈ 1.1 to give 12 m2.
Solutions to problems
P19C.2 �e rate law for a unimolecular decomposition occurring on a surface is given
by [19C.1–841]
υ = krαp
1 + αp
HI is adsorbed strongly on gold, implying that αp ≫ 1, and so the rate law
reduces to υ = kr: that is, zeroth order and thus independent of pressure. On
platinum, absorption is weaker implying that αp≪ 1. In this limit the rate law
becomes υ = krαp: that is, �rst order in the pressure.