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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 81
(ii)
∆rS−○ = 12S−○m(CO2 , (g)) + 11S−○m(H2O, (l)) − S−○m(sucrose, (s)) − 12S−○m(O2 , (g))
= 12 × (213.74 JK−1mol−1) + 11 × (69.91 JK−1mol−1)
− (360.2 JK−1mol−1) − 12 × (205.138 JK−1mol−1)
= +512.0 JK−1mol−1 .
E3C.3(b) Consider chemical equation
N2(g) + 1
2O2(g)Ð→ N2O(g)
Standard reaction entropy is given by [3C.3b–94], ∆rS−○ = ∑J νJS−○m(J), where νJ
are singed stoichiometric coe�cients for a given reaction equation.�erefore,
using data from the Resource section
∆rS−○ = nS−○m(N2O, (g)) − 1
2nS
−○
m(O2 , (g)) − nS−○m(N2 , (g))
= (1.00 mol) × (219.85 JK−1mol−1)
− ( 12 × 1.00 mol) × (205.138 JK−1mol−1)
− (1.00 mol) × (191.61 JK−1mol−1)
= −74.33 JK−1 .
Solutions to problems
P3C.2 Consider the process of determining the calorimetric entropy from zero to
the temperature of interest. Assuming that the Debye extrapolation is valid,
the constant-pressure molar heat capacity at the lowest temperatures is of a
form Cp ,m(T) = aT3. �e temperature dependence of the entropy is given
by [3C.1a–92], S(T2) = S(T1) = ∫
T2
T1 (Cp ,m/T)dT . �us for a given (low)
temperature T the change in molar entropy from zero is
Sm(T) − Sm(0) = ∫
T
0
Cp ,m
T ′
dT ′ = ∫
T
0
aT ′3
T ′
dT ′
= a∫
T
0
T ′2dT ′ = a
3
T3 = 1
3Cp ,m(T)
Hence
S−○m(10 K) − S−○m(0) = 1
3 × (4.64 JK−1mol−1) = 1.54... J K−1mol−1
�e increase in entropy on raising the temperature to themelting point is S−○m(234.4K)−
S−○m(10 K) = 57.74 JK−1mol−1. �e entropy change of a phase transition is
given by [3C.1b–92], ∆trsS(Ttrs) = ∆trsH(Ttrs)/Ttrs.�us
∆fusS−○m(234.4 K) =
2322 Jmol−1
234.4 K
= 9.90... J K−1mol−1