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80 3 THE SECOND AND THIRD LAWS
(b) Tf = 500 ○C
Sm(773 K) = (192.45 JK−1mol−1)
+ (29.75 JK−1mol−1) × ln(773.15 K
298 K
)
+ (25.1 × 10−3 J K−2 mol−1) × (773.15 K − 298 K)
− −1.55 × 105 J K mol−1
2
× ( 1
(773.15 K)2
− 1
(298 K)2
)
= (192.45 JK−1mol−1) + (+28.3... J K−1mol−1)
+ (+11.9... J K−1mol−1) − (+0.743... J K−1mol−1)
= +232.00 JK−1mol−1 .
3C Themeasurement of entropy
Answer to discussion question
Solutions to exercises
E3C.1(b) Assuming that the Debye extrapolation is valid, the constant-pressure molar
heat capacity is 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 . For a given temper-
ature T the change in molar entropy from zero temperature is therefore
Sm(T) − Sm(0) = ∫
T
0
Cp ,m
T ′
dT ′ = ∫
T
0
aT ′3
T ′
dT ′
= a∫
T
0
T ′2dT ′ = a
3
T3
Hence
Sm(10 K) − Sm(0) =
1.956 × 10−4 J K−4 mol−1
3
× (10 K)3
= 6.5 × 10−2 JK−1mol−1 .
E3C.2(b) �e standard reaction entropy is given by [3C.3b–94], ∆rS−○ = ∑J νJS−○m(J),
where νJ are the signed stoichiometric numbers.
(i)
∆rS−○ = S−○m(Zn2+ , (aq)) + S−○m(Cu, (s)) − S−○m(Zn, (s)) − S−○m(Cu2+ , (aq))
= (−112.1 JK−1mol−1) + (33.150 JK−1mol−1)
− (41.63 JK−1mol−1) − (−99.6 JK−1mol−1)
= −21.0 JK−1mol−1 .