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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 203
�e relationship between ∆rS−○ and E−○cell is given by [6C.6–222], dE
−○
cell/dT =
∆rS−○/νF.�e expression for E−○cell/V is di�erentiated to give
d(E−○cell/V)
d(θ/○C)
= −4.8564 × 10−4 − 6.841 × 10−6(θ/○C) + 1.7607 × 10−8(θ/○C)2
To relate this to dE−○cell/dT , note that because θ/○C = T/K− 273.15, d(θ/○C) =
d(T/K). Noting further that d(T/K) = dT/K and that d(E−○cell/V) = dE−○cell/V
gives
d(E−○cell/V)
d(θ/○C)
=
dE−○cell/V
dT/K
= K
V
dE−○cell
dT
hence
dE−○cell
dT
=
d(E−○cell/V)
d(θ/○C)
× V
K
Substituting this into [6C.6–222], dE−○cell/dT = ∆rS−○/νF, and rearranging for
∆rS−○ gives
∆rS−○ = νF ×
d(E−○cell/V)
d(θ/○C)
× V
K
= νF×[ − 4.8564 × 10−4 − 6.841 × 10−6(θ/○C) + 1.7607 × 10−8(θ/○C)2]×V
K
= 1 × (96485Cmol−1) × [ − 4.8564 × 10−4 − 6.841 × 10−6 × (24.85)
+ 1.7607 × 10−8 × (24.85)2] × (V/K) = −62.2... J K−1mol−1
Finally ∆rH−○ is calculated from [3D.9–100], ∆rG−○ = ∆rH−○ − T∆rS−○
∆rH−○ = ∆rG−○ + T∆rS−○
= (−21.4... × 103 Jmol−1) + (298 K) × (−62.2... J K−1mol−1)
= −40.0... kJmol−1
�e overall cell reaction is
AgCl(s) + 1
2H2(g)→ Ag(s) +Cl
−(aq) +H+(aq)
�ese values of ∆rG−○ , ∆rS−○ , and ∆rH−○ are used with data from the Resource
section to calculate ∆fG−○(Cl−), ∆fH−○(Cl−), and S−○m(Cl−)
Noting from Section 3D.2(a) on page 101 that ∆fG−○(H+ , aq) = 0, and also that
elements in their reference states have ∆fG−○ = 0, the standard reaction Gibbs
energy for the reaction is given by
∆rG−○ = ∆fG−○(Cl− , aq) − ∆fG−○(AgCl, s)
Hence ∆fG−○(Cl− , aq) = ∆rG−○ + ∆fG−○(AgCl, s)
= (−21.4... kJmol−1) + (−109.79 kJmol−1) = −131.3 kJmol−1
Similarly, noting from Section 2C.2 on page 51 that ∆fH−○(H+ , aq) = 0, and
that elements in their reference states have ∆fH−○ = 0, the standard reaction
enthalpy is
∆rH−○ = ∆fH−○(Cl− , aq) − ∆fH−○(AgCl, s)