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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 91
�e standard reaction Gibbs energy is de�ned in [3D.9–100], ∆rG−○ = ∆rH−○ −
T∆rS−○ , thus
∆fG−○(CH3COOC2H5 ,(l)) = (−5.32... × 102 kJmol−1)
− (298.15 K)(−0.535... kJK−1mol−1)
= −373 kJmol−1 .
Solutions to problems
P3D.2 (a) �e standard reaction Gibbs energy is de�ned in [3D.9–100], ∆rG−○ =
∆rH−○ − T∆rS−○ .�is is rearranged to give the reaction entropy change
∆rS−○ =
∆rH−○ − ∆rG−○
T
= (−20 × 103 Jmol−1) − (−31 × 103 Jmol−1)
310 K
= +35 JK−1mol−1 .
(b) �e number of moles that are hydrolysed each second is n = N/NA.�us
the amount of non-expansionwork that is done ∣wnon−exp∣ = (N/NA)∣∆rG−○ ∣.
Assuming that the cell is spherical, its volume is given by 4
3πr3. �e
power density is the work that is done in a time interval for a unit volume.
�erefore
Power density = (N/NA)∣∆rG−○ ∣
4
3πr3
= (1.0 × 106)/(6.0221 × 1023mol−1) × ∣ − 31 × 103 Jmol−1∣
4
3π(10 × 10−6)3
= 12 Wm−3 .
For a computer battery it is
Power density (battery) = 15 W
100 × 10−6 m3
= 1.5 × 105Wm−3
�us the power density of the cell is much smaller.
(c) �e non-expansion work needed to produce 1 mol of glutamine is given
14.2 kJ.�us
n =
∣wnon−exp∣
∣∆rG−○ ∣
= 14.2 kJ
31 kJmol−1
= 0.46 mol .
P3D.4 �e standard reaction Gibbs energy is given by [3D.9–100], ∆rG−○ = ∆rH−○ −
T∆rS−○ .�e standard reaction entropy is [2C.5b–55], ∆rS−○ = ∑J νJS−○m(J), where
νJ are the signed stoichiometric numbers.�erefore
∆rS−○1 = S−○m(Na+ ,(g)) + S−○m(Cl− ,(g)) − S−○m(NaCl ,(s))
= (148 JK−1mol−1) + (154 JK−1mol−1) − (72.1 JK−1mol−1)
= 229.9 JK−1mol−1