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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 181
Using ∆rG−○ = −RT lnK to substitute for K1 and setting lnK2 = ln 1 = 0 (the
crossover point) gives
∆rG−○(T1)
RT1
= −∆rH
−○
R
( 1
T2
− 1
T1
)
Rearranging for T2 gives
T2 =
T1∆rH−○
∆rH−○ − ∆rG−○(T1)
= (1120 K) × (+125 kJmol−1)
(+125 kJmol−1) − (+22 kJmol−1)
= 1.4 × 103 K
E6B.4(b) �e van ’t Ho� equation [6B.2–214], d lnK/dT = ∆rH−○/RT2, is rearranged to
obtain an expression for ∆rH−○
∆rH−○ = RT2 d lnK
dT
= RT2 d
dT
(A+ B
T
+ C
T3
) = RT2 (− B
T2
− 3C
T4
) = −R (B + 3C
T2
)
= −(8.3145 JK−1mol−1) × ((−1176 K) + 3 × (2.1 × 107 K3)
(450 K)2
)
= 7.19... × 103 Jmol−1 = 7.2 kJmol−1
�e standard reaction entropy is obtained by �rst �nding an expression for
∆rG−○ using [6A.15–208]
∆rG−○ = −RT lnK = −RT (A+ B
T
+ C
T3
) = −R (AT + B + C
T2
)
�e equation ∆rG−○ = ∆rH−○ − T∆rS−○ [3D.9–100] is then rearranged to �nd
∆rS−○
∆rS−○ =
∆rH−○ − ∆rG−○
T
= 1
T
⎛
⎜⎜⎜⎜
⎝
−R (B + 3C
T2
)
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
∆rH−○
+R (AT + B + C
T2
)
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
−∆rG−○
⎞
⎟⎟⎟⎟
⎠
= R (A− 2C
T3
) = (8.3145 JK−1mol−1) × (−2.04 − 2 × (2.1 × 107 K2)
(450 K)3
)
= −21 JK−1mol−1
An alternative approach to �nding ∆rS−○ is to use the variation of G with T
which is given by [3E.8–107], (∂G/∂T)p = −S.�is implies that d∆rG−○/dT =
−∆rS−○ where the derivative is complete (not partial) because ∆rG−○ is indepen-
dent of pressure. Using the expression for ∆rG−○ from above it follows that
∆rS−○ = −
d∆rG−○
dT
= − d
dT
⎛
⎜⎜⎜⎜
⎝
−R (AT + B + C
T2
)
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
∆rG−○
⎞
⎟⎟⎟⎟
⎠
= R (A− 2C
T3
)
which is the same expression obtained above.