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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 173
E6A.7(b) �e relationship between∆rG−○ and the equilibriumconstant is given by [6A.15–
208], ∆rG−○ = −RT lnK. �e ratio of the equilibrium constants for the two
reactions is
K1
K2
= e
−∆rG−○1 /RT
e−∆rG−○2 /RT
= exp(−∆rG
−○
1 − ∆rG−○2
RT
)
= exp(−(−200 × 103 Jmol−1) − (+30 × 103 Jmol−1)
(8.3145 JK−1mol−1) × (300 K)
) = 1.1 × 1040
E6A.8(b) �e reaction Gibbs energy at an arbitrary stage is given by [6A.11–207], ∆rG =
∆rG−○ + RT lnQ. In this case ∆rG−○ = −4.73 kJmol−1. �e values of ∆rG for
each value of Q are:
(i) At Q = 0.10
∆rG = (−4.73 × 103 Jmol−1) + (8.3145 JK−1mol−1)×(298 K)×ln(0.10)
= −1.04... × 104 Jmol−1 = −10 kJmol−1
(ii) At Q = 1.0
∆rG = (−4.73 × 103 Jmol−1) + (8.3145 JK−1mol−1)×(298 K)×ln(1.0)
= −4.73... × 103 Jmol−1 = −4.7 kJmol−1 (= ∆rG−○)
(iii) At Q = 10
∆rG = (−4.73 × 103 Jmol−1) + (8.3145 JK−1mol−1)×(298 K)×ln(10)
= +9.75... × 102 Jmol−1 = +0.98 kJmol−1
(iv) At Q = 100
∆rG = (−4.73 × 103 Jmol−1) + (8.3145 JK−1mol−1)×(298 K)×ln(100)
= +6.68... × 103 Jmol−1 = +6.7 kJmol−1
�e equilibrium constant K is the value of Q for which ∆rG = 0. From the
above values, K will therefore be somewhere between 1.0 and 10. To �nd
exactly where by linear interpolation, note that according to ∆rG = ∆rG−○ +
RT lnQ, a plot of ∆rG against lnQ should be a straight line. Consider the two
points on either side of zero, that is, (ii) and (iii). �e point ∆rG = 0 occurs a
fraction (4.73...)/(0.975... + 4.73...) = 0.829... of the way between points (ii)
and (iii), so is at
lnK = ln 1 + (0.822...)×(ln 10 − ln 1) = 1.90...
Hence K = e1.90... = 6.75
�e value is calculated directly by setting ∆rG = 0 and Q = K in ∆rG = ∆rG−○ +
RT lnQ and rearranging for K
K = e−∆rG
−○/RT = exp(− −4.73 × 103 Jmol−1
(8.3145 JK−1mol−1)×(298 K)
) = 6.75
which is the same result as obtained from the linear interpolation.