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642 18 REACTION DYNAMICS
�is expression is then rearranged for ∆ER
−4RT ln ket,2/ket,1 − 2(∆rG−○2 − ∆rG−○1 ) =
(∆rG−○2 )2 − (∆rG−○1 )2
∆ER
∆ER =
(∆rG−○2 )2 − (∆rG−○1 )2
−4RT ln ket,2/ket,1 − 2(∆rG−○2 − ∆rG−○1 )
∆ER =
(∆rG−○1 )2 − (∆rG−○2 )2
4RT ln ket,2/ket,1 + 2(∆rG−○2 − ∆rG−○1 )
�e data are given in eV so it is convenient to express 4RT in eV with the help
of the conversion factors inside the front cover
4RT = 4×(8.3145 JK−1mol−1)×(298 K)× 1 eV
96.485 × 103 Jmol−1
= 0.102... eV
With the data given
∆ER =
(∆rG−○1 )2 − (∆rG−○2 )2
4RT ln ket,2/ket,1 + 2(∆rG−○2 − ∆rG−○1 )
= (−0.665 eV)2 − (−0.975 eV)2
(0.102... eV) × ln (3.33×106 s−1)
(2.02×105 s−1) + 2[(−0.975 eV) − (−0.665 eV)]
= 1.53... eV = 1.53 eV
Using the data for the �rst value of the rate constant
∆‡G = (∆rG−○ + ∆ER)2
4∆ER
= [(−0.665 eV) + (1.53... eV)]2
4 × (1.53... eV)
= 0.122... eV
�e expression for kr given by [18E.5–812] is then rearranged to �nd Het(d)2
Het(d)2 = hket (
RT∆ER
π3
)
1/2
e∆
‡G/RT
As the values of ∆ER and ∆‡G have already been found in eV, it is convenient
to express the term RT also in eV; following the same process as above gives
RT = 0.0256... eV.
Het(d)2 = (6.6261 × 10−34 J s) × (2.02 × 105 s−1)
× ((0.0256... eV) × (1.53... eV)
π3
)
1/2
e(0.122... eV)/(0.0256... eV)
= 5.59... × 10−28 J eV = 8.97... × 10−47 J2
On the �nal line the units are converted using 1 eV = 1.6021 × 10−19 J. Hence
Het(d) = 9.47 × 10−24 J .