Tuning Reactivity of Platinum(II) Complexes
Tuning Reactivity of Platinum(II) Complexes Tuning Reactivity of Platinum(II) Complexes
the concentration of one of the reactants is at least ten-fold excess, e.g. [B]0 >> [A]0. The equation can be treated as a reversible reaction and the rate of formation of C can be given as: d [A ] - = - d t d [B ] d t = d [C ] d t = k 2 [A ] t [B ] t - k -2 [C ] t Where k2 = second-order rate constant = k 2 [A ] t [B ] t - k -2 [A ] 0 - [A ] t = k 2 [A ] t [B ] t - k -2 [A ] t - k -2 [A ] 0 k-2 = observed first-order rate constant for the reverse reaction. By applying a mass balance for a given stoichiometry of 1:1:1, at time, t0 and t [A ] t = [A ] 0 - [C ] t , a n d [B ] t = [B ] 0 - [C ] t At equilibrium, [A ] e q = [A ] 0 - [C ] e q , a n d [B ] e q = [B ] 0 - [C ] e q = [B ] 0 - [A ] 0 + [A ] e q 8 (2 .5 ) (2 .6 ) (2 .7 ) Thus, at equilibrium the rates of the forward and reverse reactions are equal resulting in -d[A ] d t = k 2 [A ] eq [B ] eq + k -2 [A ] eq - k -2 [A ] 0 = 0 k -2 [A ] 0 = k 2 [A ] eq [B ] eq + k -2 [A ] eq (2 .8 ) Substitution of k-2[A]t from Equation 2.8 into Equation 2.4 gives -d[A] dt = k 2 [A] t [B] t - k -2 {(k 2 [A] eq [B] eq ) +k -2 [A] eq } - k -2 [A] t (2.9) Substitution of [ B ] t and [ B ] eq in Equations 2.6 and 2.7 into (2.9), and approximating k2[A]t[A]o ≈ k2[A]eq[A]o and k2[A] 2 t ≈ k2[A] 2 eq leads to − d dt [ A] = k2[A]t[B]o - k2[A]eq[B]o +k-2[A]t - k-2[A]eq
= (k2[B]o + k-2) ([A]t- [A]eq) (2.10) Separation of variables and integration gives: This results in ln [ ] d[ A] ( [ A] − [ A] ) t ( [ ] ) dt A t ∫ [ A] = − 2 B 0 + k− [ A ] t - [ A ] e q [ A ] 0 - [ A ] e q k 2 ∫ 0 0 (2.11) t eq = - ( k 2 [ B ] 0 + k -2 ) t = -k o b s t w h e r e , k o b s = k 2 [ B ] 0 + k -2 9 ( 2 .1 2 ) Under pseudo first-order conditions, the experimental first order rate constant, kobs, is given by: -d [M L 3 X ] d t w ith = k o bs [M ] k o b s = k 2 [Y ] + k -2 (2 .1 3 ) A plot of kobs versus the initial concentration of the nucleophile, [B]0 = [Y]0, is linear with a slope equal to the second-order rate constant, k2, and the y-intercept value equal to the first-order rate constant, k-2. Typical kinetic plots are shown in Figure 2.3. A plot which passes through zero implies that the forward reaction is irreversible and directly goes to completion, whereas a non-zero y-intercept signifies a back reaction in which the nucleophile Y is being substituted from the metal centre by the solvolysis process.
- Page 17 and 18: Figure 2.2 Potential energy profile
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- Page 25 and 26: Table 6.4 Summary of rate constants
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- Page 34 and 35: toxic potential. The most well-know
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- Page 70 and 71: k obs , s -1 0.00030 0.00025 0.0002
- Page 72 and 73: k = Ae -Ea/RT 2.14 lnk = lnA - E a
- Page 74 and 75: = 23.76 + R Hence, a plot of ln ⎛
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- Page 92 and 93: PEt 3 PEt 3 R PEt 3 Pt Pt Y Y Cl R
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- Page 100 and 101: References 1 (a) J. Reedijk, Chem.
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- Page 114 and 115: 84% (34.7 mg, 0.0618 mmol). 1 H NMR
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the concentration <strong>of</strong> one <strong>of</strong> the reactants is at least ten-fold excess, e.g. [B]0 >> [A]0. The<br />
equation can be treated as a reversible reaction and the rate <strong>of</strong> formation <strong>of</strong> C can be<br />
given as:<br />
d [A ]<br />
- = -<br />
d t<br />
d [B ]<br />
d t<br />
=<br />
d [C ]<br />
d t<br />
= k 2 [A ] t [B ] t - k -2 [C ] t<br />
Where k2 = second-order rate constant<br />
= k 2 [A ] t [B ] t - k -2 [A ] 0 - [A ] t<br />
= k 2 [A ] t [B ] t - k -2 [A ] t - k -2 [A ] 0<br />
k-2 = observed first-order rate constant for the reverse reaction.<br />
By applying a mass balance for a given stoichiometry <strong>of</strong> 1:1:1, at time, t0 and t<br />
[A ] t = [A ] 0 - [C ] t , a n d [B ] t = [B ] 0 - [C ] t<br />
At equilibrium,<br />
[A ] e q = [A ] 0 - [C ] e q , a n d [B ] e q = [B ] 0 - [C ] e q = [B ] 0 - [A ] 0 + [A ] e q<br />
8<br />
(2 .5 )<br />
(2 .6 )<br />
(2 .7 )<br />
Thus, at equilibrium the rates <strong>of</strong> the forward and reverse reactions are equal resulting in<br />
-d[A ]<br />
d t<br />
= k 2 [A ] eq [B ] eq + k -2 [A ] eq - k -2 [A ] 0 = 0<br />
k -2 [A ] 0 = k 2 [A ] eq [B ] eq + k -2 [A ] eq (2 .8 )<br />
Substitution <strong>of</strong> k-2[A]t from Equation 2.8 into Equation 2.4 gives<br />
-d[A]<br />
dt = k 2 [A] t [B] t - k -2 {(k 2 [A] eq [B] eq ) +k -2 [A] eq } - k -2 [A] t (2.9)<br />
Substitution <strong>of</strong> [ B ] t and [ B ] eq in Equations 2.6 and 2.7 into (2.9), and approximating<br />
k2[A]t[A]o ≈ k2[A]eq[A]o and k2[A] 2 t ≈ k2[A] 2 eq leads to<br />
− d<br />
dt<br />
[ A]<br />
= k2[A]t[B]o - k2[A]eq[B]o +k-2[A]t - k-2[A]eq