A solution and solid state study of niobium complexes University of ...
A solution and solid state study of niobium complexes University of ...
A solution and solid state study of niobium complexes University of ...
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Chapter 3<br />
conditions, one <strong>of</strong> the concentrations remain constant while the other varies, e.g.<br />
[A]>>>[B]. This simplifies Equation 3.17 to:<br />
Rate = kobs[A] x <strong>and</strong> then kobs = k[B] y (3.18)<br />
kobs is the observed pseudo-first order rate constant. By varying the concentration <strong>of</strong><br />
A, the rate constant <strong>of</strong> the reaction can be determined. The rate law for a second<br />
order reaction (where x = y = 1) is given by:<br />
a[A] + b[B] c[C] (3.19)<br />
Rate = k1[A][B] + k2[A] (3.20)<br />
kobs under pseudo-first order conditions ([B]>>>[A]) is given by:<br />
kobs = k1[B] + k2 (3.21)<br />
The equilibrium constant for an equilibrium reaction, where k1 represents the forward<br />
reaction <strong>and</strong> k2 represents the reverse reaction, is given by:<br />
Keq = <br />
<br />
Integration <strong>of</strong> Equation 3.22 from t = 0 to t yields Equation 3.23:<br />
ln <br />
<br />
53<br />
(3.22)<br />
= kobs or [C]t = [C]0 e kobst (3.23)<br />
[C]t <strong>and</strong> [C]0 represent the concentration change <strong>of</strong> the reactant at time = 0 <strong>and</strong> t<br />
respectively. Through the basic principles <strong>of</strong> the Beer-lambert law, expressing<br />
absorbance in terms <strong>of</strong> concentration, <strong>and</strong> a bit <strong>of</strong> mathematical manipulation,<br />
Equation 3.23 can be given as:<br />
At = A∞ - (A∞ - A0)e kobst (3.24)<br />
where At = absorbance after time t <strong>and</strong> A∞ = absorbance at infinite time (when<br />
reaction is complete). Absorbance versus time data can be used in a least-squares<br />
fit to give kobs for the reaction. The half-life (t1/2) for a first order reaction, meaning<br />
the time needed for the reactant concentration to decay by 50%, will be:<br />
k1<br />
k2