NucWik - Theoretical Background to Isotopic Exchange Reactions

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In order to demonstrate how measurements of radioactive nuclide isotopic exchange can provide an evaluation of the total reaction rate, it is necessary to analyze the process quantitatively. Consider an exchange reaction involving the interchange of a common atom X between the compounds AX and BX where some of the X-atoms in the compound BX is the tagged X*. The overall reaction is:

AX+BX* ⇄ AX* + BX

eqn (1)

Let '''r''' be the rate of exchange of all X-atoms between AX and BX (not just exchange of X*). The dimension is mol/L•s. Further, let the total concentration of AX and BX be [AX]T and [BX]T. Thus;
[AX]T=[AX]+[AX*] =const

and

[BX]T=[BX]+[BX*]=const

eqn(2)

For the duration of the experiment. The instantaneous rate of formation of [AX*] is then d[AX*]/dt=rate of formation of AX* -rate of destruction of AX*

$\frac{d[AX]}{dt}=\left(\frac{[BX^]}{[BX]_T}\right)\cdot r-\left(\frac{[AX^*]}{[AX]_T}\right)$
$=r\cdot\left(\frac{[BX^]}{[BX]_T}-\frac{[AX^]}{[AX]_T}\right)$	eqn(3)

The rate of formation of AX* is dependent on the total rate of exchange r and the labeled fraction BX* of the total concentration of the BX-molecules, as only exchange involving BX* will produce AX*. Similarly, the rate of destruction of AX* molecules will depend on r and the fraction at any time of AX* of the total concentration of the AX-molecules.
Once a specific distribution of X* atoms has been achieved, the reaction appears to stop as the net rate of exchange of X* becomes zero (i.e. d[AX*]/dt = 0). At this exchange equilibrium we obtain from Eqn.4;

$\frac{[AX^*]_{infty}}{[AX]_T}=\frac{[BX^*]_{\infty}}{[BX]_T}$

eqn(4)

where AX* is the concentration of AX* at equilibrium (t = ∞) and [BX*]∞ is the concentration of BX* at equilibrium. Since there is no difference in the behavior of the radio labelled and the non-radio labelled molecule, we also have that the fraction at equilibrium of X* in the AX-molecule, i.e. [AX*]∞/([AX*]∞ + [BX*]∞, equals the fraction of the total X in the AX-molecule, i.e.[AX]T/([AX]T + [BX]T) Hence:

$\frac{[AX^*]_{infty}}{[AX^*]_{infty}+[BX^*]_{\infty}}=\frac{[AX]_T}{[Ax]_T+[BX]_T}$

eqn(5)

where AX* is the concentration of AX* at equilibrium (t = ∞) and [BX*]∞ is the concentration of BX* at equilibrium. Since there is no difference in the behaviour of the radiolabelled and the non-radiolabelled molecule, we also have that the fraction at equilibrium of X* in the AX-molecule, i.e. [AX*]∞/([AX*]∞ + [BX*]∞, equals the fraction of the total X in the AX-molecule, i.e.[AX]T/([AX]T + [BX]T) Hence:

$\frac{[AX^{\star}]_{\infty}}{[AX^{\star}]_{\infty}+[BX^{\star}]_{\infty}}= \frac{[AX]_{T}}{[AX]_{T}+[BX]_{T}}$

eqn(6)

Since [AX*]∞ + [BX*]∞ equals the sum of concentrations of AX* and BX* at any time (and is a constant), we can write:

$\frac{[AX^{\star}]}{[AX^{\star}]+[BX^{\star}]}=\frac{[AX]_{T}}{[AX]_{T}+[BX]_{T}}$

eqn(7)

Solving Eqn.7 with respect to [BX*] gives:

$[BX^{\star}]=[AX^{\star}_{\infty}]\cdot\left(\frac{[AX]_{T}+[BX]_{T}}{[AX]_{T}}\right)-[AX^{\star}]$

eqn(8)

Introduction into Eqn.4 gives:

$\frac{d[AX^{\star}]}{dt}=r\cdot\left(\frac{[AX^{\star}]+[BX^{\star}]}{[AX^{\star}]_{T}[BX^{\star}]_{T}}\right)\left([AX^{\star}]_{\infty}-[AX^{\star}]\right)$

eqn(9)

In Eqn.9 the only variable is the [AX*]. By setting a = [AX]T, b = [BX]T, x = [AX*] and x∞ = [AX*]∞, Eqn.9 takes the following form:

$\frac{dx}{dt}=r\left(\frac{a+b}{a\cdot b}\right)(x_{\infty}-x)$

eqn(10)

Rearrangement and integration from x0 (at t = 0) to x∞ yields:

$\int^{x_{\infty}}_{x_{o}}\frac{dx}{x_{\infty}-x}=r\left(\frac{a+b}{ab}\right)\int^{\infty}_{o}dt$

eqn(11)

The solution is:

$\ln\left(\frac{x_{\infty}-x_{o}}{x_{\infty}-x}\right)=r\cdot t\cdot\frac{a+b}{a\cdot b}$

eqn(12)

Assuming that x0=0 at
t=0, i.e. AX is non-radioactive initially, gives:

$\ln\left(1-\frac{x}{x_{\infty}}\right)=-r\cdot t\cdot\frac{a+b}{a\cdot b}$

eqn(13)

Here, x and x∞ can be substituted by the corrected counting rates R and R∞. From Eqn.13 it is evident that r can be obtained from the slope of a plot of ln(1- x/x∞) as a function of time t. The quantity x∞, which is the activity of AX* at exchange equilibrium, can be obtained experimentally by allowing the reaction to proceed until no further change takes place. x∞ can also be found analytically from Eqn.8 by using the short-form notations above where a = [AX] + [AX*] and b = [BX] + [BX*] and introducing that y0 = x + y. This gives:

${\infty}=\left(\frac{a}{a+b}\right)\cdot y_{o}$

eqn(14)

The reaction rate r is given by;

$r=k\cdot [NaI]\cdot [RI]$

eqn(15)

where RI denotes an organic iodide compound. The rate constant k can be determined by measuring r as a function of the concentration of one reactant while keeping the other constant. Note that reaction rates are strongly temperature dependant so that provision must be made to prevent temperature variations during the experiment.