Difference between revisions of "Theoretical Bakground to Isotopic Exchange Reactons"
(→Written and developed by Prof. Tor Bjørnstad (IFE/UiO) ) |
(→Written and developed by Prof. Tor Bjørnstad (IFE/UiO) ) |
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| Eqn. 6 | | Eqn. 6 | ||
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| − | Since [AX*] | + | Since [AX*]<sub><big>∞</big></sub> + [BX*]<sub><big>∞</big></sub> equals the sum of concentrations of AX* and BX* at any time (and is a constant), we can write: |
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| <math>[BX^{\star}]=[AX^{\star}_{\infty}]\cdot\left(\frac{[AX]_{T}+[BX]_{T}}{[AX]_{T}}\right)-[AX^{\star}]</math> | | <math>[BX^{\star}]=[AX^{\star}_{\infty}]\cdot\left(\frac{[AX]_{T}+[BX]_{T}}{[AX]_{T}}\right)-[AX^{\star}]</math> | ||
| Eqn. 8 | | Eqn. 8 | ||
| + | |} | ||
| + | Introduction into Eqn.4 gives: | ||
| + | {| width="200" cellspacing="0" cellpadding="0" border="0" | ||
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| + | | <math>\frac{d[AX^{\star}]}{dt}=\bold{r}\cdot\left(\frac{[AX^{\star}]+[BX^{\star}]}{[AX^{\star}]_{T}[BX^{\star}]_{T}}\right)\left([AX^{\star}]_{\infty}-[AX^{\star}]\right)</math> | ||
| + | | Eqn. 9 | ||
|} | |} | ||
Revision as of 12:51, 20 June 2012
Written and developed by Prof. Tor Bjørnstad (IFE/UiO)
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/Ls. Further, let the total concentration of AX and BX be [AX]T and [BX]T. Thus:
| [AX]T=[AX]+[AX]* (=const.) | Eqn. 2 |
and
| [BX]T=[BX]+[BX]* (=const) | Eqn. 3 |
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*
| |
| Eqn. 4 |
The rate of formation of AX* is dependant 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:
| 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:
| Eqn. 6 |
Since [AX*]∞ + [BX*]∞ equals the sum of concentrations of AX* and BX* at any time (and is a constant), we can write:
| Eqn. 7 |
Solving Eqn.7 with respect to [BX*] gives
| Eqn. 8 |
Introduction into Eqn.4 gives:
| Eqn. 9 |
