What is the unit for energy transfer rate constant?(adsbygoogle = window.adsbygoogle || []).push({});

I am confused because of the order of reaction of energy transfer. Consider the case for following energy transfer A^{*}+ C → A + C^{*}. The star represent excited state. Then the rate equation for A^{*}would be as follows:

[itex]\frac{d[A^{*}]}{dt} = \frac{d[C]}{dt} = k_{C^{*}\rightarrow A}[A][C^{*}] -k_{A^{*}\rightarrow C}[A^{*}][C][/itex]

and you can see that energy transfer in this case is a second-order reaction.

If [itex][A^{*}][/itex] is in unit of concentration M (mol l^{-1}), obviously the unit for [itex]k_{A^{*}\rightarrow C}[/itex] would be M^{-1}s^{-1}. As so, units are different from typical photophysical process like fluorescence which is usually first order reaction. However, I have seen papers that treats energy transfer as first-order reaction (like this one), while I've seen ones that treats it as second-order reaction (like this one). Which one is right? I think that the latter is right, but then if you think about for example Forster transfer equation:

[itex]k_{A^{*}\rightarrow C} = \frac{9000c^{4}ln10}{128\pi ^{5}n^{4}N_{A}\tau _{0}^{a}}\cdot \frac{\kappa ^{2}}{R^{6}}\int f_{a}(\upsilon )\varepsilon _{b}(\upsilon )\frac{d\upsilon }{\upsilon ^{4}}[/itex]

it is obvious that the unit is given as first-order reaction. Do you have to change the units into second-order reaction? If so, then how do you do that?

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# Energy Transfer and rate equation

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