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Topic: Determining kinetic constants from quantum yield  (Read 1946 times)

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slyasafax

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Determining kinetic constants from quantum yield
« on: January 04, 2023, 07:48:04 PM »
Hi, trying to help my niece with physical chemistry but I'm not a chemist. And I don't really have a clue for this one in terms of dimensional analysis, at least not without biacetyl concentration.

Irradiating biacetyl at 426.5 nm gives the triplet state with high yield (φISC=0.97). Determine kinetic constants of triplet production (kISC), of fluorescence (kF), phosphorescence (kPh) and Non-radiative deactivation of the triplet state (kNR(T1)), knowing that the quantum yield of phosphorescence is 0.15 and the lifetime of the singlet and of the excited triplet are 24 ns and 1.5 ms, respectively.

I think you get kPh from this equation I found in her professor's notes,

φPh = φISC * kPh * τPh

And i would guess τPh is 1.5 ms (the average life time of the triplet state determines fluorescence).

But kF, kISC, kNR are a complete mistery to me!

Corribus

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Re: Determining kinetic constants from quantum yield
« Reply #1 on: January 05, 2023, 10:56:50 AM »
Hi, trying to help my niece with physical chemistry but I'm not a chemist.

You've sure set yourself up for a challenge

First, you don't need any concentration information, since photophysical rate constants are intrinsic properties of molecules. The only dimensions you need worry about here are time (for lifetimes) and inverse time (for rate constants). Quantum yields are unitless.

Second, it would be helpful to start with a Jablonski diagram, so you can visualize all the radiative and nonradiative processes that you need to keep track of and how they're related. Since it would be hopeless to try to reproduce that here, I encourage you to seek one out either in the textbook or online.

Third, you should write out in equation form the definitions of the quantum yields and lifetimes. Quantum yields can be defined in a lot of ways but the one that is needed here is this: the quantum yield of a process is the ratio of the rate constant for that process divided by the sum of the rate constants for all possible processes. So, the quantum yield of fluorescence Φf is:

$$\Phi_f = \frac{k_f}{k_f + k_{nr} + k_{isc}}$$

And the quantum yield of the (S1-T1) nonradiative intersystem crossing Φisc is:

$$\Phi_{isc} = \frac{k_{isc}}{k_f + k_{nr} + k_{isc}}$$

Here, kf, knr, and kisc are the respective rate constants of fluorescence (radiative deactivation of the singlet state, S1 S0), nonradiative internal conversion (S1 S0), and nonradiative intersystem crossing (S1 T1).

(Note here that there is some sloppiness in the problem, as there are actually two nonradiative intersystem crossing processes, so don't get them confused.)

The denominator in both quantum yield expressions is the same and is called the overall rate constant for singlet state deactivation. Since the inverse of any rate constant is the photophysical lifetime of the process, the inverse of the rate constant for singlet state deactivation is the intrinsic lifetime of the singlet state, τS1. So therefore the quantum yield expressions can be written as:

$$\Phi_f = k_f \tau_{S1}$$
$$\Phi_{isc} = k_{isc} \tau_{S1}$$

Where

$$\tau_{S1}= (k_f + k_{nr} + k_{isc})^{-1}$$

Quantum yields for phosphorescence Φp are formulated in more or less the same way except that you must recall that the quantum yield of phosphorescence is usually understood to be the fraction of all excited molecules that phosphoresce, not the fraction of triplet states that deactivate radiatively. So although we use the same rate constant formulation, it has to be scaled by the quantum yield of intersystem crossing to account for the fact that not all excited molecules end up in the triplet state.

I.e.,

$$\Phi_p = \Phi_{isc} \frac{k_p}{k_p + k_{nr}} = \Phi_{isc} k_p \tau_t$$

$$\Phi_{nr} = \Phi_{isc} \frac{k_{nr}}{k_p + k_{nr}} = \Phi_{isc} k_{nr} \tau_t$$

Where $$\tau_t$$ is the triplet state lifetime. Do note that, for the triplet state, knr is also formally an intersystem crossing process.

With these equations and the provided information you should be able to calculate all the parameters. However I noted that they did not give you a quantum yield of fluorescence. I didn't run through the problem myself but I suspect this will make it impossible to determine the rate constant for fluorescence (since it won't be possible to separate radiative from S1-S0 nonradiative deactivation of the singlet state). You may be expected to assume that all excited singlet that do not cross to the excited triplet deactivate radiatively (i.e., the fluorescence yield is 3%). Generally this is not a great assumption, but you could do it now to complete the exercise and then follow up with the instructor.

Do make sure you express the lifetime values in the same time unit, otherwise you will get an erroneous result.

Hope that helps.

« Last Edit: January 05, 2023, 11:12:14 AM by Corribus »
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slyasafax

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Re: Determining kinetic constants from quantum yield
« Reply #2 on: January 06, 2023, 03:42:06 PM »

You've sure set yourself up for a challenge
I know statistical physics very well so she seeked help for that part, I also managed some photochemistry problems that where much harder than this one... but I didn't know the definitions for much of this stuff! So that made it hard to navigate between the topics.

But I must say I still enjoy learning the theory behind chemistry (and seeing problems through completion!).

In the end we solved it with a combination of her instructor's (dense) notes and chemistry stack exchange.

We found that

$$k_{F} = \frac {1}{\tau_{S}}$$

$$k_{Ph} = \frac {1}{\tau_{T}}$$

where by τS and τT, respectively, I mean the given expected lifetime of singlet and triplet state (I know my notation is probably not standard but, yep, not a chemist... sorry ).

then, we found

$$\Phi_{ISC} = \frac {k_{ISC}}{k_{s}} = \frac {\tau_{s}}{\tau_{ISC}} = k_{ISC}\tau_{s}$$

so we did set

$$k_{ISC} = \frac {\Phi_{ISC}}{\tau_{s}}.$$

for determining $$k_{NR(T1)}$$ we used the equation linking the efficiency of phosphorescence to the efficiency of triplet production,

$$\Phi_{Ph} = \Phi_{ISC} \frac {k_{Ph}}{k_{Ph} + k_{NR(T1)}}$$

to get

$$k_{NR(T1)} = k_{Ph} \frac {\Phi_{ISC}}{\Phi_{Ph} } - k_{Ph}.$$

Corribus

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Re: Determining kinetic constants from quantum yield
« Reply #3 on: January 06, 2023, 04:21:47 PM »
I would just be careful with your first two equations because the singlet state lifetime is not equal to the inverse of the radiative rate for fluorescence, which is often designated kr but sometimes kf or k0. The singlet state lifetime incorporates both radiative and nonradiative deactivation channels, so it is equal to the inverse of the sum of all the respective deactivation rates. Similarly, the triplet state lifetime is not equal to the inverse of the radiative rate for phosphorescence.

The inverse of the radiative rate for fluorescence is often called the natural or intrinsic radiative/fluorescence lifetime; it is how long the singlet state would persist in the event there was no nonradiative decay and all excited states relaxed radiatively. Sometimes it is designated τ0 or τr. Analogous terms are used for phosphorescence.

As you probably are see, the terminology here isn't very well standardized, which can lead to a lot of confusion. For instance, I have seen kf used to represent the radiative rate for fluorescence (as I have above) but in other cases I have seen it used to represent the overall singlet deactivation rate as measured by time-resolved fluorescence. In the latter case, kf would be equal to the inverse of the singlet state lifetime. Since I don't know what terminology the instructor is using, it's hard for me to make any definitive judgments. The important thing is knowing exactly what we mean when we write "kf" or "kr" for any particulate problem and proceed accordingly.
« Last Edit: January 06, 2023, 04:38:12 PM by Corribus »
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman