Stern-Volmer plot and calculation of rate constant for quenching

I did an experiment irradiating benzophenone in UV light and worked out the rate constant for the reaction. Then I repeated the experiment using various concentrations of a quencher and worked out their respective rate constants. Now I need to calculate the rate constant for quenching $k\_\mathrm{q}$ using a Stern-Volmer plot. How would I do this?


The equation I have used before for the Stern-Volmer relationship is:


$$\frac{1}{I\_\mathrm{f}} = \frac{1}{I\_\mathrm{abs}}\left(1 + \frac{k\_\mathrm{q}[\ce{Q}]}{k\_\mathrm{lum}}\right)$$


where the *x*-axis is $[\ce{Q}]$, the concentration of the quencher, and the *y*-axis is $1/I\_\mathrm{f}$, and to work out $k\_\mathrm{q}$ you calculate the gradient/*y*-intercept = $k\_\mathrm{q}/k\_\mathrm{lum}$. 


I think I would plot a graph of concentration of quencher $[\ce{Q}]$ on the $x$-axis against 1/rate constant on the y axis. But how would I work out $k\_\mathrm{q}$?




Benzophenone is a notorious case where prompt fluorescence, phosphorescence and delayed fluorescence can mix. Have a look at *Modern Molecular Photochemistry* by N.J. Turro for some general information and [*Prompt and delayed fluorescence from benzophenone*](http://dx.doi.org/10.1016/0009-2614(72)87176-8) by 
R.E. Brown and L.A. Singer for the original reference.


\[
\begin{align\*}
\ce{BP &->[h\nu]\ ^{1}BP^{\ast}}\\ 
\ce{^{1}BP^{\ast} &-> BP + h\nu'\quad &(prompt\ fluorescence)}\\
\ce{^{1}BP^{\ast} &->[ISC]\ ^{3}BP^{\ast} &(intersystem\ crossing)}\\
\ce{^{3}BP^{\ast} &-> BP + h\nu''\quad &(phosphorscence)}\\
\ce{^{3}BP^{\ast} &->\ ^{1}BP^{\ast} &(repopulation\ of\ S\_1)}\\
\ce{^{1}BP^{\ast} &-> BP + h\nu'\quad &(delayed\ fluorescence)}\\
\end{align\*}
\]


Let's ignore all this (for a moment).


Fluorescence quenching, i.e the observed decrease of fluorescence intensity in the presence of a quencher, may occcur through two different pathways:


### Static fluorescence quenching


In this case, **formation of a ground state complex between the fluorophore $\ce{F}$ a quencher $\ce{Q}$** decreases the concentration of $\ce{F}$, hence the concentration of the excited state $\ce{^1F^{\ast}}$ and the fluorescence intensity. **This has no impact on the fluorescence lifetimes!** These are the same in the absence ($\tau\_0$) and presence ($\tau$) of a quencher.


\[
\frac{\tau\_0}{\tau} = 1
\] 


### Dynamic fluorescence quenching


Here, **quenching occurs from the excited state $\ce{^1F^{\ast}}$**. With other words, the presence of a quencher introduces an additional, non-radiative pathway for the deactivation of the excited singlet state. Consequently, **the fluorescence lifetime $\tau$ is shortened!**


Let's try to derive the *Stern-Volmer* equation and see if that leads to some insights: 


(*You probably know all of this already, but I'm just in the mood...*)


In an experiment on dynamic fluorescence quenching, the following processes and their rate constants have to be considered:


\[
\begin{align\*}
\ce{F + h\nu &->[k\_{abs}]\ ^1F^{\ast}\quad &\mathrm{(excitation)} } \\
\ce{^1F^{\ast} &->[k\_{ic}]\ F + heat\quad &(internal\ conversion)}\\
\ce{^1F^{\ast} &->[k\_{f}]\ F + h\nu'\quad &(fluorescence)}\\
\ce{^1F^{\ast} + Q &->[k\_{q}]\ F + Q'\quad &(quenching)}\\
\end{align\*}
\]


The fluorescence quantum yield $\Phi$ is given as the quotient of emitted ($I\_{em}$) and absorbed ($I\_{abs}$) photons:


\[
\Phi = \frac{I\_{em}}{I\_{abs}}
\]


The number of emitted photons depends on the concentration of the fluorophore in its excited singlet state $\ce{^1F^{\ast}}$ and the rate constant $k\_{f}$:


\[
I\_{em} = k\_{f}\cdot[^1F^{\ast}]
\]


The number of absorbed photons depends on the concentration of the fluorophore $\ce{F}$ and the rate constant $k\_{abs}$:


\[
I\_{abs} = k\_{abs}\cdot[F]
\]


The rate of absorption processes is difficult to measure, they're bloody fast. Fortunately, we don't have to! **Under steady-state conditions, population of the excited singlet state and its deactivation balances out.**


Consequently, we can express $I\_{abs}$ as: 


\[
I\_{abs} = k\_{f}\cdot[^1F^{\ast}] + k\_{ic}\cdot[^1F^{\ast}] +k\_{q}\cdot[^1F^{\ast}]\cdot[Q]
\]


The expression for the fluorescence quantum yield thus is:


\[
\begin{align\*}
\Phi &= \frac{k\_{f}\cdot[^1F^{\ast}]}{k\_{f}\cdot[^1F^{\ast}] + k\_{ic}\cdot[^1F^{\ast}] +k\_{q}\cdot[^1F^{\ast}]\cdot[Q]}\\
&= \color{\red}{\frac{k\_{f}}{k\_{f} + k\_{ic} +k\_{q}\cdot[Q]}}
\end{align\*}
\]


The latter is pretty convenient since we don't have to think about the pesky $[^1F^{\ast}]$ anymore! **YAY!**


If we do perform a fluorescence measurement in the absence of the quencher, $[\ce{Q}] = 0$, $\Phi\_0$, the fluorescence quantum yield under this condition is:


\[
\Phi\_0= \frac{k\_{f}}{k\_{f} + k\_{ic}}
\]


The effect of a quencher $\ce{Q}$ on the fluorescence is expressed as:


\[
\begin{align\*}
\color{red}{\frac{\Phi\_0}{\Phi}} &= \frac{k\_{f}}{k\_{f} + k\_{ic}} \cdot \frac{k\_{f} + k\_{ic} +k\_{q}\cdot[Q]}{k\_{f}}\\
&= \color{\red}{1+ \frac{k\_{q}\cdot[Q]}{k\_{f} + k\_{ic}}}
\end{align\*}
\]


Too bad that we don't know all these rate constants and using just a steady-state spectrometer, we won't be able to measure them! And for a lab chemist, there's too much stuff in the equation anyway! **Bummer!**


But we can argue about what influences the **fluorescence lifetimes**, which, btw, can be measured!


Lifetimes are pretty much the inverse of the (sum of the) rate constants.


Without the quencher, the fluorescence lifetime $\tau\_0$ is:


\[
\tau\_0 = \frac{1}{k\_{f} + k\_{ic}}
\]


In the presence of a quencher, the fluorescence lifetime $\tau$ is:


\[
\tau = \frac{1}{k\_{q}\cdot[Q] + k\_{f} + k\_{ic}}
\]


If we plug that in again and rearrange a bit, we get:


\[
\begin{align\*}
\frac{\Phi\_0}{\Phi} &= 1 + \tau\_0\cdot k\_q\cdot[\ce{Q}]\\
&= 1+ {K\_{SV}}\cdot[\ce{Q}]
\end{align\*}
\]


**We name $K\_{SV}$ the Stern-Volmer constant!**


### Summary


1. Plotting $\frac{\Phi\_0}{\Phi}$ or the quotient of the areas of the fluorescence spectra vs the concentration of the quencher should give a straight line, that cuts the $y$ axis at 1.
2. The slope of the curve is the Stern-Volmer constant $K\_{SV}$.
3. The real rate constant $k\_q$ for the reaction of the excited state $\ce{^1F^{\ast}}$ with the quencher can only be obtained from $K\_{SV}$ if the fluorescence lifetime $\tau\_0$ has been obtained from a **time-resolved fluorescence measurement**!


 5

If you plot such a graph you should obtain a straight line with some slope $a$ and intercept $b$: $y=ax+b$. You can see that also Stern Volmer relation is a linear function (of concentration). Thus, $k\_q$ can be calculated by equating value of slope you obtained ($a$) to $k\_q/klum$. Such procedure is simple and important to understand as can be used to determine many quantities which occur in linear equations.


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