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 $k_\mathrm{q}$?


2 Answers 2


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 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!


  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!

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.

  • $\begingroup$ I don't know what klum is. Is there a way to work it out? $\endgroup$
    – Jane
    Mar 11, 2015 at 21:39
  • $\begingroup$ I find that the commonly used definition of quantum yield as $\Phi = I_{em}/I_{abs} $ is not very intuitive. It is easier to realise that the yield is the $\Phi =$ (rate of fluorescence)/(sum of rate for all other processes) which leads directly to the first equation above in red. $\endgroup$
    – porphyrin
    Dec 14, 2016 at 17:35

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