In an experiment to investigate the decay of photoexcited tryptophan, the observed lifetime of the excited state was measured for various concentrations of oxygen, O₂, which acted as a quencher. By constructing a Stern-Volmer plot of the following data, determine the rate constant for fluorescence quenching.
$\begin{array}{llllll}{\left[\mathrm{O}_{2}\right] / \mathrm{mol} \mathrm{dm}^{-3}} & 0.02 & 0.04 & 0.06 & 0.08 & 0.10 \\\tau_{\text {pbs }}\left(\mathrm{O}_{2}\right) / \mathrm{ns} & 1.64 & 1.18 & 0.91 & 0.74 & 0.63\end{array}$

Question

In an experiment to investigate the decay of photoexcited tryptophan, the observed lifetime of the excited state was measured for various concentrations of oxygen, O₂, which acted as a quencher. By constructing a Stern-Volmer plot of the following data, determine the rate constant for fluorescence quenching.

\begin{array}{llllll}{\left[\mathrm{O}_{2}\right] / \mathrm{mol} \mathrm{dm}^{-3}} & 0.02 & 0.04 & 0.06 & 0.08 & 0.10 \\\tau_{\text {pbs }}\left(\mathrm{O}_{2}\right) / \mathrm{ns} & 1.64 & 1.18 & 0.91 & 0.74 & 0.63\end{array}

Quizplus · Accepted Answer

The Stern-Volmer equation is:$$\frac{\tau_0}{\tau}=1+k_q\tau_0[Q]$$where $\tau_0$ is the lifetime in the absence of quencher, $\tau$ is the lifetime in the presence of quencher, $k_q$ is the quenching rate constant, and $[Q]$ is the quencher concentration.Plotting $\frac{\tau_0}{\tau}$ against $[Q]$ gives a straight line with slope $k_q\tau_0$.From the data, we can calculate:$$\frac{\tau_0}{\tau}=\frac{1.64}{1.18}=1.39$$$$\frac{\tau_0}{\tau}=\frac{1.64}{0.91}=1.80$$$$\frac{\tau_0}{\tau}=\frac{1.64}{0.74}=2.22$$$$\frac{\tau_0}{\tau}=\frac{1.64}{0.63}=2.60$$Plotting these values against $[Q]$ gives a straight line with slope $k_q\tau_0=12.3\times10^{-9}\text{ mol}^{-1}\text{ dm}^3\text{ s}^{-1}$.

In an Experiment to Investigate the Decay of Photoexcited Tryptophan