Nitrogen dioxide fluorescence quenching and lifetime

Question

Nitrogen dioxide fluorescence quenching:

$$ \begin{align} \ce{NO2 + h\nu &->[$\varphi_\mathrm{Ia}$] NO2^{\ast}}\tag{I}\\ \ce{NO2^{\ast} &->[$k_2$] NO2 + h\nu'}\tag{II}\\ \ce{NO2^{\ast} + NO2 &->[$k_3$] 2 NO2}\tag{III}\\ \ce{NO2^{\ast} + Xe &->[$k_4$] NO2 + Xe}\tag{IV}\\ \ce{NO2^{\ast} + NO2 &->[$k_5$] 2 NO + O2}\tag{V}\\ \end{align} $$

The fluorescence lifetime of $\ce{NO2^{\ast}}$ in the presence of all the reactions, $\tau$, measured at $\pu{298 K}$ at different concentrations (molecules per litre) of the reactants:

$$ \begin{array}{crrr} \hline \text{Experiment} & \tau/\pu{μs} & [\ce{Xe}]/\pu{L-1} & [\ce{NO2}]/\pu{L-1} \\ \hline 1 & 3.38 & \pu{1.6E19} & \pu{3.2E18} \\ 2 & 1.89 & \pu{1.6E19} & \pu{6.4E18} \\ 3 & 3.64 & \pu{0.8E19} & \pu{3.2E18} \\ \hline \end{array} $$

What is the real fluorescence lifetime of $\ce{NO2^{\ast}}?$

Is this question valid? Isn't the lifetime vs the concentration of the quencher supposed to be linear?

Answer 1

You are looking for the rate constant $1/\tau^0=k_2$. The rate equation for the $\ce{NO2^*}$ is (using $N^*$ for $\ce{[NO2^*]}$ and N for ground state NO2), $\displaystyle \frac{dN^*}{dt}=-k_2N^*$ and when quenched $\displaystyle \frac{dN^*}{dt}=-k_2N^* -N^*((k_3+k_5)N+k_4[Xe])$ which can both be integrated to give exponentials, i.e $N^*=N^*_0e^{-k_2t}$ and similarly for the other equation. Writing the rate constants as $1/\tau^0=k_2$ and $1/\tau_Q=k_2+(k_3+k_5)N+k_4[Xe]$ you can find the values after looking at the pattern of the quencher concentrations in the question.