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For phenanthrene, the measured lifetime of the triplet state \(\tau_{p}\) is \(3.3 \mathrm{s}\), the fluorescence quantum yield is \(0.12,\) and the phosphorescence quantum yield is 0.13 in an alcohol-ether glass at \(77 \mathrm{K}\). Assume that no quenching and no internal conversion from the singlet state occurs. Determine \(k_{p}, k_{i s c}^{T},\) and \(k_{i s c}^{S} / k_{f}\)

Short Answer

Expert verified
Answer: The rate constants for phosphorescence (kp), intersystem crossing from the triplet state (kisc^T), and the ratio of the rate constants for intersystem crossing from the singlet state (kisc^S) and fluorescence (kf) for phenanthrene are approximately 0.303 s^(-1), 1.581 s^(-1), and 7.33, respectively.

Step by step solution

01

Calculate the phosphorescence rate constant

We are given the lifetime of the triplet state, \(\tau_{p}=3.3 s\), and asked to find the phosphorescence rate constant, \(k_{p}\). To do this, we use the relation: $$k_{p} = \frac{1}{\tau_{p}}$$ Plug in the value of \(\tau_{p}\): $$k_{p} = \frac{1}{3.3 \, \textrm{s}}$$ Calculate \(k_{p}\): $$k_{p} \approx 0.303 \, \mathrm{s}^{-1}$$
02

Calculate the intersystem crossing rate constant from the triplet state

We are given the phosphorescence quantum yield, \(\Phi_{p} = 0.13\), and asked to find the rate constant for intersystem crossing from the triplet state, \(k_{i s c}^{T}\). To do this, we use the relation: $$k_{i s c}^{T} = \frac{k_p(1 - \Phi_{p})}{\Phi_{p}}$$ Plug in the values of \(k_p\) and \(\Phi_p\): $$k_{i s c}^{T} = \frac{0.303 \, \mathrm{s}^{-1}(1 - 0.13)}{0.13}$$ Calculate \(k_{i s c}^{T}\): $$k_{i s c}^{T} \approx 1.581 \, \mathrm{s}^{-1}$$
03

Calculate the ratio of the intersystem crossing rate constant from the singlet state and fluorescence rate constant

Lastly, we are given the fluorescence quantum yield, \(\Phi_{f} = 0.12\), and asked to find the ratio \(\frac{k_{i s c}^{S}}{k_{f}}\). To do this, we use the relation: $$\frac{k_{i s c}^{S}}{k_{f}} = \frac{1 - \Phi_{f}}{\Phi_{f}}$$ Plug in the value of \(\Phi_{f}\): $$\frac{k_{i s c}^{S}}{k_{f}} = \frac{1 - 0.12}{0.12}$$ Calculate the ratio: $$\frac{k_{i s c}^{S}}{k_{f}} \approx 7.33$$ The calculated values are: $$k_{p} \approx 0.303 \, \textrm{s}^{-1}$$ $$k_{i s c}^{T} \approx 1.581 \, \textrm{s}^{-1}$$ $$\frac{k_{i s c}^{S}}{k_{f}} \approx 7.33$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quantum Yield
Quantum yield is a key concept in photochemistry, which refers to the efficiency of a photochemical process. It is defined as the ratio of the number of molecules that undergo a particular photochemical process to the number of photons absorbed. In essence, it's a measure of how effectively absorbed light leads to the desired chemical reaction or transition.
For example, in the context of the exercise, the fluorescence quantum yield (\(\Phi_{f}\)) is 0.12. This means that for every 100 photons absorbed, approximately 12 lead to fluorescence. Similarly, the phosphorescence quantum yield (\(\Phi_{p}\)) is 0.13, indicating that 13 out of every 100 absorbed photons result in phosphorescence.
  • The greater the quantum yield, the more efficient the process.
  • Quantum yield values range between 0 (no efficiency) and 1 (maximum efficiency).
Understanding quantum yield is crucial because it helps scientists and engineers design efficient photochemical processes and materials.
Triplet State
The triplet state is a quantum state of a molecule where two electrons have parallel spins. This state is important in photochemistry and photophysics because it often has longer lifetimes than the singlet state, where electrons have opposite spins.
For example, in the problem, the triplet state of phenanthrene has a lifetime of 3.3 seconds. This long lifetime allows processes like phosphorescence to occur, where energy is emitted as light over an extended period.
  • The triplet state is more stable than the singlet excited state, leading to longer-lived emissions.
  • Triplet states are integral to understanding phosphorescence, a phenomenon involving the gradual release of light over time.
Recognizing the differences between singlet and triplet states helps explain why some materials glow in the dark even after the light source is removed.
Phosphorescence Rate Constant
The phosphorescence rate constant (\(k_{p}\)) quantifies how quickly phosphorescence occurs. It is calculated as the inverse of the triplet state's lifetime. In the example, where the triplet state's lifetime (\(\tau_{p}\)) is 3.3 seconds, \(k_{p}\) is approximately 0.303 \(\textrm{s}^{-1}\).
The rate constant is crucial because it helps determine how fast the total energy relaxation process occurs through phosphorescence. This insight is vital for applications that rely on controlled light emission, such as glow-in-the-dark materials and phosphorescent lighting.
  • Higher \(k_{p}\) values indicate faster phosphorescence.
  • The rate constant can be affected by environmental factors like temperature and the presence of quenching agents.
Understanding how to calculate and manipulate \(k_{p}\) is essential for optimizing materials for specific photophysical applications.
Intersystem Crossing
Intersystem crossing is a process where an excited electron in a molecule transitions between states of different multiplicity (such as from a singlet to a triplet state). This process is non-radiative, meaning it doesn't emit a photon, but is crucial for enabling other processes like phosphorescence.
In the context of phenanthrene, we calculate the intersystem crossing rate constant from the triplet state (\(k_{i s c}^{T}\)) and from the singlet state (\(k_{i s c}^{S}\)). These values provide insight into how efficiently the molecule transitions between these states.
  • \(k_{i s c}^{T}\) informs us about the efficiency of transitioning from the triplet state back to the ground state.
  • \(\frac{k_{i s c}^{S}}{k_{f}} \) helps us understand the competition between intersystem crossing and fluorescence.
Mastering the concept of intersystem crossing is essential for designing materials with desired photochemical properties, such as those used in organic light-emitting diodes (OLEDs).

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Most popular questions from this chapter

The Kermack-McKendrick model was developed to explain the rapid rise and fall in the number of infected people during epidemics. This model involves the interaction of susceptible (S), infected (I), and recovered (R) people through the following mechanism: \\[ \begin{array}{l} \mathrm{S}+\mathrm{I} \stackrel{k_{1}}{\longrightarrow} \mathrm{I}+\mathrm{I} \\\ \mathrm{I} \stackrel{k_{2}}{\longrightarrow} \mathrm{R} \end{array} \\] a. Write down the differential rate expressions for \(S,\) I, and \(R\) b. The key quantity in this mechanism is called the epidemiological threshold defined as the ratio of \([\mathrm{S}] k_{1} / k_{2}\). When this ratio is greater than 1 the epidemic will spread; however, when the threshold is less than 1 the epidemic will die out. Based on the mechanism, explain why this behavior is observed.

The reaction of nitric oxide \((\mathrm{NO}(g))\) with molecular hydrogen \(\left(\mathrm{H}_{2}(g)\right)\) results in the production of molecular nitrogen and water as follows: \\[ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightarrow \mathrm{N}_{2} \mathrm{O}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \\] The experimentally-determined rate-law expression for this reaction is first order in \(\mathrm{H}_{2}(g)\) and second order in \(\mathrm{NO}(g)\) a. Is the reaction as written consistent with the experimental order dependence for this reaction? b. One potential mechanism for this reaction is as follows: \\[ \begin{array}{l} \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \stackrel{k_{1}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\ \mathrm{H}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \end{array} \\] Is this mechanism consistent with the experimental rate law? c. An alternative mechanism for the reaction is \\[ \begin{array}{l} 2 \mathrm{NO}(g) \frac{k_{1}}{\sum_{k-1}} \mathrm{N}_{2} \mathrm{O}_{2}(g) \text { (fast) } \\ \mathrm{H}_{2}(g)+\mathrm{N}_{2} \mathrm{O}_{2}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\ \mathrm{H}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \stackrel{k_{3}}{\longrightarrow} \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \end{array} \\] Show that this mechanism is consistent with the experimental rate law.

The enzyme fumarase catalyzes the hydrolysis of fumarate: Fumarate \((a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{L}\) -malate \((a q)\). The turnover number for this enzyme is \(2.5 \times 10^{3} \mathrm{s}^{-1},\) and the Michaelis constant is \(4.2 \times 10^{-6} \mathrm{M}\). What is the rate of fumarate conversion if the initial enzyme concentration is \(1 \times 10^{-6} \mathrm{M}\) and the fumarate concentration is \(2 \times 10^{-4} \mathrm{M} ?\)

a. For the hydrogen-bromine reaction presented in Problem P36.7 imagine initiating the reaction with only \(\mathrm{Br}_{2}\) and \(\mathrm{H}_{2}\) present. Demonstrate that the rate law expression at \(t=0\) reduces to \\[ \left(\frac{d[\mathrm{HBr}]}{d t}\right)_{t=0}=2 k_{2}\left(\frac{k_{1}}{k_{-1}}\right)^{1 / 2}\left[\mathrm{H}_{2}\right]_{0}\left[\mathrm{Br}_{2}\right]^{1 / 2} \\] b. The activation energies for the rate constants are as follows: $$\begin{array}{cc} \text { Rate Constant } & \Delta \boldsymbol{E}_{a}(\mathbf{k J} / \mathbf{m o l}) \\ \hline k_{1} & 192 \\ k_{2} & 0 \\ k_{-1} & 74 \end{array}$$ c. How much will the rate of the reaction change if the temperature is increased to \(400 .\) K from \(298 \mathrm{K} ?\)

Determine the expression for fractional coverage as a function of pressure for the dissociative adsorption mechanism described in the text in which adsorption is accompanied by dissociation:$$R_{2}(g)+2 M(\text {surface}) \stackrel{k_{a}}{\rightleftharpoons_{k}} 2 R M(\text {surface})$$

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