/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 The quantum yield for the photos... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 30

The quantum yield for the photosubstitution reaction \\[\mathrm{Cr}(\mathrm{CO})_{6}+\mathrm{NH}_{3}+h v \longrightarrow \mathrm{Cr}(\mathrm{CO})_{5} \mathrm{NH}_{3}+\mathrm{CO}\\] in octane solution at room temperature is 0.71 for a photolysis wavelength of \(308 \mathrm{nm} .\) How many \(\mathrm{Cr}(\mathrm{CO})_{6}\) molecules are destroyed per second when the solution is irradiated by a continuous laser with an output radiant power of \(1.00 \mathrm{mW}\) at \(308 \mathrm{nm} ?\) If you wanted to produce one mole of \(\mathrm{Cr}(\mathrm{CO})_{5} \mathrm{NH}_{3}\) per minute of exposure, what would the output radiant power of the laser need to be? (For both questions, assume the sample is sufficiently concentrated so that all the incident light is absorbed.)

Short Answer

Expert verified
1) Destroyed molecules per second: \(2.73 \times 10^{15}\), 2) Required power for mole/min: \(85.2 \text{ W}\).

Step by step solution

01

Calculate the Energy of Each Photon

To find the energy of each photon, we use the formula \(E = \frac{hc}{\lambda}\), where \(h\) is Planck's constant \(6.626 \times 10^{-34} \text{ J s}\), \(c\) is the speed of light \(3.00 \times 10^8 \text{ m/s}\), and \(\lambda\) is the wavelength \(308 \times 10^{-9} \text{ m}\). Substitute in these values to calculate the energy per photon.
02

Calculate the Number of Photons Per Second

The radiant power \(P\) is the energy per second. With \(P = 1.00 \text{ mW} = 1.00 \times 10^{-3} \text{ W}\), divide \(P\) by the energy per photon (from Step 1) to get the number of photons hitting the solution per second.
03

Determine the Number of Molecules Destroyed per Second

The quantum yield \(\Phi = 0.71\) indicates that 71% of absorbed photons lead to the reaction. Multiply the number of photons (from Step 2) by 0.71 to find the number of \(\text{Cr(CO)}_6\) molecules destroyed per second.
04

Calculate Photons Needed for Mole per Minute Production

One mole corresponds to \(6.022 \times 10^{23}\) molecules. To produce one mole per minute, divide \(6.022 \times 10^{23}\) by 60 seconds to find the molecules needed per second. Then, divide this number by the quantum yield (0.71) to find the required photons per second.
05

Determine New Output Radiant Power

Multiply the required number of photons per second (from Step 4) by the energy per photon (from Step 1) to find the new radiant power necessary to produce one mole per minute.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Photosubstitution Reaction
Photosubstitution reactions involve the replacement of a ligand in a complex molecule through the absorption of light. In the given example, the reaction involves the substitution of a carbonyl ligand (\(\mathrm{CO}\)) with an ammonia molecule (\(\mathrm{NH}_3\)) in the chromium hexacarbonyl compound \(\mathrm{Cr}( ext{CO})_6\). This type of reaction is significant in photochemistry because it relies on the energy provided by photons to trigger chemical changes.

The process typically involves:
  • The absorption of light by the reactant molecule, which results in an excited state.
  • A ligand substitution that occurs due to the energy fill from the absorbed light.
  • Returning the molecule to a stable state with the new ligand attached.
Understanding photosubstitution reactions is crucial for fields like materials science and pharmaceuticals, where light-induced changes are utilized for developing innovative materials and drugs.
Photolysis
Photolysis refers to the chemical decomposition of a compound caused by photons. In simpler terms, it is the breakdown of molecules under the influence of light. This process is widespread in nature and is a fundamental mechanism in various chemical and biological processes.

Key aspects of photolysis include:
  • Initiation by light waves that provide energy to break chemical bonds within a molecule.
  • It often results in radical species that are highly reactive due to unpaired electrons.
  • Photolysis is crucial in the Earth's atmosphere 鈥� for example, it helps decompose ozone, which protects living organisms from harmful UV radiation.
In the context of the given exercise, photolysis acts as the pivotal step that enables the photosubstitution reaction to proceed by providing the necessary energy to overcome activation barriers.
Radiant Power
Radiant power, also known as radiant flux, refers to the total energy emitted by a source per unit time. It鈥檚 a crucial parameter in photochemical experiments, as it defines the availability of photon energy to induce reactions. In our exercise, the radiant power provides valuable insight into how many photons strike the solution and participate in the chemical reaction.

Important points about radiant power are:
  • It is measured in watts (W), indicating joules per second, thus quantifying energy delivery over time.
  • In photochemical terms, higher radiant power means more photons available to cause reactions, potentially increasing the rate of conversion of reactants to products.
  • Adjusting the radiant power of a light source can control the progress and efficiency of photo-induced reactions.
The concept of radiant power is integral to designing experiments that require controlled light exposure and accurate prediction of reaction rates.
Planck's Constant
Planck's constant is a fundamental quantity in quantum mechanics denoted by \(h\). It is crucial in the relationship between the energy of a photon and its frequency. This constant enables the conversion of light's frequency into an energy value, allowing for the calculation of photon energy involved in photochemical processes like photosubstitution and photolysis.

Consider these points about Planck's constant:
  • The value of \(h\) is approximately \(6.626 \times 10^{-34} \, \text{Js}\), providing a link between the macroscopic and microscopic worlds.
  • It is instrumental in the energy equation \(E = hf\), which connects photon energy \(E\) to frequency \(f\).
  • This constant sets the scale for quantum mechanics, making it possible to describe atomic and subatomic systems accurately.
Planck's constant not only aids in computing the energies involved in light-driven reactions but also underscores the dual nature of light as both wave and particle.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The \(X^{1} A_{1} \rightarrow \tilde{A}\) electronic excitation of \(\mathrm{CH}_{3} \mathrm{I}(\mathrm{g})\) at \(260 \mathrm{nm}\) results in the following two competing photodissociation reactions: \\[\begin{aligned}\mathrm{CH}_{3} \mathrm{I}(\mathrm{g})+h v & \longrightarrow \mathrm{CH}_{3}(\mathrm{g})+\mathrm{I}(\mathrm{g})\left(^{2} \mathrm{P}_{3 / 2}\right) \\ & \longrightarrow \mathrm{CH}_{3}(\mathrm{g})+\mathrm{I}^{*}(\mathrm{g})\left(^{2} \mathrm{P}_{1 / 2}\right) \end{aligned}\\] The energy difference between the excited \(^{2} P_{1 / 2}\) state and the ground \(^{2} P_{3 / 2}\) state of \(I(g)\) is \(7603 \mathrm{cm}^{-1}\). The total quantum yield for dissociation is 1.00 with \(31 \%\) of the excited molecules producing I' \((\mathrm{g}) .\) Assuming that \(\mathrm{I}^{*}(\mathrm{g})\) relaxes by only radiative decay, calculate the number of photons emitted per second by a \(\mathrm{CH}_{3} \mathrm{I}(\mathrm{g})\) sample that absorbs \(10 \%\) of the light generated by a 1.00 -mW 260 -nm laser.

The decadic absorbance A of a sample is defined by \(A=\log \left(I_{0} / I\right),\) where \(I_{0}\) is the light intensity incident on the sample and \(I\) is the intensity of the light after it has passed through the sample. The decadic absorbance is proportional to \(c,\) the molar concentration of the sample, and \(l\), the path length of the sample in meters, or in an equation \\[A=\varepsilon c l\\] where the proportionality factor \(\varepsilon\) is called the molar absorption coefficient. This expression is called the Beer-Lambert law. What are the units of \(A\) and \(\varepsilon ?\) If the intensity of the transmitted light is \(25.0 \%\) of that of the incident light, then what is the decadic absorbance of the sample? At \(200 \mathrm{nm},\) a \(1.42 \times 10^{-3} \mathrm{M}\) solution of benzene has decadic absorbance of \(1.08 .\) If the pathlength of the sample cell is \(1.21 \times 10^{-3} \mathrm{m},\) what is the value of \(\varepsilon ?\) What percentage of the incident light is transmitted through this benzene sample? (It is common to express \(\varepsilon\) in the non SI units \(\mathrm{L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{cm}^{-1}\) because \(l\) and \(c\) are commonly expressed in \(\mathrm{cm}\) and \(\mathrm{mol} \cdot \mathrm{L}^{-1},\) respectively. This difference in units leads to annoying factors of 10 that you need to be aware of.)

There are nonlinear optical materials that can sum two laser beams at frequencies \(v_{1}\) and \(v_{2}\) and thereby generate light at frequency \(v_{3}=v_{1}+v_{2} .\) Suppose that part of the output from a krypton ion laser operating at \(647.1 \mathrm{nm}\) is used to pump a rhodamine 700 dye laser that produces laser light at 803.3 nm. The dye laser beam is then combined with the remaining output from the krypton ion laser in a nonlinear optical material that sums the two laser beams. Calculate the wavelength of the light created by the nonlinear optical material. The following four problems examine how the intensity of absorption lines are quantified.

The Einstein coefficients can also be derived using quantum mechanics. If the ground state and the excited state have a degeneracy of \(g_{1}\) and \(g_{2}\), respectively, the Einstein \(A\) coefficient is given by $$ A=\frac{16 \pi^{3} v^{3} g_{1}}{3 \varepsilon_{0} h c^{3} g_{2}}|\mu|^{2} $$ where \(|\mu|\) is the transition dipole moment (see Section 13-11). Now consider the \(1 s \rightarrow 2 p\) absorption of \(\mathrm{H}(\mathrm{g})\), which is observed at \(121.8 \mathrm{~nm}\). The radiative lifetime (see Example 15-3) of the triply degenerate excited \(2 p\) state of \(\mathrm{H}(\mathrm{g})\) is \(1.6 \times 10^{-9} \mathrm{~s}\). Determine the value of the transition dipole moment for this transition.

Which laser pulse contains more photons, a 10 -ns, 1.60 -mJ pulse at 760 nm or a 500 -ms, \(1.60-\mathrm{mJ}\) pulse at \(532 \mathrm{nm} ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Chemical Reactions

Read Explanation

Making Measurements

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.