Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The dissociation energy of a carbon-bromine bond is typically about 276 \(\mathrm{kJ} / \mathrm{mol}\) . (a) What is the maximum wave-length of photons that can cause \(\mathrm{C}-\) Br bond dissociation? (b) Which kind of electromagnetic radiation-ultraviolet, visible, or infrared-does the wavelength you calculated in part (a) correspond to?

Short Answer

Expert verified
To find the maximum wavelength of photons that can cause the C-Br bond dissociation, we first convert the dissociation energy to Joules per photon: Energy per photon (J) = (276 kJ per mol) * (1000 J per kJ) / (Avogadro's number, \(6.022\times10^{23}\) mol−1). Then, we use the Planck's equation \(\lambda = \frac{hc}{E}\) and plug in the values for \(E\), \(h\), and \(c\) to calculate the maximum wavelength \(\lambda\). Finally, we compare the calculated wavelength with the ranges of ultraviolet (10 nm - 400 nm), visible (400 nm - 700 nm), and infrared (700 nm - 1 mm) radiation to identify the type of radiation it corresponds to.

Step by step solution

01

Convert energy into Joules per photon

We are given the dissociation energy of the carbon-bromine bond as 276 kJ/mol, but we need it in Joules per photon. Since 1 mol consists of Avogadro's number of particles, we need to divide the given energy by Avogadro's number to get the energy per photon. We must also convert kJ to Joules by multiplying by 1000. Energy per photon (J) = (276 kJ per mol) * (1000 J per kJ) / (Avogadro's number, \(6.022\times10^{23}\) mol−1)
02

Calculate the maximum wavelength using Planck's equation

Planck's equation relates the energy of a photon to its wavelength: \(E = h\frac{c}{\lambda}\), where \(E\) is the energy of the photon, \(h\) is Planck's constant (\(6.626\times10^{-34}\) Js), \(c\) is the speed of light (\(3 \times 10^8\) m/s), and \(\lambda\) is the wavelength. To find the maximum wavelength, we need to rearrange Planck's equation for \(\lambda\): \(\lambda = \frac{hc}{E}\) Now, plug in the values for \(E\), \(h\), and \(c\), and calculate the maximum wavelength \(\lambda\).
03

Identify the corresponding electromagnetic radiation

Based on the calculated wavelength, determine whether it corresponds to ultraviolet, visible, or infrared radiation: - Ultraviolet (UV) radiation: 10 nm - 400 nm - Visible light: 400 nm - 700 nm - Infrared (IR) radiation: 700 nm - 1 mm Compare the calculated wavelength with these ranges to identify the type of radiation.

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 Vaia!

Key Concepts

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

Dissociation Energy
Dissociation energy refers to the amount of energy required to break a chemical bond between atoms within a molecule. It is an important concept in chemistry,
as it helps us understand how much energy is needed to separate the atoms completely. For example, in the case of a carbon-bromine bond, we note that the dissociation energy is 276 kJ/mol.
This metric indicates how much energy is necessary to disrupt the bond for all molecules present in one mole of this compound. To use this energy value in calculations relating to a single photon, we must convert it from a per mole basis (kJ/mol) to per photon (J/photon). This conversion involves dividing the total energy by Avogadro's number, as 1 mol corresponds to approximately 6.022 x 10^23 particles.
By performing this calculation, we convert the bond’s dissociation energy into a suitable form for analyzing photon interactions.
Electromagnetic Radiation
Electromagnetic radiation is energy that travels through space in waves and is a fundamental concept in fields like physics and chemistry. It includes a broad range of wavelengths and frequencies,
from short-wavelength gamma rays to long-wavelength radio waves. Different portions of this spectrum interact with matter in unique ways. In the context of breaking chemical bonds like the carbon-bromine bond,
we consider specific portions of the electromagnetic spectrum that can supply the required energy to initiate the dissociation process. Depending on where the wavelength of the impacting photon falls in the spectrum, it can cause dissociation or possibly other effects.
  • Ultraviolet (UV) Radiation: Ranges from 10 nm to 400 nm.
    These photons have enough energy to break many chemical bonds, including some carbon-bromine bonds.
  • Visible Light: Occupies the range from 400 nm to 700 nm.
    This range is typically less energetic but significant for various chemical processes.
  • Infrared (IR) Radiation: Ranges from 700 nm to 1 mm.
    Though typically involved in heating processes, it may not provide enough energy for dissociation of stronger bonds like C-Br.
These interactions show how understanding electromagnetic radiation is crucial for predicting and explaining chemical reactivity.
Planck's Equation
Planck's Equation is a fundamental equation that bridges the energy of photons with their wavelength. It is given by \[E = h\frac{c}{\lambda}\]where:
  • \(E\) is the energy of the photon in joules.
  • \(h\) is Planck's constant, approximately 6.626 x 10^-34 Js.
  • \(c\) is the speed of light, roughly 3 x 10^8 m/s.
  • \(\lambda\) is the wavelength of the photon in meters.
This equation shows how small changes in wavelength can have substantial impacts on the energy.
To find the maximum wavelength that can cause a bond to dissociate, we can rearrange this equation to solve for \(\lambda\):\[\lambda = \frac{hc}{E}\]Inputting the known values for Planck's constant, the speed of light,
and the dissociation energy (after conversion to J/photon), allows us to calculate the wavelength accurately.
Understanding Planck's equation provides insight into why certain portions of the electromagnetic spectrum have enough energy to break specific chemical bonds.

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

Do the reactions involved in ozone depletion involve changes in oxidation state of the O atoms? Explain.

The enthalpy of fusion of water is 6.01 \(\mathrm{kJ} / \mathrm{mol} .\) Sunlight striking Earth's surface supplies 168 \(\mathrm{W}\) per square meter \((1 \mathrm{W}=1\) watt \(=1 \mathrm{J} / \mathrm{s})\) . (a) Assuming that melting of ice is due only to energy input from the Sun, calculate how many grams of ice could be melted from a 1.00 square meter patch of ice over a 12 -h day. (b) The specific heat capacity of ice is 2.032 \(\mathrm{J} / \mathrm{g}^{\circ} \mathrm{C} .\) If the initial temperature of a 1.00 square meter patch of ice is \(-5.0^{\circ} \mathrm{C},\) what is its final temperature after being in sunlight for 12 \(\mathrm{h}\) , assuming no phase changes and assuming that sunlight penetrates uniformly to a depth of 1.00 \(\mathrm{cm} ?\)

The pH of a particular raindrop is 5.6 . (a) Assuming the major species in the raindrop are \(\mathrm{H}_{2} \mathrm{CO}_{3}(a q), \mathrm{HCO}_{3}^{-}(a q)\) and \(\mathrm{CO}_{3}^{2-}(a q),\) calculate the concentrations of these species in the raindrop, assuming the total carbonate concentration is \(1.0 \times 10^{-5} \mathrm{M}\) . The appropriate \(K_{a}\) values are given in Table \(16.3 .\) (b) What experiments could you do to test the hypothesis the rain also contains sulfur-containing species that contribute to its pH? Assume you have a large sample of rain to test.

The rate of solar energy striking Earth averages 168 watts per square meter. The rate of energy radiated from Earth's surface averages 390 watts per square meter. Comparing these numbers, one might expect that the planet would cool quickly, yet it does not. Why not?

An impurity in water has an extinction coefficient of \(3.45 \times 10^{3} M^{-1} \mathrm{cm}^{-1}\) at \(280 \mathrm{nm},\) its absorption maximum (A Closer Look, p. 576\() .\) Below 50 \(\mathrm{ppb}\) , the impurity is not a problem for human health. Given that most spectrometers cannot detect absorbances less than 0.0001 with good reliability, is measuring the absorbance of a water sample at 280 \(\mathrm{nm}\) a good way to detect concentrations of the impurity above the 50 -ppb threshold?

See all solutions

Recommended explanations on Chemistry Textbooks

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.

Sign-up for free