A CI-CI bond has a bond-dissociation energy of 243 kJ/mol. What is the longest wavelength of electromagnetic radiation (in nm) which can cleave this bond? O 492 nm 817 nm O 136 nm O2031 nm
A CI-CI bond has a bond-dissociation energy of 243 kJ/mol. What is the longest wavelength of electromagnetic radiation (in nm) which can cleave this bond? O 492 nm 817 nm O 136 nm O2031 nm
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![### Understanding Bond Dissociation Energy and Electromagnetic Radiation
**Question:**
> A Cl—Cl bond has a bond-dissociation energy of 243 kJ/mol. What is the longest wavelength of electromagnetic radiation (in nm) which can cleave this bond?
**Choices:**
- ○ 492 nm
- ○ 817 nm
- ○ 136 nm
- ○ 2031 nm
### Explanation
To find the longest wavelength of electromagnetic radiation that can cleave a Cl—Cl bond, we can use the relationship between energy \( E \) and wavelength \( \lambda \). The energy of a single photon is given by the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \( E \) is the energy,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength.
First, convert the bond-dissociation energy from kJ/mol to J/photon.
\[ 243 \, \text{kJ/mol} \times \frac{1000 \, \text{J}}{1 \, \text{kJ}} \times \frac{1 \, \text{mol}}{6.022 \times 10^{23} \, \text{photons}} = 4.037 \times 10^{-19} \, \text{J/photon} \]
Rearrange the equation \( E = \frac{hc}{\lambda} \) to solve for wavelength \( \lambda \):
\[ \lambda = \frac{hc}{E} = \frac{(6.626 \times 10^{-34} \, \text{J} \cdot \text{s})(3 \times 10^8 \, \text{m/s})}{4.037 \times 10^{-19} \, \text{J}} \approx 4.92 \times 10^{-7} \, \text{m} \]
Convert meters to nanometers (\(1 \, \text{m} = 10^9 \, \text{nm}\)):
\[ \lambda](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F277ae00b-d056-42ed-a729-386f4fca5973%2F0dd94f4b-298c-46be-a80c-ce0f29f8b91e%2Ffcdj4dk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Understanding Bond Dissociation Energy and Electromagnetic Radiation
**Question:**
> A Cl—Cl bond has a bond-dissociation energy of 243 kJ/mol. What is the longest wavelength of electromagnetic radiation (in nm) which can cleave this bond?
**Choices:**
- ○ 492 nm
- ○ 817 nm
- ○ 136 nm
- ○ 2031 nm
### Explanation
To find the longest wavelength of electromagnetic radiation that can cleave a Cl—Cl bond, we can use the relationship between energy \( E \) and wavelength \( \lambda \). The energy of a single photon is given by the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \( E \) is the energy,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength.
First, convert the bond-dissociation energy from kJ/mol to J/photon.
\[ 243 \, \text{kJ/mol} \times \frac{1000 \, \text{J}}{1 \, \text{kJ}} \times \frac{1 \, \text{mol}}{6.022 \times 10^{23} \, \text{photons}} = 4.037 \times 10^{-19} \, \text{J/photon} \]
Rearrange the equation \( E = \frac{hc}{\lambda} \) to solve for wavelength \( \lambda \):
\[ \lambda = \frac{hc}{E} = \frac{(6.626 \times 10^{-34} \, \text{J} \cdot \text{s})(3 \times 10^8 \, \text{m/s})}{4.037 \times 10^{-19} \, \text{J}} \approx 4.92 \times 10^{-7} \, \text{m} \]
Convert meters to nanometers (\(1 \, \text{m} = 10^9 \, \text{nm}\)):
\[ \lambda
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