QuestionAugust 4, 2025

The radical is generated when chlorine radicals react with ozone. Calculate the maximum wavelength (in nanometers) required to break a Cl-Cl bond in Cl_(2) (the precursor to chlorine radicals)Round your answer to 3 significant digits. Note: Reference the Bond energies and Fundamental Constants tables for additional information. square nm

The radical is generated when chlorine radicals react with ozone. Calculate the maximum wavelength (in nanometers) required to break a Cl-Cl bond in Cl_(2) (the precursor to chlorine radicals)Round your answer to 3 significant digits. Note: Reference the Bond energies and Fundamental Constants tables for additional information. square nm
The radical is generated when chlorine radicals react with ozone. Calculate the maximum wavelength (in nanometers) required to break a Cl-Cl bond in Cl_(2) (the precursor to chlorine radicals)Round your answer to 3 significant digits. Note: Reference the Bond energies and Fundamental Constants tables for additional information. square nm

Solution
4.0(224 votes)

Answer

493 nm Explanation 1. Identify the bond energy The bond energy for a Cl-Cl bond is approximately 243 kJ/mol. 2. Convert bond energy to energy per molecule Use E = \frac{\text{Bond Energy}}{N_A}, where N_A = 6.022 \times 10^{23} \text{ mol}^{-1}. E = \frac{243 \times 10^3 \text{ J/mol}}{6.022 \times 10^{23} \text{ mol}^{-1}} = 4.034 \times 10^{-19} \text{ J/molecule} 3. Calculate wavelength using energy formula Use E = \frac{hc}{\lambda}, rearrange to \lambda = \frac{hc}{E}. h = 6.626 \times 10^{-34} \text{ J s}, c = 3.00 \times 10^8 \text{ m/s}. \lambda = \frac{(6.626 \times 10^{-34} \text{ J s})(3.00 \times 10^8 \text{ m/s})}{4.034 \times 10^{-19} \text{ J}} = 4.93 \times 10^{-7} \text{ m} 4. Convert meters to nanometers 1 \text{ m} = 10^9 \text{ nm}. \lambda = 4.93 \times 10^{-7} \text{ m} \times 10^9 \text{ nm/m} = 493 \text{ nm}

Explanation

1. Identify the bond energy<br /> The bond energy for a Cl-Cl bond is approximately 243 kJ/mol.<br />2. Convert bond energy to energy per molecule<br /> Use $E = \frac{\text{Bond Energy}}{N_A}$, where $N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$.<br /> $E = \frac{243 \times 10^3 \text{ J/mol}}{6.022 \times 10^{23} \text{ mol}^{-1}} = 4.034 \times 10^{-19} \text{ J/molecule}$<br />3. Calculate wavelength using energy formula<br /> Use $E = \frac{hc}{\lambda}$, rearrange to $\lambda = \frac{hc}{E}$.<br /> $h = 6.626 \times 10^{-34} \text{ J s}$, $c = 3.00 \times 10^8 \text{ m/s}$.<br /> $\lambda = \frac{(6.626 \times 10^{-34} \text{ J s})(3.00 \times 10^8 \text{ m/s})}{4.034 \times 10^{-19} \text{ J}} = 4.93 \times 10^{-7} \text{ m}$<br />4. Convert meters to nanometers<br /> $1 \text{ m} = 10^9 \text{ nm}$.<br /> $\lambda = 4.93 \times 10^{-7} \text{ m} \times 10^9 \text{ nm/m} = 493 \text{ nm}$
Click to rate:

Similar Questions