Problem 48 - Chemistry An Atoms First Approach

Question

X rays have wavelengths on the order of \(1 \times 10^{-10} \mathrm{m}\). Calculate the energy of \(1.0 \times 10^{-10} \mathrm{m}\) X rays in units of kilojoules per mole of X rays. (1 mol X rays \(=6.022 \times 10^{23}\) X rays.) AM radio waves have wavelengths on the order of \(1 \times 10^{4} \mathrm{m}\). Calculate the energy of \(1.0 \times 10^{4} \mathrm{m}\) radio waves in units of kilojoules per mole of radio waves. Consider that the bond energy of a carbon- carbon single bond found in organic compounds is 347 kJ/mol. Would X rays and/or radio waves be able to disrupt organic compounds by breaking carbon- carbon single bonds?

Step-by-Step Solution

Answer

The energy of X rays with a wavelength of \(1.0 \times 10^{-10} \mathrm{m}\) is 119.7 kJ/mol, and the energy of AM radio waves with a wavelength of \(1.0 \times 10^{4} \mathrm{m}\) is 1.197 \(\times 10^{-6}\) kJ/mol. Since both of these energies are smaller than the bond energy of carbon-carbon single bonds (347 kJ/mol), neither X rays nor AM radio waves would be able to disrupt organic compounds by breaking carbon-carbon single bonds.

Step 1: Frequency of X rays

Using the relationship between wavelength and frequency, we can calculate the frequency of X rays: \(\nu_{xray} = \dfrac{c}{\lambda_{xray}} = \dfrac{3.0 \times 10^8 \mathrm{m/s}}{1.0 \times 10^{-10} \mathrm{m}} = 3.0 \times 10^{18} \mathrm{s}^{-1}\)

Step 2: Frequency of AM radio waves

Similarly, we can calculate the frequency of AM radio waves: \(\nu_{radio} = \dfrac{c}{\lambda_{radio}} = \dfrac{3.0 \times 10^8 \mathrm{m/s}}{1.0 \times 10^4 \mathrm{m}} = 3.0 \times 10^4 \mathrm{s}^{-1}\) Step 2: Calculate the energy of X rays and AM radio waves

Step 3: Energy of X rays

Now, using the energy of a photon formula, we can calculate the energy of X rays: \(E_{xray} = h \times \nu_{xray} = (6.626 \times 10^{-34} \mathrm{Js}) \times (3.0 \times 10^{18} \mathrm{s}^{-1}) = 1.988 \times 10^{-15} \mathrm{J}\)

Step 4: Energy of AM radio waves

Similarly, we can calculate the energy of AM radio waves: \(E_{radio} = h \times \nu_{radio} = (6.626 \times 10^{-34} \mathrm{Js}) \times (3.0 \times 10^4 \mathrm{s}^{-1}) = 1.988 \times 10^{-29} \mathrm{J}\) Step 3: Convert the energy to kilojoules per mole

Step 5: Energy of X rays in kJ/mol

We can now convert the energy of X rays to kilojoules per mole using Avogadro's constant: \(E_{xray} \times \dfrac{6.022 \times 10^{23} \,\text{photons}}{1\,\text{mol}} \times \dfrac{1\,\text{kJ}}{10^3\,\text{J}} = 119.7\,\text{kJ/mol}\)

Step 6: Energy of AM radio waves in kJ/mol

Similarly, we can convert the energy of AM radio waves to kilojoules per mole using Avogadro's constant: \(E_{radio} \times \dfrac{6.022 \times 10^{23} \,\text{photons}}{1\,\text{mol}} \times \dfrac{1\,\text{kJ}}{10^3\,\text{J}} = 1.197 \times 10^{-6}\,\text{kJ/mol}\) Step 4: Compare the energies with the bond energy of carbon-carbon single bonds

Step 7: Compare with carbon-carbon single bond energy

The bond energy of a carbon-carbon single bond is 347 kJ/mol. The energy of X rays is 119.7 kJ/mol, and the energy of AM radio waves is 1.197 \(\times 10^{-6}\) kJ/mol. Since the energies of X rays and AM radio waves are both smaller than the bond energy of carbon-carbon single bonds, neither X rays nor AM radio waves would be able to disrupt organic compounds by breaking carbon-carbon single bonds.

Key Concepts

Wavelength-Frequency Relationship

Understanding the relationship between wavelength and frequency is essential for topics in physics and chemistry that involve waves, like X rays. The equation that links these two properties is given by the formula \( u = \frac{c}{\lambda} \) where \( u \) is the frequency, \( c \) is the speed of light in a vacuum (approximately \( 3.0 \times 10^8 \mathrm{m/s} \) ), and \( \lambda \) is the wavelength. This equation shows that the frequency is inversely proportional to the wavelength: as the wavelength decreases, the frequency increases and vice versa. When considering X rays, knowing either the frequency or the wavelength allows us to calculate the other, provided we know the speed of light.

Energy of a Photon Formula

The energy of a photon is calculated using the formula \( E = h \times u \), where \( E \) is the energy of the photon, \( h \) is Planck’s constant (which is \( 6.626 \times 10^{-34} \mathrm{Js} \) ), and \( u \) is the frequency of the photon. This equation shows that the energy of a photon is directly proportional to its frequency: photons with higher frequency, such as X rays, carry more energy than those with lower frequency, such as radio waves. This concept is particularly important when determining the potential effects of electromagnetic radiation on matter, such as its ability to break chemical bonds.

Avogadro's Constant

Avogadro's constant, often represented by \( N_A \), is a fundamental value in chemistry that indicates the number of constituent particles, usually atoms or molecules, that are contained in one mole of a substance. Its value is approximately \( 6.022 \times 10^{23} \) particles per mole. When used in calculations, Avogadro's constant allows us to relate the microscopic world of individual particles to macroscopic quantities that can be measured in the laboratory. For example, when expressing the energy of electromagnetic radiation, such as X rays, per mole, Avogadro's constant enables us to scale the energy of a single photon to an amount of substance that chemists can work with.

Bond Energy of Carbon-Carbon Single Bonds

In chemistry, bond energy is a measure of the strength of a chemical bond. Specifically, for carbon-carbon single bonds, often found in organic compounds, the bond energy is typically around 347 kJ/mol. This value quantifies the amount of energy required to break one mole of bonds in the gas phase. When comparing the energy of electromagnetic radiation with bond energies, we gain insight into the radiation's ability to cause chemical changes, such as breaking these bonds. For example, an electromagnetic wave with energy higher than 347 kJ/mol could potentially disrupt a molecule by breaking its carbon-carbon single bonds, while a photon with less energy would not have the same effect.