Question

It takes to break an carbon-chlorine single bond. Calculate the maximum wavelength of light for which an carbon-chlorine single bond could be broken by absorbing a single photon. Round your answer to significant digits.

Answer 1

Answer:

It takes approximately $5.43\times 10^{-19}\; \rm J$ of energy to break one $\rm C-Cl$ single bond.

The maximum wavelength of a photon that can break one such bond is approximately $3.66\times 10^{-7}\; \rm m$ (in vacuum.) That's the same as $3.66 \times 10^{2}\; \rm nm$ (rounded to three significant figures.)

Explanation:

Energy per bond

The standard bond enthalpy of $\rm C-Cl$ single bonds is approximately $\rm 327\; \rm kJ \cdot mol^{-1}$ (note that the exact value can varies across sources.) In other words, it would take approximately $327\; \rm kJ$ of energy to break one mole of these bonds.

The Avogadro Constant $N_A \approx 6.023\times 10^{23}\; \rm mol^{-1}$ gives the number of $\rm C-Cl$ bonds in one mole of these bonds. Based on these information, calculate the energy of one such bond:

$\begin{aligned}& E(\text{one \mathrm{C-Cl} bond}) \\ &= \frac{E(\text{one mole of \mathrm{C-Cl} bonds})}{N_A} = \frac{327\; \rm kJ\cdot mol^{-1}}{6.023\times 10^{23}\; \rm mol^{-1}} \\ &\approx 5.429\times 10^{-22}\; \rm kJ = 5.429\times 10^{-19}\; \rm J \end{aligned}$.

Therefore, it would take approximately $5.43\times 10^{-19}\; \rm J$ of energy to break one $\rm C-Cl$ single bond.

Minimum frequency and maximum wavelength

The Einstein-Planck Relation relates the frequency $f$ of a photon to its energy $E$:

$E = h \cdot f$.

The $h$ here represents the Planck Constant:

$h \approx 6.63 \times 10^{-34}\; \rm J \cdot s$.

A photon that can break one $\rm C-Cl$ single bond should have more than $5.43\times 10^{-19}\; \rm J$ of energy. Apply the Einstein-Planck Relation to find the frequency of a photon with exactly that much energy:

$\begin{aligned}f &= \frac{E}{h}\\ &\approx \frac{5.43\times 10^{-19}\; \rm J}{6.63 \times 10^{-34}\; \rm J\cdot s} \\ &\approx 8.19 \times 10^{14}\; \rm s^{-1} = 8.19 \times 10^{14}\; \rm Hz\end{aligned}$.

What would be the wavelength $\lambda$ of a photon with a frequency of approximately $8.19 \times 10^{14}\; \rm Hz$? The exact answer to that depends on the medium that this photon is travelling through. To be precise, the exact answer depends on the speed of light in that medium:

$\displaystyle \lambda = \frac{(\text{speed of light})}{f}$.

In vacuum, the speed of light is $c \approx 2.998\times 10^{8}\; \rm m \cdot s^{-1}$. Therefore, the wavelength of that $8.19 \times 10^{14}\; \rm Hz$ photon in vacuum would be:

$\begin{aligned} \lambda &= \frac{c}{f} \\ & \approx \frac{2.998\times 10^{8}\; \rm m \cdot s^{-1}}{8.19\times 10^{14}\; \rm s^{-1}} \\ &\approx 3.66 \times 10^{-7}\; \rm m = 3.66 \times 10^{2}\; \rm nm\end{aligned}$.

(Side note: that wavelength corresponds to a photon in the ultraviolet region of the electromagnetic spectrum.)