Question

it takes 338 kj/mol 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.
be sure your answer has the correct number of significant digits.

nm

Explanation:

Step1: Calculate energy per photon

Given energy per mole \(E_{mol}=338\space kJ/mol = 338\times10^{3}\space J/mol\).
Using \(E = \frac{E_{mol}}{N_{A}}\), where \(N_{A}=6.022\times 10^{23}\space mol^{-1}\).
\(E=\frac{338\times 10^{3}\space J/mol}{6.022\times 10^{23}\space mol^{-1}}\)
\(E\approx5.613\times 10^{-19}\space J\)

Step2: Use the formula \(E = h

u=\frac{hc}{\lambda}\) to find \(\lambda\)
We know \(h = 6.626\times 10^{-34}\space J\cdot s\) and \(c = 3\times 10^{8}\space m/s\).
From \(\lambda=\frac{hc}{E}\), substituting the values:
\(\lambda=\frac{6.626\times 10^{-34}\space J\cdot s\times3\times 10^{8}\space m/s}{5.613\times 10^{-19}\space J}\)
\(\lambda=\frac{19.878\times 10^{-26}}{5.613\times 10^{-19}}\space m\)
\(\lambda\approx 3.54\times 10^{-7}\space m\)

Step3: Convert meters to nanometers

Since \(1\space m = 10^{9}\space nm\)
\(\lambda=3.54\times 10^{-7}\times10^{9}\space nm\)
\(\lambda = 354\space nm\)

Answer:

\(354\space nm\)