Question

it takes 163 kj/mol to break an nitrogen - nitrogen single bond. calculate the maximum wavelength of light for which an nitrogen - nitrogen 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 \(E_{mol}=163\space kJ/mol = 163\times10^{3}\space J/mol\).
Using \(E = \frac{E_{mol}}{N_{A}}\), where \(N_{A}=6.022\times 10^{23}\space mol^{-1}\).
\(E=\frac{163\times 10^{3}}{6.022\times 10^{23}}\space J\)
\(E = 2.707\times 10^{-19}\space J\)

Step2: Use the formula \(E=\frac{hc}{\lambda}\) to find \(\lambda\)

We know that \(h = 6.626\times 10^{-34}\space J\cdot s\) and \(c=3\times 10^{8}\space m/s\)
From \(E=\frac{hc}{\lambda}\), we can solve for \(\lambda\): \(\lambda=\frac{hc}{E}\)
Substitute the values:
\(\lambda=\frac{6.626\times 10^{-34}\times3\times 10^{8}}{2.707\times 10^{-19}}\space m\)
\(\lambda=\frac{19.878\times 10^{-26}}{2.707\times 10^{-19}}\space m\)
\(\lambda = 7.34\times 10^{-7}\space m\)

Step3: Convert meters to nanometers

Since \(1\space m = 10^{9}\space nm\)
\(\lambda=7.34\times 10^{-7}\times10^{9}\space nm\)
\(\lambda = 734\space nm\)

Answer:

\(734\space nm\)