Question

it takes 463. kj/mol to break an oxygen-hydrogen single bond. calculate the maximum wavelength of light for which an oxygen-hydrogen 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 the energy per photon

We know that \(1\space mol = 6.022\times10^{23}\) photons.
The energy per mole \(E_{mol}=463\space kJ/mol = 463\times10^{3}\space J/mol\)
The energy per photon \(E=\frac{E_{mol}}{N_{A}}\), where \(N_{A} = 6.022\times 10^{23}\space mol^{-1}\)
\(E=\frac{463\times 10^{3}\space J/mol}{6.022\times 10^{23}\space mol^{-1}}\)
\(E\approx7.69\times 10^{-19}\space J\)

Step2: Use the formula \(E = h

u=\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 \(h = 6.626\times 10^{-34}\space J\cdot s\), \(c = 3\times 10^{8}\space m/s\) and \(E = 7.69\times 10^{-19}\space J\)
\(\lambda=\frac{6.626\times 10^{-34}\times3\times 10^{8}}{7.69\times 10^{-19}}\space m\)
\(\lambda=\frac{19.878\times 10^{-26}}{7.69\times 10^{-19}}\space m\)
\(\lambda\approx2.58\times 10^{-7}\space m\)
Convert to nanometers: \(1\space m = 10^{9}\space nm\), so \(\lambda=2.58\times 10^{-7}\times10^{9}\space nm = 258\space nm\)

Answer:

\(258\space nm\)