Question

it takes 242. kj/mol to break a chlorine-chlorine single bond. calculate the maximum wavelength of light for which a chlorine-chlorine single bond could be broken by absorbing a single photon. be sure your answer has the correct number of significant digits.

Explanation:

Step1: Recall the energy - photon - wavelength relationships

The energy of a photon is given by \(E = h
u\), and the relationship between frequency \(
u\) and wavelength \(\lambda\) is \(c=
u\lambda\) (where \(c = 3.00\times10^{8}\space m/s\) is the speed of light, \(h = 6.626\times10^{-34}\space J\cdot s\) is Planck's constant). So we can combine these two formulas to get \(E=\frac{hc}{\lambda}\).

First, we need to find the energy per photon. The bond energy is given as \(242\space kJ/mol\). We know that 1 mole of photons contains \(N_A=6.022\times 10^{23}\) photons (Avogadro's number). So we convert the bond energy from \(kJ/mol\) to \(J/photon\).

Step2: Convert bond energy to energy per photon

The bond energy \(E_{molar}=242\space kJ/mol = 242\times10^{3}\space J/mol\).

The energy per photon \(E=\frac{E_{molar}}{N_A}\)

\(E=\frac{242\times 10^{3}\space J/mol}{6.022\times 10^{23}\space mol^{-1}}\)

\(E=\frac{242000}{6.022\times 10^{23}}\space J\approx4.0186\times 10^{-19}\space J\)

Step3: Solve for wavelength \(\lambda\)

From \(E = \frac{hc}{\lambda}\), we can re - arrange to get \(\lambda=\frac{hc}{E}\)

Substitute \(h = 6.626\times 10^{-34}\space J\cdot s\), \(c = 3.00\times 10^{8}\space m/s\) and \(E = 4.0186\times 10^{-19}\space J\) into the formula:

\(\lambda=\frac{6.626\times 10^{-34}\space J\cdot s\times3.00\times 10^{8}\space m/s}{4.0186\times 10^{-19}\space J}\)

First, calculate the numerator: \(6.626\times 10^{-34}\times3.00\times 10^{8}=1.9878\times 10^{-25}\space J\cdot m\)

Then, divide by the energy: \(\lambda=\frac{1.9878\times 10^{-25}}{4.0186\times 10^{-19}}\space m\approx4.946\times 10^{-7}\space m\)

Step4: Convert meters to nanometers

Since \(1\space m = 10^{9}\space nm\), we multiply the wavelength in meters by \(10^{9}\) to get the wavelength in nanometers.

\(\lambda = 4.946\times 10^{-7}\space m\times10^{9}\space nm/m = 495\space nm\) (rounded to three significant figures, since the given bond energy has three significant figures)

Answer:

\(495\)