Question

The energy from radiation can be used to cause the rupture of chemical bonds. A minimum energy of 242kJ/mol is required to break the chlorine–chlorine bond in Cl2. What is the longest wavelength of radiation that possesses the necessary energy to break the bond? What type of electromagnetic radiation is this?

Answer 1

Answer:

The longest wavelength of radiation that passesses the necessary energy for breaking the Cl- Cl bond (in Cl2) is approximately 494.2 nm, which corresponds to the visible spectrum.

Explanation:

In order to answer this question we need to recall that the energy of a photon is given by:

E = hc/lambda,   where

E = energy

h = Planck's constant

c = speed of light in vacuum

lambda = associated photon wavelength

In order to perform the calculations, first we need to change the units of 242kJ/mol to J. For doing this, we to divide by Avogadro's number and multiply by a 1000:

242kJ/mol = (242kJ/mol)*(1mol/6.022x10^23 particles)*(1000J/1kJ)= 4.0186x10^-19 J

Now, we simply solve for lambda and substitute the appropriate values in the energy equation:

lambda = hc/E = (6.626x10^-34 J s)*(3x10^8 m/s)/(4.0186x10^-19 J) = (1.986x10^-25 J m)/(4.0186x10^-19 J) = 4.942x10^-7 m = 494.2x10^-9 m = 494.2 nm

Therefore, the wavelength for a photon to break the Cl-Cl bond in a Cl2 molecule should be 494.2 nm at most, which corresponds to the visible spectrum (The visible spectrum includes wavelengths between 400 nm and 750 nm).