Photodissociation of ozone (O3) can be described as producing an oxygen atom with heat of formation 247.5 kJ/mol. However, in reality photodissociation yields an oxygen atom that is temporarily in an excited state (O*) with a standard enthalpy of 438 kJ/mol. O3 + hv → O2+ O^+
H^o: (142.9) (0) 438(KJ/mol)

Required:
What maximum wavelength could cause this photodissociation?

0.2193 μm

Explanation:

The reaction showing the Photodissociation of ozone (O3) is given below as:

           O₃         +        hv   >      O₂         +       O⁺  

H°        (142.9)                                                         (0)                  (438kJ/mol).

The first thing to do here is to determine the change in the enthalpy of the total reaction, this can be done by subtracting the change in the enthalpy of the reactant from the change in enthalpy in the product. Hence, we have:

ΔH° = [438 kJ/mol + 247.5 kJ/mol] - (142.9) = 542.6 KJ/mol.

This value, that is 542.6 KJ/mol will then be used in the determination of the value for the maximum wavelength that could cause this photodissociation.

Therefore, the maximum wavelength could cause this photodissociation ≤ h × c/ E = [ 1.199 × 10⁻⁴]/ 542.6 = 2.193 × 10⁻⁷ =  0.2193 μm

there are no statements provided but here is why:

sulfur belongs to group 16 because it can be found in nature whether it is in a free or combined state.

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