Respuesta :

Demiurgos
I attached the full question.
We need to calculate the enthalpy change first: 
[tex]\Delta H=$(247kJ)-(142.3kJ)=+105.2kJ [/tex]
Keep in mind this is the energy per mole, we need energy per atom:
[tex]E=\frac{105200}{6.02 \cdot 10^{23}} [/tex]
We can see that this reaction needs the energy to get started (enthalpy change is positive).
The energy of a photon is given with this formula:
[tex]E=\frac{hc}{\lambda}[/tex]
We need this photon to have the same energy (or higher) as our enthalpy change:
[tex]105.2=\frac{6.62\cdot 10^{-34}\cdot 3\cdot 10^ 8}{\lambda}\\ \lambda=\frac{6.62\cdot 10^{-34}\cdot 3\cdot 10^ 8}{105.2/(6.02\cdot 10^{23})}
\\\lambda=1.138\mu m[/tex]
This wavelenght is clasified as near infrared.


Ver imagen Demiurgos

The maximum wavelength a photon which can dissociate Ozone atom into molecular oxygen and atomic oxygen is 1138 nm.

The complete question is what is the maximum wavelengtha photon have to split an Ozone atom into molecular oxygen and atomic oxygen?

The reaction is

[tex]\rm \bold { O_3\rightarrow O_2+ O}[/tex]

The Enthalpy of the reaction,

[tex]\rm \bold{ \Delta H= +105.2kJ/mol}[/tex]

The energy to split one molecule is,

[tex]\rm \bold{ \Delta H = \frac{105.02}{6.02\times 10^2^3} }[/tex]

The energy of photon is,

[tex]\rm \bold{ E= \frac{hc}{\lambda } }[/tex]

The energy of photon will be equal to the enthalpy change,

[tex]\rm \bold{ E = \Delta H}\\\\\rm \bold{\frac{ (6.626\times 10^-^3^4 J)(2.99 \times 10^8 m / s)}{\lambda} = \frac{105.02}{6.02\times 10^2^3} } }\\\\\rm \bold{ \lambda = 1138 nm }[/tex]

In electromagnetic spectrum 1138 nm comes in Infrared region.

Hence, we can conclude that The maximum wavelength a photon which can dissociate Ozone atom into molecular oxygen and atomic oxygen is 1138 nm.

To know more about photon energy, refer to the link:

https://brainly.com/question/2393994?referrer=searchResults

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