The reaction between `H_(2)` and `Br^(2)` to form `HBr` in presence of light is initiated by the photo decomposition of `Br_(2)` into free Br atoms (free radicals) by absorption of light. The bond dissociation energy of `Br_(2)` is `192kJ//"mole"`.What is the longest wavelength of the photon that would initiate the reaction?

The reaction between `H_(2)` and `Br^(2)` to form `HBr` in presence of

1.	The reaction between `H_(2)` and `Br^(2)` to form `HBr` in presence of light is initiated by the photo decomposition of `Br_(2)` into free Br atoms (free radicals) by absorption of light. The bond dissociation energy of `Br_(2)` is `192kJ//"mole"`.What is the longest wavelength of the photon that would initiate the reaction?
Answer» Correct Answer - `6235A^(@)` `H_(2)+Br_(2)overset(hv)rarr2HBr` `Br_(2)overset(hv)rarr2Br` `BE=192kJ//"mole"` `(192)/(93.368)eV//"mole"=(hc)/(hv)` or `(192)/(96.368)=(1240)/(lamda(nm))` `lambda=6235Å`

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