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$O_2$ undergoes photochemical dissociation into one normal oxygen atom and one oxygen atom, 1.967 eV more energetic than normal. The dissociation of $O_2$ into two normal atoms of oxygen requires 498 KJ$mol^{-1}$.what is the maximum wavelength effective for photochemical dissociation of $O_2$ ?


1 Answer

$O_2 \rightarrow O_{Normal}+O_{Excited}$
$O_2 \rightarrow O_{Normal}+O_{Normal}$
Energy required for simple dissociation of $O_2$ into two normal atoms
$=498\times 10^3 J/mol$
$=\large\frac{498\times10^3}{6.023\times10^{23}}$ J/Molecule
If one atom is excited state has more energy i.e 1.967 eV
$=1.967\times 1.602\times 10^{-19} J$
Energy required for photochemical dissociation of $O_2$
$=\large\frac{498\times 10^3}{6.023\times 10^{23}} + 1.967\times1.602\times 10^{-19}$
$= 82.68\times 10^{-20} + 31.51\times 10^{-20}$
$= 114.19\times 10^{-20} J$
$\therefore E = \large\frac{hc}{\lambda}$
$114.19\times 10^{-20} = \large\frac{6.625\times10^{-34}\times3.0\times 10^8}{\lambda}$
$\lambda=1740.2\times 10^{-10} m$
$\lambda = 1740.2 A^{\large\circ}$
answered Feb 1, 2014 by sharmaaparna1

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