Bond dissociation energy of a molecule is 300 kilo joule per mole the number of moles of photon of wavelength 6625 angstrom required to disassociate 3 molecules of a b molecules
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First convert 300kj/mol to joules ie
(300*10^3)/(6.022*10^23)=0.498*10^-18------[1]
Now we need to find the no of moles photons for 3 moles of AB molecule, so
3*E=nhc/λ {where n=no of moles of photons; E=[1]=0.498*10^-18
and λ=6625*10^-10 m}
=> 3*0.498*10^-18=(n*6.625*10^-34*3*10^8)/6625*10^-10
Solving the above equation we obtain
n=4.98≈5 mol
(300*10^3)/(6.022*10^23)=0.498*10^-18------[1]
Now we need to find the no of moles photons for 3 moles of AB molecule, so
3*E=nhc/λ {where n=no of moles of photons; E=[1]=0.498*10^-18
and λ=6625*10^-10 m}
=> 3*0.498*10^-18=(n*6.625*10^-34*3*10^8)/6625*10^-10
Solving the above equation we obtain
n=4.98≈5 mol
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