Chemistry, asked by Jessica3143, 1 year ago

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

Answers

Answered by sidtintin
62
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
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