Answer to Question #62720 in Physical Chemistry for vinu

Question

In a photochemical decomposition of acetone using 313 nm light, 7.57  10
6
mol of
carbon monoxide is formed in 20 minutes. If the light absorbed corresponds to
2.41  10
3
J s
1, calculate the quantum efficiency for the formation for carbon monoxide

Answer 1

The quantum efficiency or quantum yield here should be calculated as

fi = (Number of molecules produced) / (Number of photons absorbed)

N(mol) - number of molecules produced
N(phot) - number of photons absorbed
Both of them will be in mols.

N(phot) = Energy of absorbed light / (Energy of the single photon * Avogadro number)

Energy of the single photon, E(phot) = h*c/lambda = 6.62*10^(-34) * 3*10^(8) / 313*10^(-9) = 6.34*10^(-19) J

N(phot) = 2.41*10^(-3)*20*60 / ( 6.34*10^(-19) * 6.02*10^(23) ) = 7.58*10(-8) mol

Therefore, quantum yield is

fi = 7.57*10^(-6) / 7.58*10^(-8) = 100 (or 10000%)