The reactant 1,3-cyclohexadiene can be photochemically converted to cis hexatriene. In an experiment, 2.5 mmol of cyclohexadiene are converted to cis-hexatriene when irradiated with 100. W of 280. nm light for 27.0 s. All of the light is absorbed by the sample. What is the overall quantum

Question

The reactant 1,3-cyclohexadiene can be photochemically converted to cis-hexatriene. In an experiment, 2.5 mmol of cyclohexadiene are converted to cis-hexatriene when irradiated with 100. W of 280. nm light for 27.0 s. All of the light is absorbed by the sample. What is the overall quantum yield for this photochemical process?

Accepted Answer

First, the total photon energy absorbed by the sample, [latex]E_{abs}[/latex],is

[latex]E_{abs}=\left( power \right)\Delta t=\left( 100.Js^{-1} \right)\left( 27.0 s \right)=2.70 ×10^{3} J[/latex]

Next, the photon energy at 280. nm is

[latex]E_{ph}=\frac{hc}{\lambda}=\frac{\left( 6.626 ×10^{-34} J s \right)\left( 2.998 ×10^{8} ms^{-1} \right)}{2.80×10^{-7} m}=7.10×10^{-19} J[/latex]

The total number of photons absorbed by the sample is therefore

[latex]\frac{E_{abs}}{E_{ph}}=\frac{2.70 ×10^{3}J}{7.10 ×10^{-19}J photon^{-1}}=3.80×10^{21} photons[/latex]

Dividing this result by Avogadro’s number results in [latex]6.31×10^{-3}[/latex] Einsteins or moles of photons. Therefore, the overall quantum yield is

[latex]\phi=\frac{moles_{react}}{moles_{photon}}=\frac{2.50×10^{-3} mol}{6.31×10^{-3} mol}=0.396\approx 0.40[/latex]

Question 19.6: The reactant 1,3-cyclohexadiene can be photochemically conve...

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