The reactant 1,3-cyclohexadiene can be photochemically converted to cis-hexatriene. In an experiment, $2.5 \mathrm{mmol}$ of cyclohexadiene are converted to cis-hexatriene when irradiated with 100. W of 280. nm light for $27.0 \mathrm{~s}$. All of the light is absorbed by the sample. What is the overall quantum yield for this photochemical process?
First, the total photon energy absorbed by the sample, $E_{a b s}$, is
$$
E_{a b s}=(\text { power }) \Delta t=\left(100 . \mathrm{Js}^{-1}\right)(27.0 \mathrm{~s})=2.70 \times 10^3 \mathrm{~J}
$$
Next, the photon energy at $280 . \mathrm{nm}$ is
$$
\mathrm{E}_{p h}=\frac{h c}{\lambda}=\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(2.998 \times 10^8 \mathrm{~ms}^{-1}\right)}{2.80 \times 10^{-7} \mathrm{~m}}=7.10 \times 10^{-19} \mathrm{~J}
$$
The total number of photons absorbed by the sample is therefore
$$
\frac{E_{a b s}}{E_{p h}}=\frac{2.70 \times 10^3 \mathrm{~J}}{7.10 \times 10^{-19} \mathrm{~J} \text { photon }^{-1}}=3.80 \times 10^{21} \text { photons }
$$
Dividing this result by Avogadro's number results in $6.31 \times 10^{-3}$ Einsteins or moles of photons. Therefore, the overall quantum yield is
$$
\phi=\frac{\text { moles }_{\text {react }}}{\text { moles }_{\text {photon }}}=\frac{2.50 \times 10^{-3} \mathrm{~mol}}{6.31 \times 10^{-3} \mathrm{~mol}}=0.396 \approx 0.40
$$