Many chemical reactions can be brought about by the absorption of light of suitable wavelength. Such reactions are known as photochemical reactions. In a photochemical reaction, only the first step is photochemical and the rest are thermal. The energy of the absorbed photon is used for the chemical conversions. For the photochemical combination of hydrogen and chlorine to form hydrogen chloride, the following mechanism was suggested.
$C l_{2} (h v) - - \to 2 C l^{∙}$ . . . . . . (i)
$C l^{∙} + H_{2} k_{1} \to H C l + H^{∙}$ . . . . . . (ii)
$H^{∙} + C l_{2} k_{2} \to H C l + C l^{∙}$ . . . . . . (iii)
$2 C l^{∙} k_{3} \to C l_{2}$ . . . . . . (iv)
In the photochemical reaction of $H_{2}$ and $C l_{2}, C l_{2}$ absorbs at wavelength of 488 nm.When the $C l_{2}$ gas is irradiated with a 60 watt lamp that emits radiation with 80% efficiency, 4.1 mmol of $H C l$ were formed in 10 sec. Calculate the quantum yield for the reaction. (write the value to the nearest integer)

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Solution

$H_{2} + C l_{2} \to 2 H C l$
Energy of one photon $E = \frac{h c}{λ} = \frac{6.626 \times 10^{- 34} \times 3 \times 10^{8}}{488 \times 10^{- 9}}$
$E = 0.040733 \times 10^{- 17} J$
Energy of one mol of photon $= 0.040733 \times 10^{- 17} \times 6.023 \times 10^{23} = 0.24533 \times 10^{6} J / m o l = 2.4533 \times 10^{5} J / m o l$
Energy of one mmol of photon $= 245.33 J / m m o l$
Power of 60 watt corresponds to 60 J per second. For 10 sec, 600 J of energy is emitted. This corresponds to
$\frac{600}{245.33} = 2.4456$ mmol of photons.
For 80 % efficiency, the number of mmol of photon $= \frac{2.4456 \times 80}{100} = 1.9565 m m o l$
1 mmol of photon should theoretically give 2 mmol of $H C l$
Quantum yield $= \frac{4.1}{2 \times 1.9565} = 1.05$

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