Photochemical Rate Law

The photochemical decomposition reaction occurs by the absorption of a quantum energy hν;

\[\displaystyle A\underset{{}}{\overset{{hv}}{\longrightarrow}}products\]

If a beam of intensity I₀ is the Incident on the reaction width b and cross-sectional area S, then the number of Einstein absorbed per second is (I₀-I)

Here, I = intensity of transmitted radiation

The intensity of radiation is expressed in einsteins cm^-2S^-1

According to Beer-Lambert law,

\[\displaystyle \ln \frac{I}{{{{I}_{0}}}}=-ab\left[ A \right]---(1)\]

\[\displaystyle I={{I}_{0}}{{e}^{{-ab\left[ A \right]}}}---(2)\]

Number of einsteins absorbed per second =S(I₀-I)

Now, substitute the value I from equation (2) we obtain,

\[\displaystyle S\left( {{{I}_{0}}-I} \right)=S{{I}_{0}}\left( {1-{{e}^{{^{{-ab\left[ A \right]}}}}}} \right)\]

\[\displaystyle =S{{I}_{0}}\left( {ab\left[ A \right]} \right)\]

e^-x=1-x when x is very small

The number of einsteins absorbed per second = SI₀ab[A]

The quantum yield,

\[\displaystyle \phi =\frac{{\text{Number of moles of A decomposed per second}}}{{\text{Number of einsteins absorbed per second }}}\]

\[\displaystyle =\frac{{\text{Number of moles of A decomposed per second}}}{{\text{S}{{I}_{0}}\text{ab}\left[ A \right]\text{ }}}---(3)\]

Sb is the volume of the reaction cell; the number of moles of A decomposed per second divided by the cell volume gives the concentration of A in moles per unit volume;

\[\displaystyle \phi =\frac{{\text{change in }\left[ A \right]\text{per Second }}}{{{{I}_{0}}a\left[ A \right]}}\]

\[\displaystyle =\frac{{\frac{{-d\left[ A \right]}}{{dt}}}}{{{{I}_{0}}a\left[ A \right]}}---(4)\]

\[\displaystyle -\frac{{d\left[ A \right]}}{{dt}}=\phi {{I}_{0}}a\left[ A \right]---(5)\]

It is the potential rate law, the photochemical decomposition of A in the first order in A.