Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 4 / a

The decadic Beer-Lambert law can be writtenHere $l$ is the path length in metres, $c$ is concentration in $\mathrm{mol}{\mathrm{L}}^{-1}$ or M, and $\epsilon$ is the decadic molar attenuation coefficient in ${\mathrm{M}}^{-1}{\mathrm{m}}^{-1}$. The intensity ratio and the exponent are dimensionless. Equivalently $I=I_0{e}^{-\kappa lc}$ with $\kappa =(\ln 10)\epsilon$ in the same units. Defining the logarithm convention avoids a missing factor of $\ln 10$.

The absorbance is $\boxed{A={\log }_{10}(I_0/I)=-{\log }_{10}(I/I_0)}$. It is dimensionless. It differs from the absorbed fraction $1-I/I_0$.

For a first-order reaction with no appreciable reverse contribution in this decay experiment,With fixed path length and attenuation coefficient, and absorbance attributable to ONOO after blank/background correction, the Beer-Lambert law gives $A(t)=\epsilon l{c}_0{e}^{-kt}$. ThereforeThe gradient is $-k$, in ${\mathrm{s}}^{-1}$ when time is measured in seconds. Constant illumination wavelength, homogeneous sampling and negligible product absorbance at that wavelength are required for this direct interpretation.