= Solution
For a <first-order reaction> with no appreciable reverse contribution in this decay experiment,
$$
\frac{dc}{dt}=-kc,\qquad c(t)=c_0e^{-kt}.
$$
With fixed path length and attenuation coefficient, and absorbance attributable to ONOO after blank/background correction, the <Beer-Lambert law> gives $A(t)=\varepsilon lc_0e^{-kt}$. Therefore
$$
\boxed{\ln A(t)=\ln(\varepsilon lc_0)-kt.}
$$
\b[The 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.
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