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In second order, the time dependent rate coefficient is the time derivative of the inverse of the survival function[insert citation]  \begin{equation}  \frac{dS(t)^{-1}}{dt}=k(t)   \end{equation}  Now consider a population of some species A decaying into B following the reaction $A+A\rightarrow B$. The second order integrated rate law is [cite]  \begin{equation}  [A_t]=\frac{[A_0]}{1+\omega t[A_0]}  \end{equation}  The second order survival function is  \begin{equation}  S(t)=\frac{[A_t]}{[A_0]}=\frac{1}{1+\omega t[A_0]}  \end{equation}  The time dependent rate coefficient for a second order reaction is  \begin{equation}  k(t)=\left(\frac{dS(t)^{-1}}{dt}\right)=\frac{d}{dt}\left[1+\omega t[A_0]\right]=\omega[A_0]  \end{equation}  S(t) has been changed to fit a second order model of irreversible decay. From this definition of k(t), we define a statistical distance . The statistical distance represents the distance between two different probability distributions, which can be applied to survival functions and rate coefficients.[cite] Integrating the arc length of the survival curve , $\frac{1}{S(t)}$, gives the statistical length.  \begin{equation}  \mathcal{L}(\Delta{t})=\int_{t_i}^{t_f}k(t)dt  \end{equation}  Which is equal to $\frac{1}{S(t)}\big|_{S_(t_f)}^{S_(t_i)}$, which measures the cumulative rate coefficient, same as in first order irreversible decay. As seen in first order, the statistical length is also dependent on the time interval, with the statistical length being infinite in an infinite time interval. To form an inequality, another quantity called the Fisher divergence is calculated, which is defined as [cite]  \begin{equation}  \frac{\mathcal{J}(\Delta{t})}{\Delta{t}}=\int_{t_i}^{t_f}k(t)^{2}dt  \end{equation}  It has been shown that in first order kinetics the inequality between the statistical length squared and Fisher divergence determines when a rate coefficient is constant, which is only when the inequality turns into an equality.[cite] A similar inequality is found in second order kinetics. The inequality between statistical length squared and Fisher divergence is  \begin{equation}  \mathcal{L}(\Delta{t})^2\leq \mathcal{J}(\Delta{t})  \end{equation}  Which also means that  \begin{equation}  \mathcal{J}(\Delta{t})-\mathcal{L}(\Delta{t})^2\geq 0  \end{equation}  \begin{equation}  \Delta{t}\int_{t_i}^{t_f}k(t)^{2}dt-\left[{\int_{t_i}}^{t_f}k(t)dt\right]^{2}dt\geq0  \end{equation}  Putting in the time dependent rate coefficient for a second order irreversible decay, the inequality becomes  \begin{equation}  \omega^2[A_0]^2\Delta{t}^2-\left(\omega[A_0]\Delta{t}\right)^2\geq0  \end{equation}  This result is very similar to that of first order irreversible decay($\omega^2\Delta{t}^2-\left(\omega\Delta{t}\right)^2\geq0$), the only difference being a dependence on the initial concentration of the reactant. This initial concentration dependence serves to cancel the concentration units in the second order rate coefficient, making the statistical length and Fisher divergence dimensionless. When there is a time independent rate coefficient and there is no static disorder, the equality holds $\omega^2[A_0]^2\Delta{t}^2=\left(\omega[A_0]\Delta{t}\right)^2$.