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The second form of the inequality is the most useful, as the difference in $\mathcal{J}(\Delta t)$ and $\mathcal{L}(\Delta t)^2$ measures the variation in the rate coefficient. When the difference between these two quantities is zero, the rate coefficient is constant.
This inequality is derived from the first order rate law and survival function. In traditional kinetics, irreversible decay is only dependent on one rate coefficient, $\omega$, and the mechanism of the reaction. The rate law of a first order reaction of A irreversibly decaying into B is
\begin{equation}
\frac{-dC_A(t)}{dt}=\omega[A] \frac{-dC_A(t)}{dt}=\omega[C_A]
\end{equation}
The time dependent rate coefficient, k(t), is determined by integrating the rate law of the reaction and forming a survival function from the integrated rate law. The survival function is simply a measure of a the cocentration of species at some time compared to its initital concentration.
\begin{equation}
S(t) =
\frac{[A]_t}{[A]_0} \frac{C_A(t)}{[C_A(0)}
\end{equation}
From the survival function, the time dependent rate coefficient is determined by taking various time derivatives of the survival function, depending on the total order of reaction. For first order irreversible decay reactions, the time dependent rate coefficient is the negative time derivative of the natural log of the survival function[insert citation]
\begin{equation}
...
In order to determine the time dependent rate coefficient in higher order reactions, it is useful to use the survival function, but is not necessary. The survival function of any reaction involving any number of the same molecule can be derived using the integrated rate laws of reactions. For example, the $n^{th}$ order integrated rate law is
\begin{equation}
\frac{1}{[A_t]^{n-1}}=\frac{1}{[A_0]^{n-1}}+(n-1)\omega \frac{1}{[C_A(t)]^{n-1}}=\frac{1}{[C_A(0)]^{n-1}}+(n-1)\omega t
\end{equation}
From the integrated rate law, we get the survival function
\begin{equation}
S(t)=\frac{[A_t]}{[A_0]}=\sqrt[n-1]{\frac{1}{1+(n-1)\omega S(t)=\frac{[C_A(t)}{[C_A(0)}=\sqrt[n-1]{\frac{1}{1+(n-1)\omega t[A_0]^{n-1}}}
\end{equation}
From the survival function, the time dependent rate coefficient is
\begin{equation}
...
\end{equation}
Taking our definitions of the integrated rate law, survival function, and time dependent rate coefficient, we are able to apply them to to second order irreversible decay. The integrated rate law in second order is
\begin{equation}
[A_t]=\frac{[A_0]}{1+\omega t[A_0]} [[C_A(t)]=\frac{[[C_A(0)]}{1+\omega t[[C_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]} S(t)=\frac{[C_A(t)}{[C_A(0)}=\frac{1}{1+\omega t[[C_A(0)]}
\end{equation}
From this survival function, taking the inverse of the survival function and then the time derivative yields the time dependent rate coefficient.
\begin{equation}
...
\end{equation}
When it comes to mixed second order reactions, the rate law does not allow a survival function to be obtained for the process because the rate law only allows the survival function of one reactant to be looked at. The rate law of a mixed second order reaction is
\begin{equation}
\int\frac{dx}{([A_0]-x)([B_0]-x)}=k(t) \int\frac{dx}{([C_A(0)]-x)([C_B(0)-x)}=k(t)
\end{equation}
Taking the time derivative of the left hand side gives
\begin{equation}
\int\frac{\frac{dx}{dt}}{{([A_0]-x)([B_0]-x)}}dt=k(t) \int\frac{\frac{dx}{dt}}{{([C_A(0)-x)([C_B(0)-x)}}dt=k(t)
\end{equation}
And from that definition of k(t), again, we can arrive at the inequality.
