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\begin{equation}  k(t) = \left(\frac{dS(t)^{-1}}{dt}\right)  \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}{(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}}{{(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.