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Jonathan Nichols added Mixed Second Order Reactions.tex
over 9 years ago
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Second order reactions can also occur in the form of two different molecules combining into products irreversibly. Determining when the rate coefficient of a mixed second order reaction is constant is possible, and is able to be done without survival functions. Survival functions were not used in the case of mixed second order decay because the integrated rate law only allows the survival function of one reactant to be looked at at a time. This is true as long as the initial concentrations of both reactants are not equal. If the initial concentration of both reactants is equal, then the concentration of both reactants will be equal to each other at all times, and the mixed second order reaction can be treated as a regular second order reaction, leading to the inequality in equation 11. The rate law for a mixed second order reaction is
\begin{equation}
\int\frac{dx}{([A_0]-x)([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)
\end{equation}
However, $\frac{dx}{dt}=\omega([A_0]-x)([B_0]-x)$
Filling in $\frac{dx}{dt}$ in the integral and evaluating, we get that k(t)=$\omega$t, which goes on to provide the same results that are seen in first order, except that the statistical length and the divergence are no longer dimensionless since the second order rate constant has units of $\frac{1}{[concentration][time]}$. The inequality this result leads to has already been seen in first order kinetics as
\begin{equation}
\omega^2(\Delta{t})^2-(\omega\Delta t)^2\geq0
\end{equation}
