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You and a co-worker have developed a molecule that has shown potential as cobra anti-venom ($\ce{AV}$). This anti-venom works by binding to the venom ($\ce{V}$), thereby rendering it non-toxic. This reaction can be described by the rate law, $\text{rate} = k\left[\ce{AV}\right]^1\left[\ce{V}\right]^1$.

You have been given the following data from your co-worker: $\left[\ce{V}\right]_0=0.20\ \mathrm M$, $\left[\ce{AV}\right]_0=1.0\times10^{-4}\ \mathrm M$. A plot of $\ln\left[\ce{AV}\right]$ versus $t$ (s) gives a straight line with a slope of $-0.32\ \mathrm s^{-1}$. What is the value of the rate constant ($k$) for this reaction?

My logic tends to fail me on this question. So to start off, the overall reaction order is 2, thus the equation is: $${A}^{-1}=kt+{a}^{-1}$$ However, we're only given a $k$ value for the law: $\text{rate}=k\left[\ce{AV}\right]$. I'm lost as to how to proceed next. Given the k we can calculate the initial rate of $\left[\ce{AV}\right]$ but how is this going to help me find the $k$ for $rate=k\left[\ce{AV}\right]^1\left[\ce{V}\right]^1$? I suspect something tiny that I missed in this problem, any directions or hints would be appreciated!

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You can regard the reaction to be pseudo-first order, as the concentration of $\left[\ce{V}\right]$ is much higher than the concentration of $\left[\ce{AV}\right]$. Therefore, $k\cdot\left[\ce{V}\right]_0=\text{const.}=k_{2}$. The rate law of this reaction would change to $r=k_{2}\cdot\left[\ce{AV}\right]$. Integrating this would lead to the expression: $\ln\left[\ce{AV}\right]=\ln\left[\ce{AV}\right]_0-k_{2}\cdot t$. The dependence of $\ln\left[\ce{AV}\right]$ on time has the slope $-k_{2}=-0.32\ \mathrm s^{-1}$, so you can easily calculate $k$ from $k_{2}$ and $\left[\ce{V}\right]_0$.

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