Question

You and a coworker have developed a molecule that has shown potential as cobra anti-venom (AV). This anti-venom works by binding to the venom (V), thereby rendering it nontoxic. This reaction can be described by the rate law $\mathbf{Rate}\mathbf{=}\mathbf{k}{\left[\mathrm{AV}\right]}^{\mathbf{1}}{\left[V\right]}^{\mathbf{1}}$. You have been given the following data from your coworker:

${\left[V\right]}_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{20}\mathbf{M}$

${\left[\mathrm{AV}\right]}_{\mathbf{0}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{M}$

A plot of ln [AV] versus time gives a straight line with a slope of $\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{32}{\mathbf{s}}^{\mathbf{-}\mathbf{1}}$. What is the value of the rate constant (k) for this reaction?

Short answer

The value of the rate constant(k) for this reaction is $1.6{\mathrm{M}}^{-1}{\mathrm{s}}^{-1}$.

Step-by-step solution

Data are given in the above question

The data given is as follows

$$\begin{gathered} {\left[V\right]}_0=0.20M \\ {\left[AV\right]}_0=1.0\times {10}^{-4}M \end{gathered}$$

A plot of ln [AV] vs time(t) gives a straight line with a slope is $-0.32{\mathrm{s}}^{-1}$.

Describing the molecularity of the reaction

The given reaction is shown as a bimolecular reaction as the [AV] and [V] have an index one each. But when we check the values of initial concentrations of molecules. The concentration of [V] is higher than the concentration of [AV]. This assumption expresses the reaction pseudo-first-order reaction.

Therefore,$\mathrm{Rate}=\mathrm{k}\left[\mathrm{AV}\right]\left[\mathrm{V}\right]=\mathrm{k}\text{'}\left[\mathrm{AV}\right]$...(1)

Deducing the equation in terms of In[AV] and t

Now, Using equation (1), we get

$\mathrm{Rate}=-\frac{\mathrm{d}\left[\mathrm{AV}\right]}{\mathrm{dt}}={\mathrm{k}}^{\text{'}}\left[\mathrm{AV}\right]$

$-\frac{d\left[AV\right]}{dt}=k^{\text{'}}\left[AV\right]$

$\frac{d\left[AV\right]}{\left[AV\right]}=-k^{\text{'}}t$

By integrating, we get

$$\begin{gathered} \int {}_{\left[AV\right]0}{}^{\left[AV\right]}\frac{d\left[AV\right]}{\left[AV\right]}=-k^{\text{'}}\underset{0}{\overset{t}{\int }}t \\ \because ln\left[AV\right]=ln{\left[AV\right]}_0--k\text{'}t \end{gathered}$$

Describing the calculation of rate constant (k)

Here, By comparing the above equation with the line of equation $\mathrm{y}=\mathrm{mx}+\mathrm{c}$, we get

$k\text{'}=0.32s^{-1}$

Thus,

From equation (1)

$\begin{array}{rcl}{k}^{\text{'}}& =& k\left[V\right]\\ 0.32& =& k\left(0.2\right)\\ \frac{0.32}{0.2}& =& k\\ k& =& 1.6M^{-1}{s}^{-1}\end{array}$

Therefore, the value of k for the reaction is .$1.6{\mathrm{M}}^{-1}{\mathrm{s}}^{-1}$.