Question

A certain reaction has the following general form:

$\mathbf{aA}\mathbf{\to }\mathbf{bB}$

At a particular temperature and ${\left[A\right]}_{\mathbf{0}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{80}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{M}$, concentration v/s time data were collected for this reaction and a plot of $\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\left[A\right]$}\right.$versus time resulted in a straight line with a slope value $\mathbf{3}\mathbf{.}\mathbf{60}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}{\mathbf{Lmol}}^{\mathbf{-}\mathbf{1}}{\mathbf{s}}^{\mathbf{-}\mathbf{1}}$.

a. Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction.

b. Calculate the half-life for this reaction.

c. How much time is required for the concentration of A to decrease to $\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{M}$?

Short answer

a. Rate law will be $\mathrm{Rate}=\mathrm{K}{\left[\mathrm{A}\right]}^2$

Integrated rate law $\frac{1}{{\left[\mathrm{A}\right]}_0}-\frac{1}{\left[\mathrm{A}\right]}=-\mathrm{Kt}$

Rate constant$\mathrm{K}=3.60\times {10}^{-2}{\mathrm{Lmol}}^{-1}{\mathrm{s}}^{-1}$.

b. The half-life for this reaction is $9.22\times {10}^3\mathrm{s}$

c. The time required for the concentration of A to decrease to $7\times {10}^{-4}\mathrm{M}$is $2.98\times {10}^4\mathrm{s}$.

Step-by-step solution

Determination of the rate law 

Rate law: Rate of reaction is directly proportional to the reactant of concentration

Order of reaction: The sum of the power of the reactants that takes part in a chemical reaction

$aA\to bB$

The reaction is second order with respect to A as the slope is obtained by plotting $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\left[\mathrm{A}\right]\mathrm{vs}\mathrm{Time}$}\right.$. The slope is equal to K.

Therefore, the rate law will be

$\mathrm{Rate}=\mathrm{K}{\left[\mathrm{A}\right]}^2$

Determination of the integrated rate law and the value of the rate constant for the reaction

Second order integrated rate law derivation

We have

$\begin{array}{rcl}\mathrm{Rate}& =& K{\left[A\right]}^2\\ \frac{d\left[A\right]}{dt}& =& -K{\left[A\right]}^2\\ \underset{\left[A_0\right]}{\overset{\left[A\right]}{\int }}\frac{d\left[A\right]}{{\left[A\right]}^2}& =& -K\underset{0}{\overset{t}{\int }}dt\end{array}$

After integrating, we have

$$\begin{gathered} \frac{1}{{\left[A\right]}_0}-\frac{1}{\left[A\right]}=-Kt \\ \frac{1}{\left[A\right]}=Kt+\frac{1}{{\left[A\right]}_0} \end{gathered}$$

Therefore, the graph or plot between $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\left[A\right]$}\right.\mathrm{vs}\mathrm{time}$is found to be a straight line.

Therefore, the value of slope (K) =$3.60\times {10}^{-2}{\mathrm{Lmol}}^{-1}{\mathrm{s}}^{-1}$.

Determine the half-life for the reaction

Half-life: The time required for a reactant to react to half of its original concentration is called the half-life.

The formula of half-life for a second-order reaction

$\begin{array}{rcl}{t}_{\frac{1}{2}}& =& \frac{1}{K{\left[A\right]}_0}\\ & =& \frac{1}{\left(3.60\times {10}^{-2}\right)\left(2.80\times {10}^{-3}\right)}\\ & =& 9.22\times {10}^{3}s\end{array}$

Time required for the concentration of A to decrease to 7×10-4M

For a second-order reaction $aA\to bB$

$\frac{1}{{\left[A\right]}_0}-\frac{1}{\left[A\right]}=-Kt$ (Integrated rate law)

As time is spent concentration of reactant decreases simultaneously

We can calculate the time for the reactant concentration decreases to $7\times {10}^{-4}\mathrm{M}$.

Given data

[A₀]= initial concentration= $2.80\times {10}^{-3}\mathrm{M}$

[A]= final concentration= $7\times {10}^{-4}\mathrm{M}$

K = $3.60\times {10}^{-2}{\mathrm{Lmol}}^{-1}{\mathrm{s}}^{-1}$.

Plugging respective values in the integrated rate law equation, we get

$\begin{array}{rcl}\frac{1}{7\times {10}^{-4}}& =& \left(3.60\times {10}^{-2}\right)t+\frac{1}{2.80\times {10}^{-3}}\\ t& =& \frac{\frac{1}{7\times {10}^{-4}}-\frac{1}{2.80\times {10}^{-3}}}{3.60\times {10}^{-2}}\\ t& =& 2.98\times {10}^{4}s\end{array}$