Question

A certain reaction has the form

$\mathbf{\text{aA}}\mathbf{\to }\mathbf{\text{Products}}$

At a particular temperature, concentration versus time data was collected. A plot of$\mathbf{1}\mathbf{/}\mathbf{\left[}\mathbf{\text{A}}\mathbf{\right]}$ versus time (in seconds) gave a straight line with a slope of $\mathbf{6}\mathbf{.}\mathbf{90}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}$. What is the differential rate law for this reaction? What is the integrated rate law for this reaction? What is the value of the rate constant for this reaction? If$\mathbf{[}\mathbf{\text{A}}{\mathbf{]}}_{\mathbf{0}}$ for this reaction is $\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{\text{M}}$, what is the first half-life (in seconds)? If the original concentration (at $\mathbf{t}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$) is $\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{\text{M}}$, what is the second half-life (in seconds)?

Short answer

The differential rate law equation is $K[A{]}^2$. The integrated rate law equation is $\frac{1}{\left[\text{A}\right]}:\text{kt}+\frac{1}{{\left[\text{A}\right]}_0}$.

The Reaction rate constant is $\text{K}=6.90\times {10}^{-2}{\text{M}}^{-1}{\text{S}}^{-1}$.

The first half-life corresponds to${\text{t}}_{\text{1/2}}=1449\text{seconds}$.

The second half-life corresponds to${\text{t}}_{\text{1/2}}=2898\text{seconds}$.

Step-by-step solution

Calculating the rate law and integrated rate law equation for the chemical reaction 3:

Due to the fact plot of $\mathbf{\text{1/[A]}}$verses time gives a straight line we know we have a second-order reaction.

Differential rate law (for second-order reaction): rate $={\text{K[A]}}^{\text{2}}$

Integrated rate law (for second-order reaction)localid="1663846048449" $\frac{\text{1}}{\text{[A]}}=\text{kt +}\frac{\text{1}}{{\text{[A]}}_{\text{0}}}$

slope $=\mathrm{K}=6.90\times {10}^{-2}{\mathrm{M}}^{-1}{\mathrm{S}}^{-1}$

Calculating first half-life and second half-life for the given reaction n: 

Half-life is given by ${\text{t}}_{\text{1/2}}=\frac{\text{1}}{{\text{K[A]}}_{\text{0}}}$.

First half-life: ${\text{t}}_{\text{1/2}}=\frac{1}{{\text{K[A]}}_{\text{0}}}=\frac{1}{6.90\times {10}^{-2}{\text{M}}^{-1}{\text{S}}^{-1}\times 0.0100\text{M}}=1449\text{sec}$.

Second half-life : $t_{1/2}=\frac{1}{K{\left[A\right]}_0}=\frac{1}{6.90\times {10}^{-2}{\text{M}}^{-1}{\text{S}}^{-1}\times 0.005\text{M}}=2898\text{sec}$.