Q16.22 P - Chemistry: Molecular Nature Of Matter And Change

Question

Although the depletion of stratospheric ozone threatens life on Earth today, its accumulation was one of the crucial processes that allowed life to develop in prehistoric times: 

$\mathbf{3}{\mathit{O}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{\to }\mathbf{2}{\mathit{O}}_{\mathbf{3}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}$

(a) Express the reaction rate in terms $\left[O_2\right]$and $\left[O_3\right]$

(b) At a given instant, the reaction rate in terms of  is 2.17105 mol/L s. What is it in terms of $\left[O_3\right]$?

Step-by-Step Solution

Answer

a. The reaction rate of is, $\text{Reaction rate}=-\frac{1}{3}\frac{\Delta \left[O_2\right]}{\Delta t}$

           The reaction rate of is, $\text{Reaction rate}=\frac{1}{2}\frac{\Delta \left[O_3\right]}{\Delta t}$

b. The reaction rate in terms of data-custom-editor="chemistry" $\left[{\mathrm{O}}_3\right]$ is $1.45\times {10}^{-5}mol/L.s$

Step 1: (a) Calculate the rate of reaction in terms of [ O 2 ] and [ O 3 ]

The rate of reaction can be measured in terms of reactant disappearance or product appearance. Calculate the reaction rate of  as follows:

$\text{Reaction rate}=-\frac{1}{3}\frac{\Delta \left[O_2\right]}{\Delta t}$

In terms of data-custom-editor="chemistry" $\left[{\mathrm{O}}_3\right]$, express the reaction rate as follows:

$\text{Reaction rate}=\frac{1}{2}\frac{\Delta \left[O_3\right]}{\Delta t}$

The inverse of the fraction yields the molecules' stoichiometric coefficients. The negative sign indicates that the reactant concentration is decreasing.

Step 2: (b) The reaction rate in terms of [ O 2 ] is 2.17105 mol/L s at any given time. In terms of [ O 3 ] , what is it?

Calculate the reaction rate in terms of $\left[{\mathrm{O}}_2\right]$ :

$\begin{array}{rcl}\mathbf{-}\frac{\mathbf{1}}{\mathbf{3}}\frac{\mathbf{\Delta }\left[O_2\right]}{\mathbf{\Delta }\mathbf{t}}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{2}}\frac{\mathbf{\Delta }\left[O_3\right]}{\mathbf{\Delta }\mathbf{t}}\\ \frac{\mathbf{\Delta }\left[O_3\right]}{\mathbf{\Delta }\mathbf{t}}& \mathbf{=}& \frac{\mathbf{2}}{\mathbf{3}}\mathbf{\times }\left(-\frac{\Delta \left[O_2\right]}{\Delta t}\right)\\ & \mathbf{=}& \frac{\mathbf{2}}{\mathbf{3}}\mathbf{\times }\left(2.17\times {10}^{-5}mol/L.s\right)\\ & \mathbf{=}& \mathbf{1}\mathbf{.}\mathbf{45}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathit{m}\mathit{o}\mathit{l}\mathbf{/}\mathit{L}{\mathit{s}}^{\mathbf{-}\mathbf{1}}\\ & & \end{array}$

As a result, the reaction rate in terms of  $\left[{\mathrm{O}}_3\right]$ is $1.45\times {10}^{-5}mol/L{s}^{-1}$