Chapter DY, Problem 10PQ

Interpretation Introduction

Interpretation:

Based on the given data, the rate law for the given reaction has to be found.  The tabulated data given is:

  ${\text{CH}}_{\text{3}}{\text{OCH}}_{\text{3}}{}_{\text{(aq)}}{\text{+Br}}_{\text{2}}{}_{\text{(aq)}}{\text{+H}}_{\text{3}}{\text{O}}_{\text{(aq)}}^{\text{+}}\stackrel{}{\to }\text{products}$

    $\begin{array}{l}\text{Initial Concentrations, M;InitialRates,M}{\text{.s}}^{\text{-1}}\\ \text{-------------------------------------------------------}\\ {\text{Exp CH}}_{\text{3}}{\text{OCH}}_{\text{3}}{\text{ Br}}_{\text{2}}{\text{ H}}_{\text{3}}{\text{O}}^{\text{+}}\text{ Rate}\\ \text{1 0}\text{.30 0}\text{.050 0}\text{.050 5}\text{.8}\times {\text{10}}^{-5}\\ \text{2 0}\text{.30 0}\text{.100 0}\text{.050 5}\text{.8}\times {\text{10}}^{-5}\\ \text{3 0}\text{.30 0}\text{.050 0}\text{.100 1}\text{.2}\times {\text{10}}^{-4}\\ \text{4 0}\text{.40 0}\text{.050 0}\text{.200 3}\text{.2}\times {\text{10}}^{-4}\end{array}$

Concept Introduction:

Rate of a reaction:

Speed at which the chemical reaction happens is known as rate of a reaction.  Slower the speed of the reaction, lower will be the rate.  Concentration, temperature, pressure, catalyst are the factors affecting rate of a reaction.

Rate law – it is a relationship between reaction rates and reactant concentrations

Order of a reaction:

Order of a reaction is the power of the rate on the concentration of each reactant.  Order of a reaction is an experimentally determined one.

Explanation of Solution

Reason for correct option:

The given reaction is:

    ${\text{CH}}_{\text{3}}{\text{OCH}}_{\text{3}}{}_{\text{(aq)}}{\text{+Br}}_{\text{2}}{}_{\text{(aq)}}{\text{+H}}_{\text{3}}{\text{O}}_{\text{(aq)}}^{\text{+}}\stackrel{}{\to }\text{products}$.

Rate of the reaction is:

    ${\text{r=k[CH}}_{\text{3}}{\text{OCH}}_{\text{3}}{\text{]}}^{\text{x}}{{\text{[Br}}_{\text{2}}\text{]}}^{\text{y}}{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}^{\text{Z}}$.

From the table,

    $\begin{array}{l}{\text{r}}_1{\text{=k[CH}}_{\text{3}}{\text{OCH}}_{\text{3}}{\text{]}}^{\text{x}}{{\text{[Br}}_{\text{2}}\text{]}}^{\text{y}}{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}^z=\text{k[0}{\text{.30]}}^{\text{x}}{\text{[0}\text{.050]}}^{\text{y}}{\text{[0}\text{.050]}}^z=5.8\times {10}^{-5}------(1)\\ {\text{r}}_2{\text{=k[CH}}_{\text{3}}{\text{OCH}}_{\text{3}}{\text{]}}^{\text{x}}{{\text{[Br}}_{\text{2}}\text{]}}^{\text{y}}{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}^z=\text{k[0}{\text{.30]}}^{\text{x}}{\text{[0}\text{.100]}}^{\text{y}}{\text{[0}\text{.050]}}^z=5.8\times {10}^{-5}------(2)\\ {\text{r}}_3{\text{=k[CH}}_{\text{3}}{\text{OCH}}_{\text{3}}{\text{]}}^{\text{x}}{{\text{[Br}}_{\text{2}}\text{]}}^{\text{y}}{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}^z=\text{k[0}{\text{.30]}}^{\text{x}}{\text{[0}\text{.050]}}^{\text{y}}{\text{[0}\text{.100]}}^z=1.2\times {10}^{-4}------(3)\\ {\text{r}}_4{\text{=k[CH}}_{\text{3}}{\text{OCH}}_{\text{3}}{\text{]}}^{\text{x}}{{\text{[Br}}_{\text{2}}\text{]}}^{\text{y}}{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}^z=\text{k[0}{\text{.40]}}^{\text{x}}{\text{[0}\text{.050]}}^{\text{y}}{\text{[0}\text{.200]}}^z=3.2\times {10}^{-4}------(4)\end{array}$.

Divide the equation (2) by equation (1),

    $\begin{array}{l}{\text{r}}_{\text{1}}{\text{=k[CH}}_{\text{3}}{\text{OCH}}_{\text{3}}{\text{]}}^{\text{x}}{{\text{[Br}}_{\text{2}}\text{]}}^{\text{y}}{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}^{\text{z}}\text{=k[0}{\text{.30]}}^{\text{x}}{\text{[0}\text{.050]}}^{\text{y}}{\text{[0}\text{.050]}}^{\text{z}}\text{=5}{\text{.8×10}}^{\text{-5}}\text{------(1)}\\ {\text{r}}_{\text{2}}{\text{=k[CH}}_{\text{3}}{\text{OCH}}_{\text{3}}{\text{]}}^{\text{x}}{{\text{[Br}}_{\text{2}}\text{]}}^{\text{y}}{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}^{\text{z}}\text{=k[0}{\text{.30]}}^{\text{x}}{\text{[0}\text{.100]}}^{\text{y}}{\text{[0}\text{.050]}}^{\text{z}}\text{=5}{\text{.8×10}}^{\text{-5}}\text{------(2)}\\ {\text{r}}_{\text{2}}{\text{/r}}_{\text{1}}\text{=k[0}{\text{.30]}}^{\text{x}}{\text{[0}\text{.100]}}^{\text{y}}{\text{[0}\text{.050]}}^{\text{z}}\text{/k[0}{\text{.30]}}^{\text{x}}{\text{[0}\text{.050]}}^{\text{y}}{\text{[0}\text{.050]}}^{\text{z}}\text{=5}{\text{.8×10}}^{\text{-5}}\text{/5}{\text{.8×10}}^{\text{-5}}\text{=1}\\ {\text{1/2=(1/2)}}^{\text{0}}\\ \text{y=0}\end{array}$.

Order is 0 with respect to ${\text{Br}}_2$.

Divide the equation (1) by equation (3),

    $\begin{array}{l}{\text{r}}_{\text{1}}{\text{=k[CH}}_{\text{3}}{\text{OCH}}_{\text{3}}{\text{]}}^{\text{x}}{{\text{[Br}}_{\text{2}}\text{]}}^{\text{y}}{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}^{\text{z}}\text{=k[0}{\text{.30]}}^{\text{x}}{\text{[0}\text{.050]}}^{\text{y}}{\text{[0}\text{.050]}}^{\text{z}}\text{=5}{\text{.8×10}}^{\text{-5}}\text{------(1)}\\ {\text{r}}_3{\text{=k[CH}}_{\text{3}}{\text{OCH}}_{\text{3}}{\text{]}}^{\text{x}}{{\text{[Br}}_{\text{2}}\text{]}}^{\text{y}}{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}^{\text{z}}\text{=k[0}{\text{.30]}}^{\text{x}}{\text{[0}\text{.050]}}^{\text{y}}{\text{[0}\text{.100]}}^{\text{z}}\text{=1}{\text{.2×10}}^{\text{-4}}\text{------(3)}\\ {\text{r}}_1{\text{/r}}_2\text{=k[0}{\text{.30]}}^{\text{x}}{\text{[0}\text{.050]}}^{\text{y}}{\text{[0}\text{.050]}}^{\text{z}}\text{/k[0}{\text{.30]}}^{\text{x}}{\text{[0}\text{.050]}}^{\text{y}}{\text{[0}\text{.100]}}^{\text{z}}\text{=5}{\text{.8×10}}^{\text{-5}}\text{/1}{\text{.2×10}}^{\text{-4}}\text{=1}\\ {\text{1/2=(1/2)}}^1\\ \text{z=1}\end{array}$.

Therefore the order of the reaction is 1 with respect to ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$.

Divide the equation (3) by equation (4),

    $\begin{array}{l}{\text{r}}_3{\text{=k[CH}}_{\text{3}}{\text{OCH}}_{\text{3}}{\text{]}}^{\text{x}}{{\text{[Br}}_{\text{2}}\text{]}}^{\text{y}}{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}^z=\text{k[0}{\text{.30]}}^{\text{x}}{\text{[0}\text{.050]}}^{\text{y}}{\text{[0}\text{.100]}}^z=1.2\times {10}^{-4}------(3)\\ {\text{r}}_4{\text{=k[CH}}_{\text{3}}{\text{OCH}}_{\text{3}}{\text{]}}^{\text{x}}{{\text{[Br}}_{\text{2}}\text{]}}^{\text{y}}{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}^z=\text{k[0}{\text{.40]}}^{\text{x}}{\text{[0}\text{.050]}}^{\text{y}}{\text{[0}\text{.200]}}^z=3.2\times {10}^{-4}------(4)\\ {\text{r}}_3{\text{/r}}_4\text{=k[0}{\text{.30]}}^{\text{x}}{\text{[0}\text{.050]}}^{\text{y}}{\text{[0}\text{.100]}}^z\text{/k[0}{\text{.40]}}^{\text{x}}{\text{[0}\text{.050]}}^{\text{y}}{\text{[0}\text{.200]}}^z\text{=1}{\text{.2×10}}^{\text{-4}}\text{/3}{\text{.2×10}}^{\text{-4}}\text{=0}\text{.375}\\ 0.375=0.375\\ \text{x=1}\end{array}$.

Therefore the order of the reaction is 1 with respect to ${\text{CH}}_{\text{3}}{\text{OCH}}_{\text{3}}$.

Hence the option (B) is correct.

Reason for incorrect option:

From the above rate law explanation, it is clear that ${\text{rate=k[CH}}_{\text{3}}{\text{OCH}}_{\text{3}}{\text{]}}^{\text{1}}{{\text{[Br}}_{\text{2}}\text{]}}^{\text{0}}{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}^{\text{1}}$.

Hence the options (A), (C) and (D) are incorrect.