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# The reaction ${\text{A}}\left( g \right){\text{ + 2 B}}\left( g \right){\text{ }} \to {\text{ C}}\left( g \right){\text{ + D}}\left( g \right){\text{ }}$ is an elementary process. In an experiment, the initial partial pressures of A and B are ${P_{\text{A}}}{\text{ = 0}}{\text{.60 and }}{P_{\text{B}}}{\text{ = 0}}{\text{.80 atm}}$? When ${P_{\text{C}}}{\text{ = 0}}{\text{.2 atm}}$ the rate of the reaction relative to the initial rate is: A) 1/48B) 1/24C) 9/16D) 1/6

Last updated date: 04th Aug 2024
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Hint: For elementary reaction, the orders of reactants in the rate law expression are equal to the stoichiometric coefficients in the balanced chemical equation. The rate of the reaction is directly proportional to the partial pressures of the reactants. The partial pressures are raised to appropriate stoichiometric coefficients.

Write the balanced chemical equation
${\text{A}}\left( g \right){\text{ + 2 B}}\left( g \right){\text{ }} \to {\text{ C}}\left( g \right){\text{ + D}}\left( g \right)$
Since the reaction is an elementary reaction, you can assume that the orders of A and B in the rate law expression are equal to the stoichiometric coefficients in the balanced chemical equation. Write the rate law equation
${\text{R = k}} \times {{\text{P}}_{\text{A}}} \times {\text{P}}_{\text{B}}^2$
Here, R is the rate of the reaction and k is the rate constant.
For initial and final conditions, determine the rate law expressions separately. Then divide these two rate law expressions for initial and final conditions and then calculate the ratio of the relative reaction rates for final and initial conditions.
The initial partial pressures of A and B are ${P_{\text{A}}}{\text{ = 0}}{\text{.60 and }}{P_{\text{B}}}{\text{ = 0}}{\text{.80 atm}}$ respectively.
Substitute the values in the rate law expression.
${{\text{R}}_1}{\text{ = k}} \times {\text{0}}{\text{.60}} \times {\left( {0.80} \right)^2}$ … …(1)

Prepare a table showing initial partial pressures, change in partial pressures and equilibrium partial pressures.

 ${\text{A}}\left( g \right)$ ${\text{B}}\left( g \right)$ ${\text{C}}\left( g \right)$ ${\text{D}}\left( g \right)$ Initial partial pressure (atm) 0.60 0.80 0.0 0.0 Change in partial pressure (atm) -0.2 -0.4 0.2 0.2 Equilibrium partial pressure (atm) 0.40 0.40 0.2 0.2

When ${P_{\text{C}}}{\text{ = 0}}{\text{.2 atm}}$ calculate the final values of the partial pressures
Substitute the values in the rate law expression.
${{\text{R}}_2}{\text{ = k}} \times {\text{0}}{\text{.40}} \times {\left( {0.40} \right)^2}$ … …(2)
Divide equation (2) with equation (1)
$\dfrac{{{{\text{R}}_2}}}{{{{\text{R}}_1}}} = \dfrac{{{\text{k}} \times {\text{0}}{\text{.40}} \times {{\left( {0.40} \right)}^2}}}{{{\text{k}} \times {\text{0}}{\text{.60}} \times {{\left( {0.80} \right)}^2}}} \\ \dfrac{{{{\text{R}}_2}}}{{{{\text{R}}_1}}} = \dfrac{1}{6} \\$
The rate of the reaction relative to the initial rate is $\dfrac{1}{6}$.

Hence, the option D ) is the correct option.

Note: Do not forget to raise the partial pressures to appropriate stoichiometric coefficients in the rate law expression. This will lead to calculation errors.