Chapter 13, Problem 13.140QP

The following data were collected for the reaction $\text{A}(g)+\text{B}(g)\stackrel{}{\to }$ Products.

1. a Determine the rate law for this reaction.
2. b Calculate the rate constant.
3. c Calculate the rate when [A] = 0.200 M and [B] = 0.200 M.

Interpretation Introduction

Interpretation:

The rate law and the rate constant for the reaction along with the rates for the given concentrations have to be calculated.

Concept Introduction:

The rate of reaction is the quantity of formation of product or the quantity of reactant used per unit time.  The rate of reaction doesn’t depend on the sum of amount of reaction mixture used.

The raise in molar concentration of product of a reaction per unit time or decrease in molarity of reactant per unit time is called rate of reaction and is expressed in units of $\text{mol/(L}\text{.s)}$.

The variation in concentration of reaction or product over a certain interval of time is called average reaction rate.

The equation that relates the reaction rate to the reactants concentrations that is raised to various powers is called as rate law.

Rate law can be determined by the slow step or otherwise called as rate-determining step.

Answer to Problem 13.140QP

The rate law is Rate= $\text{k}\left[\text{A}\right]{\left[\text{B}\right]}^{\text{2}}$ .

Explanation of Solution

To determine the rate law for the reaction

The rate law for the reaction can be given as,

Rate= $\text{k}{\left[\text{A}\right]}^{\text{x}}{\left[\text{B}\right]}^{\text{y}}$

The value of x, can be calculated by finding the ratio of the rates for experiments 1 and 2.

In these experiments, $\left[\text{B}\right]$ is constant. This gives,

$\begin{array}{l}\frac{{\text{Rate}}_{\text{2}}}{{\text{Rate}}_{\text{1}}}\text{=}\frac{\text{k}{\left[\text{A}\right]}_{\text{2}}{}^{\text{x}}{\left[\text{B}\right]}_{\text{2}}{}^{\text{y}}}{\text{k}{\left[\text{A}\right]}_{\text{1}}{}^{\text{x}}{\left[\text{B}\right]}_{\text{2}}{}^{\text{y}}}\text{=}\left[\frac{{\left[\text{A}\right]}_{\text{2}}}{{\left[\text{A}\right]}_{\text{1}}}\right]\\ \\ \frac{\text{3}{\text{.0×10}}^{\text{-3}}\text{M/s}}{\text{1}{\text{.0×10}}^{\text{-3}}\text{M/s}}\text{=}{\left(\frac{\text{0}\text{.0300}\text{M}}{\text{0}\text{.0100}\text{M}}\right)}^{\text{x}}\end{array}$

This goes down to $\text{3=3x}$ or x=1

The value of y can be calculated by finding the ratio of experiments 3 and 2.

In these experiments, $\left[\text{A}\right]$ is constant. This gives,

$\begin{array}{l}\frac{{\text{Rate}}_{\text{3}}}{{\text{Rate}}_{\text{2}}}\text{=}\frac{\text{k}{\left[\text{A}\right]}_{\text{3}}{}^{\text{x}}{\left[\text{B}\right]}_{\text{2}}{}^{\text{y}}}{\text{k}{\left[\text{A}\right]}_{\text{3}}{}^{\text{x}}{\left[\text{B}\right]}_{\text{2}}{}^{\text{y}}}\text{=}\left[\frac{{\left[\text{B}\right]}_{\text{3}}}{{\left[\text{B}\right]}_{\text{2}}}\right]\\ \\ \frac{\text{2}{\text{.7×10}}^{\text{-3}}\text{M/s}}{\text{3}{\text{.0×10}}^{\text{-3}}\text{M/s}}\text{=}{\left(\frac{\text{0}\text{.0300}\text{M}}{\text{0}\text{.0100}\text{M}}\right)}^{\text{x}}\end{array}$

This goes down to ${\text{9=3}}^{\text{y}}$ or y=2

Hence, the rate law becomes

Rate= $\text{k}{\left[\text{A}\right]}^{\text{x}}{\left[\text{B}\right]}^{\text{y}}$ = $\text{k}\left[\text{A}\right]{\left[\text{B}\right]}^{\text{2}}$

Interpretation Introduction

Interpretation:

The rate law and the rate constant for the reaction along with the rates for the given concentrations have to be calculated.

Concept Introduction:

The rate of reaction is the quantity of formation of product or the quantity of reactant used per unit time.  The rate of reaction doesn’t depend on the sum of amount of reaction mixture used.

The raise in molar concentration of product of a reaction per unit time or decrease in molarity of reactant per unit time is called rate of reaction and is expressed in units of $\text{mol/(L}\text{.s)}$.

The variation in concentration of reaction or product over a certain interval of time is called average reaction rate.

The equation that relates the reaction rate to the reactants concentrations that is raised to various powers is called as rate law.

Rate law can be determined by the slow step or otherwise called as rate-determining step.

Answer to Problem 13.140QP

The rate constant is $\text{10}{\text{./M}}^{\text{2}}\text{.s}$ .

Explanation of Solution

To calculate the rate constant

Rate constant can be calculated from the data of Experiment 1.

k= $\frac{\text{Rate}}{\left[\text{A}\right]{\left[\text{B}\right]}^{\text{2}}}\text{=}\frac{\text{1}{\text{.0×10}}^{\text{-3}}\text{M/s}}{\left(\text{0}\text{.0100}\text{M}\right){\left(\text{0}\text{.100}\text{M}\right)}^{\text{2}}}\text{=10}\text{.0=10}{\text{./M}}^{\text{2}}\text{.s}$

The rate constant = $\text{10}{\text{./M}}^{\text{2}}\text{.s}$ .

Interpretation Introduction

Interpretation:

The rate law and the rate constant for the reaction along with the rates for the given concentrations have to be calculated.

Concept Introduction:

The rate of reaction is the quantity of formation of product or the quantity of reactant used per unit time.  The rate of reaction doesn’t depend on the sum of amount of reaction mixture used.

The raise in molar concentration of product of a reaction per unit time or decrease in molarity of reactant per unit time is called rate of reaction and is expressed in units of $\text{mol/(L}\text{.s)}$.

The variation in concentration of reaction or product over a certain interval of time is called average reaction rate.

The equation that relates the reaction rate to the reactants concentrations that is raised to various powers is called as rate law.

Rate law can be determined by the slow step or otherwise called as rate-determining step.

Answer to Problem 13.140QP

The rate of the reaction at given concentration is $\text{0}\text{.080}\text{M/s}$ .

Explanation of Solution

To calculate the rate for the given concentration

Concentration of $\left[\text{A}\right]\text{=}\left[\text{B}\right]\text{=0}\text{.200}\text{M}$

Rate = $\text{k}\left[\text{A}\right]{\left[\text{B}\right]}^{\text{2}}$

Rate= $\left(\text{10}{\text{./M}}^{\text{2}}\text{.s}\right)\left(\text{0}\text{.200}\text{M}\right)\left(\text{0}\text{.200}\text{M}\right)$

Rate = $\text{0}\text{.0800=0}\text{.080}\text{M/s}$

The rate of the reaction at given concentration = $\text{0}\text{.080}\text{M/s}$ .