Chapter 15, Problem 55E

(a)

Interpretation Introduction

Interpretation:

The rate constant needs to be determined.

Concept Introduction:

Rate of reaction represents the change of concentration of a reactant or a product with respect to time. It can be expressed either by reduceamount of reactant in per unit time or increase amount of product in per unit time.

Answer to Problem 55E

The value of rate constant k is 1.15 ×10² M-³s-1

Explanation of Solution

Given Information:

The initial concentration of [B]₀ is equal to the initial concentration of [C]₀ which 1.00 M and the initial concentration of the [A]₀ is 1.00×10-4 M and after 3 min. the concentration of [A] becomes 3.26×10-5 M. The given reaction is $\text{3A}\text{+}\text{B}\text{+}\text{C}\to \text{D}\text{+E}$ .

Calculation: The given rate law is shown below:

  $\text{-}\frac{\text{d}\left[\text{A}\right]}{\text{dt}}\text{=}\text{k}{\left[\text{A}\right]}^{\text{2}}\left[\text{B}\right]\left[\text{C}\right]$

The initial concentration [B]₀ = [C]₀ = 1.00 M

The initial concentration [A]₀ = 1.00×10-⁴M

  $\begin{array}{c}\text{-}\frac{\text{d}\left[\text{A}\right]}{\text{dt}}\text{=}\text{k}{\left[\text{A}\right]}^{\text{2}}\left[\text{B}\right]\left[\text{C}\right]\\ \frac{\text{d}\left[\text{A}\right]}{{\left[\text{A}\right]}^{\text{2}}}\text{=}\text{-}\text{k}\left[\text{B}\right]\left[\text{C}\right]\text{dt}\\ {\displaystyle \underset{{\left[\text{A}\right]}_{\text{0}}}{\overset{{\left[\text{A}\right]}_{\text{t}}}{\int }}\frac{\text{1}}{{\left[\text{A}\right]}^{\text{2}}}\text{d}\left[\text{A}\right]}\text{=}\text{-}\text{k}\left[\text{B}\right]\left[\text{C}\right]{\displaystyle \underset{\text{0}}{\overset{\text{t}}{\int }}\text{dt}}\\ \frac{\text{-1}}{{\left[\text{A}\right]}_{\text{t}}}\text{+}\frac{\text{1}}{{\left[\text{A}\right]}_{\text{0}}}\text{=}\text{-}\text{k}\left[\text{B}\right]\left[\text{C}\right]\text{t}\\ \text{k}\text{=}\frac{\text{1}}{\left[\text{B}\right]\left[\text{C}\right]\text{t}}\left[\frac{\text{1}}{{\left[\text{A}\right]}_{\text{t}}}\text{-}\frac{\text{1}}{{\left[\text{A}\right]}_{\text{0}}}\right]\end{array}$

Substitute the values in aboveequation which is shown below:

  $\begin{array}{c}\text{k}\text{=}\frac{\text{1}}{\left[\text{B}\right]\left[\text{C}\right]\text{t}}\left[\frac{\text{1}}{{\left[\text{A}\right]}_{\text{t}}}\text{-}\frac{\text{1}}{{\left[\text{A}\right]}_{\text{0}}}\right]\\ \text{=}\frac{\text{1}}{\left[\text{1}\text{M}\right]\left[\text{1}\text{M}\right]\left(\text{3min}\text{.}\text{×}\frac{\text{60sec}}{\text{1}\text{min}\text{.}}\right)}\left[\frac{\text{1}}{\left[\text{3}\text{.26}\text{×}{\text{10}}^{\text{-5}}\text{M}\right]}\text{-}\frac{\text{1}}{\left[\text{1}\text{×}{\text{10}}^{\text{-4}}\text{M}\right]}\right]\\ \text{=}\text{1}\text{.15}\text{×}{\text{10}}^{\text{2}}{\text{M}}^{\text{-3}}{\text{s}}^{\text{-1}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The half-life time needs to be determined.

Concept Introduction: Rate of reaction represents the change of concentration of a reactant or a product with respect to time. It can be expressed either by reduceamount of reactant in per unit time or increase amount of product in per unit time.

Answer to Problem 55E

The half-life is 87.0 s

Explanation of Solution

The integrated rate law for first A in the context of B and C and rate constant to calculate the half -life which is shown below:

  $\begin{array}{r}\frac{\text{1}}{\text{1/2}{\left[\text{A}\right]}_{\text{0}}}\text{=}{\text{kt}}_{\text{1/2}}\text{+}\frac{\text{1}}{{\left[\text{A}\right]}_{\text{0}}}\\ \text{=}\frac{\text{1}}{{\left[\text{A}\right]}_{\text{0}}}\text{=}{\text{kt}}_{\text{1/2}}\end{array}$

The half-life of the reaction is based on only the second order behavior of A in the context of C and B which is shown below:

  $\begin{array}{c}{\text{t}}_{\text{1/2}}\text{=}\frac{\text{1}}{\text{k}{\left[\text{A}\right]}_{\text{0}}}\\ \text{=}\frac{\text{1}}{\text{115}\left(\text{1}\text{.00}\text{×}{\text{10}}^{\text{-4}}\right)}\\ \text{=}\text{87}\text{.0}\text{s}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The initial concentration of [B]₀ is equal to the initial concentration of [C]₀ which 1.00 M and. predict the concentration of [A] and concentration of B after 10.0 minutes.

Concept Introduction:

Answer to Problem 55E

The concentration of [A] is 1.27×10-5 M and the concentration of [B] is 1.00 M.

Explanation of Solution

The expression of integrated second order rate law in which x is the fraction of A left after the time 10 min. this fraction will be less than 0.5 because the half-life was about 1.45 min.

  $\begin{array}{c}\frac{\text{1}}{\left[\text{A}\right]}\text{=}\text{kt}\text{+}\frac{\text{1}}{{\left[\text{A}\right]}_{\text{0}}}\\ \frac{\text{1}}{\text{x}{\left[\text{A}\right]}_{\text{0}}}\text{-}\frac{\text{1}}{{\left[\text{A}\right]}_{\text{0}}}\text{=}\text{kt}\\ \frac{\frac{\text{1}}{\text{x}}\text{-}\text{1}}{{\left[\text{A}\right]}_{\text{0}}}\text{=}\text{kt}\\ \text{kt}{\left[\text{A}\right]}_{\text{0}}\text{+1}\text{=}\frac{\text{1}}{\text{x}}\end{array}$

The fraction of A is left which is shown below:

  $\begin{array}{c}\text{x}\text{=}\frac{\text{1}}{\text{kt}{\left[\text{A}\right]}_{\text{0}}\text{+}\text{1}}\\ \text{=}\frac{\text{1}}{\text{115}\text{×}\text{600}\left(\text{1}\text{.00}\text{×}{\text{10}}^{\text{-4}}\right)\text{+}\text{1}}\\ \text{=}\text{0}\text{.1266}\end{array}$

So, the amount of A left is as follows:

  $\begin{array}{c}{\left[\text{A}\right]}_{\text{10}\text{min}\text{.}}\text{=}\text{x}{\left[\text{A}\right]}_{\text{0}}\\ \text{=}\text{0}\text{.1266}\text{×}\text{1}\text{.00}\text{×}{\text{10}}^{\text{-4}}\\ \text{=}\text{1}\text{.27}\text{×}{\text{10}}^{\text{-5}}\text{M}\end{array}$

According to the reaction, the change of A is triple times of the change of B.

  $\begin{array}{c}\frac{\left({\left[\text{A}\right]}_{\text{0}}\text{-}\left[\text{A}\right]\right)}{\text{3}}\text{=}{\left[\text{B}\right]}_{\text{0}}\text{-}\left[\text{B}\right]\\ \left[\text{B}\right]\text{=}{\left[\text{B}\right]}_{\text{0}}\text{-}\frac{\text{1}}{\text{3}}\left({\left[\text{A}\right]}_{\text{0}}\text{-}\left[\text{A}\right]\right)\\ \text{=}\text{1}\text{.00}\text{M}\text{-}\frac{\text{1}}{\text{3}}\left(\text{1}\text{.0}\text{×}{\text{10}}^{\text{-4}}\text{M}\text{-}\text{1}\text{.27}\text{×}{\text{10}}^{\text{-5}}\text{M}\right)\\ \text{=}\text{0}\text{.999}\text{M}\\ \text{=}\text{1}\text{.00}\text{M}\end{array}$