Chapter 16, Problem 16.111P

(a)

Interpretation Introduction

Interpretation:

The time taken to reduce the concentration from $\text{3}{\text{.0mol/dm}}^{\text{3}}$ to $\text{0}\text{.35}{\text{mol/dm}}^{\text{3}}$ in the spring has to be calculated.

Concept introduction:

First order reaction:

$\begin{array}{l}\text{Rate}\text{Equation:}\text{-Δ[R]/Δt}\text{=}{\text{k[R]}}^{\text{1}}\\ \\ \text{Integrated}\text{rate equation: ln}\left({\text{[R]}}_{\text{t}}{\text{/[R]}}_{\text{0}}\right)\text{= - kt}\end{array}$

Half-life: The time required for the concentration of a reactant to decrease to one half of its initial value.

$\begin{array}{l}{\text{[R]}}_{\text{t}}\text{=}\frac{{\text{[R]}}_{\text{0}}}{\text{2}}\text{,}\\ \\ \text{First}\text{order}\text{reaction:}\\ \\ {\text{ln[R]}}_{\text{t}}\text{=}\text{- kt}\text{+}{\text{ln[R]}}_{\text{0}}\\ \\ \text{ln}\frac{{\text{[R]}}_{\text{t}}}{{\text{[R]}}_{\text{0}}}\text{=}\text{- kt}\left(\text{or}\right)\text{ln}\text{2 = k}{\text{t}}_{\text{1/2}}\\ \\ {\text{t}}_{\text{1/2}}\text{=}\frac{\text{0}\text{.693}}{\text{k}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The time taken to reduce the concentration from $\text{3}{\text{.0mol/dm}}^{\text{3}}$ to $\text{0}\text{.35}{\text{mol/dm}}^{\text{3}}$ in the spring at $\text{T}\text{=}{\text{10}}^{\text{o}}\text{C}$ has to be calculated.

Concept introduction:

First order reaction:

$\begin{array}{l}\text{Rate}\text{Equation:}\text{-Δ[R]/Δt}\text{=}{\text{k[R]}}^{\text{1}}\\ \\ \text{Integrated}\text{rate equation: ln}\left({\text{[R]}}_{\text{t}}{\text{/[R]}}_{\text{0}}\right)\text{= - kt}\end{array}$

Half-life: The time required for the concentration of a reactant to decrease to one half of its initial value.

$\begin{array}{l}{\text{[R]}}_{\text{t}}\text{=}\frac{{\text{[R]}}_{\text{0}}}{\text{2}}\text{,}\\ \\ \text{First}\text{order}\text{reaction:}\\ \\ {\text{ln[R]}}_{\text{t}}\text{=}\text{- kt}\text{+}{\text{ln[R]}}_{\text{0}}\\ \\ \text{ln}\frac{{\text{[R]}}_{\text{t}}}{{\text{[R]}}_{\text{0}}}\text{=}\text{- kt}\left(\text{or}\right)\text{ln}\text{2 = k}{\text{t}}_{\text{1/2}}\\ \\ {\text{t}}_{\text{1/2}}\text{=}\frac{\text{0}\text{.693}}{\text{k}}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The rate of $\left[O_2\right]$ consumption has to be written.

Concept introduction:

Rate of the reaction is the change in the concentration of reactant or a product with time.

For a general reaction,

$aA+bB\to cC+dD$ , Here a, b, c and d are the coefficients.

$\text{Rate}\text{=}\text{-}\frac{\text{1}}{\text{a}}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}\text{=}\text{-}\frac{\text{1}}{\text{b}}\frac{\text{Δ}\left[\text{B}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{c}}\frac{\text{Δ}\left[\text{C}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{d}}\frac{\text{Δ}\left[\text{D}\right]}{\text{Δt}}.$

The negative sign indicates the reduction of concentration of reactant.

$\Delta \left[A\right],\Delta \left[B\right],\Delta \left[C\right]and\Delta \left[D\right]$ are the change in concentration of A, B, C and D over time period $\Delta t$