Chapter 16.4, Problem 16.8BFP

(a)

Interpretation Introduction

Interpretation : The time should be calculated if the initial concentration of iodine is $\text{0}\text{.075 M}$ to drop $\text{0}\text{.015 M}$ and the rate constant is $0.80\text{ L/mos}\cdot \text{s}$ at $\text{10 }°\text{C}$ .

Concept Introduction :

The integrated rate law for the second-order reaction can be shown as:

  $\frac{ 1}{{\left[A\right]}_t}=k\times t+\frac{ 1}{{\left[A\right]}_0}$

Where,

  ${\left[A\right]}_0$ = Initial concentration

  $k$ = Rate constant

  $t$ = Time

  ${\left[A\right]}_t$ = Concentration after time $t$

(b)

Interpretation Introduction

Interpretation : The initial concentration should be calculated with a half-life of $45\text{ s}$ if the initial concentration of iodine is $\text{0}\text{.075 M}$ to drop $\text{0}\text{.015 M}$ and the rate constant is $0.80\text{ L/mos}\cdot \text{s}$ at $\text{10 }°\text{C}$ .

Concept Introduction :

The order of reaction can be defined as the number raised as the power of reactant molecules in the rate law of the reaction. According to rate law, the rate of reaction is directly proportional to the concentration of reactant molecules. Based on the order of reaction, the reactions can be classified as zero-order reactions, first-order reactions, and second-order reactions. The rate constant for each type of reaction also depends on the initial concentration and half-life.

For second order reaction: $t1/2=\frac{1}{k[A]}$