Chapter 8, Problem 1P

Interpretation Introduction

Interpretation:

It should be explained that for the first order reaction, the half-life is inversely proportional to rate constant k. The constant of proportionality should be determined.

Concept introduction:

For 1^st order reaction, the rate can be determined as:

  $\begin{array}{c}\text{Rate = k }\left[\text{A}\right]\\ \text{= }-\text{d }\left[\text{A}\right]\text{ /d t}\end{array}$

Here, k is the rate constant

d [A] = change in concentration of the A

d t = change in time

Explanation of Solution

Time required for 1^st order reaction can be represented as:

  $\text{t = 2}\text{.303 /k log }\left[{\text{A}}_{\text{o}}\right]\text{ /}\left[\text{A}\right]$      .....(1)

Half-life of a substance indicates the consumption of half of the chemical substance in specific period of time thus,

  $\left[\text{A}\right]\text{ = }\left[{\text{A}}_{\text{o}}\right]\text{ / 2}$

And,

  ${\text{T = T}}_{\text{1/2}}$

Putting these values in equation (1)

  $\begin{array}{l}\begin{array}{l}\text{T = 2}\text{.303 /k log }\left[{\text{A}}_{\text{o}}\right]\text{ /}\left[\text{A}\right]\hfill \\ {\text{T}}_{\text{1/2}}\text{= }\left(\text{2}\text{.303 /k}\right)\times \left(\text{log 2}\right)\hfill \end{array}\\ {\text{T}}_{\text{1/2}}\text{= 0}\text{.6932 /k}\end{array}$

Therefore, the half-life is inversely proportional to rate constant k and constant of proportionality is 0.6932