Chapter 17, Problem 17D.1ST

Interpretation Introduction

Interpretation:

From the following data ${\text{A and E}}_{\text{a}}$ has to be determined.

  $\begin{array}{l}\text{T/K}\text{300}\text{350}\text{400}\text{450}\text{500}\\ {\text{k}}_{\text{r}}{\text{/(dm}}^{\text{3}}{\text{mol}}^{\text{-1}}{\text{s}}^{\text{-1}}\text{)}\text{7}{\text{.9×10}}^{\text{6}}\text{3}{\text{.0×10}}^{\text{7}}\text{7}{\text{.9×10}}^{\text{7}}\text{1}{\text{.7×10}}^{\text{8}}\text{3}{\text{.2×10}}^{\text{8}}\end{array}$

Concept Introduction:

Arrhenius Equation:

The temperature dependence characteristics of reaction is normally expressed mathematically by introducing two parameters, one representing the intercept and the other the slop of the straight line of a so-called ‘Arrhenius plot’ of $\text{ln k}$, against $\text{1/T}$ and writing the Arrhenius equation.

    $\text{lnk = lnA - }\frac{{\text{E}}_{\text{a}}}{\text{RT}}$

A is the parameter obtained from the intercept of the line at $\text{1/T = 0}$ is called the frequency factor.  The parameter ${\text{E}}_{\text{a}}$ is called the activation energy.  The two quantities are collectively known as Arrhenius parameters.

Figure.1

If the activation energy of a reaction is known, the value of a rate constant ${\text{k}}_{\text{2}}$ at a temperature ${\text{T}}_{\text{2}}$ can be predicted from its value ${\text{k}}_1$ at another temperature ${\text{T}}_1$

That is:

    $\frac{{\text{lnk}}_{\text{2}}}{{\text{lnk}}_{\text{1}}}\text{=}\frac{{\text{E}}_{\text{a}}}{\text{R}}\left[\frac{\text{1}}{{\text{T}}_{\text{1}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{2}}}\right]$