Chapter 6, Problem 6.24P

Interpretation Introduction

Interpretation:

The concentration of the carbanion has to be found using the steady-state approximation. The rate equation for the formation of the given product has to be derived.

Concept Introduction:

The rate of reaction is the quantity of formation of product or the quantity of reactant used per unit time. The rate of reaction doesn’t depend on the sum of amount of reaction mixture used.

The raise in molar concentration of product of a reaction per unit time or decrease in molarity of reactant per unit time is called rate of reaction and is expressed in units of $\text{mol/(L}\text{.s)}$ .

Steady state approximation: This method used to derive a rate law. The method is based on the assumption that one intermediate in the reaction mechanism is consumed as quickly as it is generated.

Example: Use the SSA approximate to calculate the concentration of the intermediate $\left(C\right)$ in the following sequence of elementary steps.

  $\begin{array}{l}A+B\underset{k_2}{\overset{k_1}{\rightleftharpoons }}C\\ C\stackrel{k_2}{\to }D\end{array}$

Where the rate constant $\left(k_r\right)$ is indicated, the overall reaction is.

$A+B\stackrel{}{\to }D$