Chapter 6, Problem 6H.4E

Interpretation Introduction

Interpretation:

The activation energy $\left(E_a\right)$ for the given reaction has to be calculated.

Concept Introduction:

• Activation energy: It is defined as the minimum energy required by the reacting species in order to undergo chemical reaction.
• Activation energy can be represented as $E_a$. The unit of activation energy is $\text{kJ/mol}$.
• Catalyst: The catalyst is a chemical substance that increases the rate of the reaction without participating in the reaction by reducing the activation energy of the reaction.

Integrated rate law for first order reaction:

Consider A as substance, that gives the product based on the equation,

    $\text{aA}\to \text{products}$

Where a= stoichiometric co-efficient of reactant A.

Consider the reaction has first-order rate law,

    $\text{Rate=-}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}\text{=k}\left[\text{A}\right]$    

The integrated rate law equation can be given as,

    $\text{ln}\frac{{\left[\text{A}\right]}_{\text{t}}}{{\left[\text{A}\right]}_{\text{o}}}\text{=-kt}$

The above expression is called integrated rate law for first order reaction.

Half-life period:

Half-life is defined as the time taken for amount of the sample to reduce to half of its starting value.  The symbol is ${\text{t}}_{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}$.

$t_{\frac{1}{2}}=\frac{\ln \left(2\right)}{k}$

Half-life for a first order reaction is, $t_{\frac{1}{2}}=\frac{0.693}{k}$