Chapter 12, Problem 12.109CHP

Interpretation Introduction

Interpretation:

The time required for the concentration of ${\text{N}}_{\text{2}}{\text{O}}_{\text{5}}$ to drop from $\text{0}\text{.030}\text{M}\text{to}\text{0}\text{.015}\text{M}$ and $\text{0}\text{.480}\text{M}\text{to}\text{0}\text{.015}\text{M}$ has to be calculated.

Concept introduction:

Integrated rate law for a first order reaction:

A first order reaction is defined as the reaction whose rate depends only on the concentration of a single reactant raised to the first power.

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

Integrated rate law for a first order reaction is given below

$\begin{array}{l}\text{ln}\frac{{\left[\text{A}\right]}_{\text{t}}}{{\left[\text{A}\right]}_{\text{0}}}\text{=}\text{-}\text{kt}\\ \text{where,}\\ \text{ln}\text{-}\text{natural}\text{logarithm}\\ {\left[\text{A}\right]}_{\text{0}}\text{-}\text{concentration}\text{of}\text{A}\text{at}\text{time}\text{t}\text{=}\text{0}\\ {\left[\text{A}\right]}_{\text{t}}\text{-}\text{concentration}\text{of}\text{A}\text{after}\text{time}\text{t}\\ \text{k}\text{-}\text{rate}\text{constant}\end{array}$