Textbook Problem

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The decomposition of nitrogen dioxide at a high temperature

NO₂(g) → NO(g) + ½ O₂(g)

is second-order in this reactant. The rate constant for this reaction is 3.40 L/mol · min. Determine the time needed for the concentration of NO₂ to decrease from 2.00 mol/L to 1.50 mol/L.

Interpretation Introduction

Interpretation: The time required for the reaction has to be calculated.

Concept introduction:

Rate law or rate equation: The relationship between the reactant concentrations and reaction rate is expressed by an equation.

$\begin{array}{l}\text{aA + bB}\to x\text{X}\\ {\text{Rate of reaction = k [A]}}^{\text{m}}{\text{[B]}}^{\text{n}}\end{array}$

Integrated rate law for second order reactions:

Taking in the example of following reaction,

    $\text{aA}\to \text{products}$

And the reaction follows second order rate law,

Then the relationship between the concentration of $\text{A}$ and time can be mathematically expressed as,

    $\frac{\text{1}}{{\left[\text{A}\right]}_{\text{t}}}\text{=kt+}\frac{\text{1}}{{\left[\text{A}\right]}_{\text{0}}}$

The above expression is called as integrated rate for second order reactions.

Explanation of Solution

The time required is calculated as,

  $\begin{array}{l}\frac{\text{-Δ[R]}}{\text{Δt}}{\text{ = k[R]}}^{\text{2}}\text{,}\\ \\ \text{The relation can be transformed into,}\\ \\ \frac{\text{1}}{{\text{[R]}}_{\text{t}}}\text{ - }\frac{\text{1}}{{\text{[R]}}_{\text{0}}}\text{ = kt}\\ \\ \text{Given:}\\ \\ {\left[{\text{NO}}_{\text{2}}\right]}_{\text{t}}\text{ = 1}\text{.50 mol/L}\\ {\left[{\text{NO}}_{\text{2}}\right]}_{\text{0}}\text{ = 2}\text{.00 mol/L}\\ \text{k = 3}\text{.40 L/mol}\text{.min }\\ \text{t = ?}\\ \\ \text{Therefore,}\\ \\ \frac{\text{1}}{{\text{[R]}}_{\text{t}}}\text{ - }\frac{\text{1}}{}\end{array}$