Problem 27 - Chemistry and Chemical Reactivity

Question

Gaseous azomethane, \(\mathrm{CH}_{3} \mathrm{N}=\mathrm{NCH}_{3},\) decomposes in a first-order reaction when heated: $$\mathrm{CH}_{3} \mathrm{N}=\mathrm{NCH}_{3}(\mathrm{g}) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+\mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{g})$$ The rate constant for this reaction at \(425^{\circ} \mathrm{C}\) is \(40.8 \mathrm{min}^{-1}\) If the initial quantity of azomethane in the flask is \(2.00 \mathrm{g}\) how much remains after 0.0500 min? What quantity of \(\mathrm{N}_{2}\) is formed in this time?

Step-by-Step Solution

Answer

0.011 moles of azomethane remain; 0.023 moles of \(\text{N}_2\) are formed.

Step 1: Determine Initial Concentration

First, calculate the initial concentration of azomethane (\(\text{CH}_3\text{N} = \text{NCH}_3\)) in moles. Convert the mass to moles using its molar mass. Molar mass of azomethane \(= 2(12.01) + 2(14.01) + 6(1.008) = 58.08 \, \text{g/mol}.\) Initial moles of azomethane \(= \frac{2.00 \, \text{g}}{58.08 \, \text{g/mol}} = 0.0344 \, \text{moles}.\)

Step 2: Apply First-order Kinetics Equation

Use the first-order kinetics equation to calculate the remaining quantity of azomethane after \(0.0500 \, \text{min}\).The first-order equation is \(\ln \left(\frac{[A]_0}{[A]}\right) = kt,\)where \([A]_0\) is the initial concentration, \([A]\) is the concentration at time \(t\), and \(k\) is the rate constant.Rearrange the formula to solve for \([A]\):\[[A] = [A]_0 \cdot e^{-kt}.\]Insert the values:\([A] = 0.0344 \, \text{moles} \times e^{-40.8 \, \text{min}^{-1} \times 0.0500 \, \text{min}}.\)

Key Concepts

First-order Reaction

A first-order reaction is characterized by a reaction rate that is directly proportional to the concentration of one reactant. In the context of the decomposition of azomethane, this means that the rate at which the reaction occurs is dependent solely on the concentration of azomethane. The mathematical representation of a first-order reaction can be expressed with the equation: \[ ext{Rate} = k[A] \]where \( k \) is the rate constant and \([A]\) is the concentration of the reactant. One critical feature of first-order reactions is that they have a constant half-life, regardless of the initial concentration of the reactant. This means that the time it takes for half of the reactant to decompose remains constant throughout the reaction. Understanding this principle helps in predicting how long a reaction will take under various conditions. It's especially crucial in our exercise where time is a key determinant of the concentration of remaining substances.

Rate Constant

The rate constant, often denoted as \( k \), is a crucial component in the realm of chemical kinetics, especially for first-order reactions. It provides insight into how fast or slow a reaction proceeds. For our azomethane decomposition problem, the rate constant at \(425^{\circ} \mathrm{C}\) is given as \(40.8 \, \mathrm{min}^{-1}\). Such a high value of \(k\) indicates a fast reaction, showing how quickly azomethane breaks down into nitrogen and ethane when heated to this temperature. Understanding the rate constant aids in predicting the outcome of reactions:

• The higher the value of \( k \), the faster the reaction occurs.
• The unit of \( k \) indicates the order of reaction. Here it is \( \mathrm{min}^{-1} \), which is typical for first-order reactions.

Knowing \( k \) helps in calculating how much reactant will remain or how much product will form over time, exactly what we did in our given step-by-step solution.

Reaction Mechanism

The reaction mechanism is a detailed step-by-step description of the pathway a chemical reaction follows from reactants to products. For the decomposition of azomethane, it involves the breaking down of the azomethane molecule into nitrogen (\(\text{N}_2\)) and ethane (\(\text{C}_2\text{H}_6\)). While the exercise simplifies the process into a single step, real-life reaction mechanisms often involve multiple intermediate stages and even the formation of short-lived species that do not appear in the overall balanced equation. Understanding these mechanisms allows chemists to:

• Predict the conditions needed to optimize the reaction speed.
• Identify possible by-products and design ways to minimize them.
• Develop catalysts that can reduce energy requirements for the reaction.

In educational exercises, focusing on the main reactants and products helps one understand the core changes occurring during the reaction.

Chemical Decomposition

Chemical decomposition is a type of chemical reaction where one substance breaks down into two or more simpler ones. In the azomethane reaction, this decomposition transforms one complex molecule into simpler molecules like nitrogen and ethane. Decomposition reactions are fundamental in chemistry, often driven by heat (as in our exercise example), radiation, or catalysts. They are pivotal in various industries for processes like refining metals, processing foods, and even in biochemical pathways. In a student's context, understanding decomposition includes:

• Recognizing the energy input needed to break bonds in the original compound.
• Predicting the resulting simpler molecules, which often require familiarity with molecular structures.
• Learning how temperature and catalysts can influence the rate of decomposition.

By acknowledging these reactions in practical exercises, we help build a foundation for more intricate chemical engineering and biochemistry studies.