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: $$\mathrm{CH}_{3} \mathrm{N}=\mathrm{NCH}_{3}(\mathrm{g}) \rightarrow \mathrm{N}_{2}(\mathrm{g})+\mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{g})$$ The rate constant for this reaction at \(600 \mathrm{K}\) is 0.0216 \(\min ^{-1} .\) If the initial quantity of azomethane in the flask is \(2.00 \mathrm{g},\) how much remains after 0.0500 hour? What quantity of \(\mathrm{N}_{2}\) is formed in this time?

Step-by-Step Solution

Answer

1.87 g of azomethane remains, and 0.0616 g of \(\mathrm{N}_2\) is formed.

Step 1: Calculate Initial Moles of Azomethane

First, determine the molar mass of azomethane \( \mathrm{CH}_{3} \mathrm{N} = \mathrm{NCH}_{3} \). The atomic masses are: C = 12.01, H = 1.008, N = 14.01. Thus, the molar mass is \( 2(12.01) + 6(1.008) + 2(14.01) = 58.08 \ \mathrm{g/mol} \). The initial amount in moles is \( \frac{2.00}{58.08} \approx 0.0344 \ \mathrm{mol} \).

Step 2: Use the First-Order Reaction Formula

For a first-order reaction, the formula is \( [A] = [A]_0 e^{-kt} \). Here, \([A]_0\) is the initial concentration, \(k = 0.0216 \ \mathrm{min^{-1}}\), and \(t = 0.0500 \ \mathrm{hours} \times 60 \ \mathrm{minutes/hour} = 3 \ \mathrm{minutes}\).

Step 3: Calculate Remaining Azomethane

Substitute into the formula: \([A] = 0.0344 \times e^{-0.0216 \times 3}\). Calculate \(e^{-0.0216 \times 3} \approx 0.935\). Thus, \([A] \approx 0.0344 \times 0.935 \approx 0.0322 \ \mathrm{mol}\).

Step 4: Convert Moles to Grams

To find the mass of remaining azomethane, multiply by its molar mass: \(0.0322 \times 58.08 \approx 1.87 \ \mathrm{g}\).

Step 5: Calculate Quantity of \(\mathrm{N}_2\) Produced

Since azomethane decomposes to form \(\mathrm{N}_2\) with a 1:1 ratio, initial moles of azomethane (\(0.0344 \ \mathrm{mol}\)) decomposed equals \(0.0344 - 0.0322 = 0.0022 \ \mathrm{mol}\). The mass of \(\mathrm{N}_2\) formed is \(0.0022 \times 28.02 \ \mathrm{g/mol} \approx 0.0616 \ \mathrm{g}\).

Key Concepts

Understanding Rate Constant in Chemical Reactions

In the world of chemical reactions, the **rate constant** plays a crucial role in determining how quickly a reaction proceeds. For first-order reactions like the decomposition of azomethane, the rate constant is a specific value that helps quantify the speed of the reaction. It is represented by the symbol \(k\) and its unit is reciprocal time, (e.g., \(\mathrm{min^{-1}}\)).

**Key Points of Rate Constant:**

• For azomethane decomposition, the given rate constant at 600 K is \(0.0216 \min^{-1}\).
• The larger the rate constant, the faster the reaction takes place.
• It is a crucial indicator for comparing the reactivity of similar substances under predefined conditions.
• Temperature impacts the rate constant, which is why it is essential to specify the conditions (like 600 K in this case) when discussing it.

The rate constant helps us predict how much of a reactant will be consumed over a period or how much product will be formed, which is vital for industrial and laboratory applications.

Exploring Reaction Kinetics of Azomethane Decomposition

**Reaction kinetics** involves studying the rates of chemical processes and understanding the factors affecting these rates. In the case of azomethane decomposition, we are dealing with a first-order reaction, which means the rate depends linearly on the concentration of azomethane.

The formula for first-order reactions is: \[ [A] = [A]_0 e^{-kt} \] where:

• \([A]_0\) is the initial concentration (or amount) of the reactant.
• \(k\) is the rate constant.
• \(t\) is the time over which the reaction occurs.

This equation helps predict the concentration of the reactant left after a certain period. For our exercise, given the initial moles of azomethane and the rate constant, we used this formula to calculate how much azomethane remained after 0.0500 hours.
The kinetics of a reaction describe not just these calculations but also the complete narrative of how reactants are transformed into products over time. Understanding kinetics allows chemists to manipulate reaction conditions to achieve desired outcomes in accordance with practical requirements.

Azomethane Decomposition and its Chemical Significance

**Azomethane decomposition** is a classic example of a first-order chemical reaction where a simple molecule breaks down into smaller molecules. In this particular reaction, azomethane (\(\mathrm{CH}_{3} \mathrm{N} = \mathrm{NCH}_{3}\)) decomposes to form nitrogen gas (\(\mathrm{N}_2\)) and ethane (\(\mathrm{C}_2\mathrm{H}_6\)).

**Why Azomethane Decomposes:**

• Like many organic compounds with weak bonds, it requires minimal energy to break apart at elevated temperatures, thus it decomposes readily when heated.
• The breaking of bonds in azomethane releases nitrogen gas – a stable and inert compound – along with ethane, a simpler hydrocarbon.
• This decomposition process is significant in understanding both environmental factors and practical uses where controlled decomposition is desired.

This reaction highlights the importance of understanding chemical decomposition, both in nature and in industry. By studying such simple reactions, researchers can learn to handle similar processes involving other organic substances. The knowledge gained from azomethane decomposition informs safety protocols, environmental considerations, and can even apply to fields like energetic materials and propellant chemistry.