Problem 26 - 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 \(0.0216 \mathrm{min}^{-1} .\) If the initial quantity of azomethane in the flask is \(2.00 \mathrm{g}\), how much remains after 0.0500 hour? What mass of \(\mathrm{N}_{2}\) is formed in this time?

Step-by-Step Solution

Answer

1.86 g of azomethane remains, and 0.0645 g of N2 is formed.

Step 1: Understand First-Order Kinetics

In a first-order reaction, the rate of reaction is directly proportional to the concentration of one reactant. The integrated rate law for a first-order reaction is given by the formula: \( [A]_t = [A]_0 \times e^{-kt} \), where \([A]_0\) is the initial concentration, \([A]_t\) is the concentration at time \(t\), \(k\) is the rate constant, and \(t\) is the time.

Step 2: Convert Mass to Moles

The initial mass of azomethane is \(2.00 \text{ g}\). The molar mass of azomethane (\(\text{CH}_3\text{N} = \text{NCH}_3\)) is 58.10 g/mol. Calculate the initial moles, \([A]_0\), using \([A]_0 = \frac{2.00 \text{ g}}{58.10 \text{ g/mol}} = 0.0344 \text{ mol}\).

Step 3: Calculate Time in Minutes

Convert the time from hours to minutes for consistency with the units of the rate constant. \(0.0500 \text{ hour} \times 60 \text{ min/hour} = 3.00 \text{ minutes}\).

Step 4: Calculate Remaining Moles

Use the first-order integrated rate law to calculate the amount of azomethane remaining after 3.00 minutes: \([A]_t = 0.0344 \text{ mol} \times e^{-0.0216 \text{ min}^{-1} \times 3.00 \text{ min}} = 0.0344 \text{ mol} \times e^{-0.0648} \approx 0.0321 \text{ mol}\).

Step 5: Convert Remaining Moles to Mass

Calculate the remaining mass of azomethane: \(\text{mass} = 0.0321 \text{ mol} \times 58.10 \text{ g/mol} = 1.86 \text{ g} \).

Step 6: Calculate Moles of N2 Formed

Since the stoichiometry of the decomposition is 1:1, the moles of \(\text{N}_2\) formed will be equal to the initial moles of azomethane minus the remaining moles: \(0.0344 \text{ mol} - 0.0321 \text{ mol} = 0.0023 \text{ mol} \text{ N}_2\).

Step 7: Convert Moles of N2 to Mass

Calculate the mass of \(\text{N}_2\) formed using its molar mass (28.02 g/mol): \(\text{mass of } \text{N}_2 = 0.0023 \text{ mol} \times 28.02 \text{ g/mol} = 0.0645 \text{ g} \).

Key Concepts

Rate Law

In the study of first-order kinetics, the rate law plays a crucial role. The rate law is an expression that relates the rate of a chemical reaction to the concentration of its reactants. For a first-order reaction like the decomposition of azomethane, the rate of reaction is directly proportional to the concentration of azomethane itself. This can be mathematically expressed as \( ext{Rate} = k[A] \), where \( k \) is the rate constant, and \( [A] \) represents the concentration of azomethane. This linear relationship ensures that as the concentration of azomethane decreases, so does the rate of the reaction, maintaining a consistent reaction speed relative to concentration changes.

Reaction Rate

The reaction rate is a measure of how quickly reactants are converted into products in a chemical process. It's typically expressed in terms of the change of concentration of a reactant or product over time. For the decomposition of azomethane, the reaction rate can be observed by measuring how fast the azomethane converts into nitrogen gas (\( ext{N}_2 \)) and ethane (\( ext{C}_2 ext{H}_6 \)). This reaction, being first-order, implies that the rate depends on the concentration of azomethane alone. As time progresses, the concentration diminishes logarithmically, reflected by the integrated rate equation \( [A]_t = [A]_0 e^{-kt} \). This equation allows you to calculate either the concentration at a particular time or the time required to reach a certain concentration.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships that exist in chemical formulas and reactions. In this exercise, the stoichiometry of the decomposition reaction \( ext{CH}_3 ext{N}= ext{NCH}_3 ightarrow ext{N}_2 + ext{C}_2 ext{H}_6 \) is straightforward since each molecule of azomethane decomposes into one molecule of nitrogen and one molecule of ethane. This 1:1 ratio simplifies calculations, as every mole of azomethane consumed results in an equivalent mole of \( ext{N}_2 \) produced. This stoichiometric relationship is crucial when using the mole method for finding the final quantities of particles post-reaction.

Molar Mass

Molar mass is an essential concept in chemistry, serving as the bridge between the mass of a substance and the amount in moles. It is the mass of one mole of a substance, usually expressed in g/mol. For azomethane, it is calculated as 58.10 g/mol, which allows for the conversion of its initial mass to moles. This conversion is critical since reaction rates and stoichiometric calculations are often mole-based. Molar mass also applies to products like \( ext{N}_2 \), which has a molar mass of 28.02 g/mol, and is employed in calculating the mass of \( ext{N}_2 \) generated in the reaction. Understanding molar mass ensures accurate translations between working quantities in laboratory settings and theoretical predictions.