Problem 100 - Chemistry The Molecular Science

Question

The rate constant for decomposition of azomethane at $$ \begin{aligned} 425^{\circ} \mathrm{C} \text { is } 0.68 \mathrm{~s}^{-1} & \\ \mathrm{CH}_{3} \mathrm{~N}=\mathrm{NCH}_{3}(\mathrm{~g}) \longrightarrow \mathrm{N}_{2}(\mathrm{~g})+\mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g}) \end{aligned} $$ (a) Based on the units of the rate constant, determine if this reaction is zeroth-, first-, or second-order. (b) If \(2.0 \mathrm{~g}\) azomethane is placed in a \(2.0-\mathrm{L}\) flask and heated to \(425^{\circ} \mathrm{C},\) calculate the mass of azomethane that remains after \(5.0 \mathrm{~s}\). (c) Calculate how long it takes for the mass of azomethane to drop from \(2.0 \mathrm{~g}\) to \(0.24 \mathrm{~g}\). (d) Calculate the mass of nitrogen that would be found in the flask after \(0.50 \mathrm{~s}\) of reaction.

Step-by-Step Solution

Answer

(a) First-order; (b) 0.0676 g; (c) 3.57 s; (d) 0.28 g of nitrogen.

Step 1: Determine the Reaction Order

The rate constant given is 0.68 s^{-1}. The units for the rate constant can help determine the order of a reaction. For a zero-order reaction, the units are mol/(L*s); for first-order, they are s^{-1}; and for second-order, they could be L/(mol*s). Since the units here are s^{-1}, this indicates a first-order reaction.

Step 2: Calculate Remaining Azomethane using First-order Kinetics

The first-order kinetics formula is \( N_t = N_0 e^{-kt} \). We know that \( k = 0.68 \text{ s}^{-1} \), \( N_0 = 2.0 \text{ g} \), and \( t=5.0 \text{ s} \). Substituting these values gives us:\[N_t = 2.0 e^{-0.68 \cdot 5}.\]Calculate:\[N_t = 2.0 e^{-3.4} \approx 2.0 \times 0.0338 = 0.0676 \text{ g}.\]

Step 3: Calculate Time for Azomethane to Reduce to 0.24g

To find the time when remaining amount \( N_t = 0.24 \text{ g} \), use the same formula:\[0.24 = 2.0 e^{-0.68t}.\]Solving for \( t \) involves taking the natural logarithm:\[ln(0.24/2.0) = -0.68t \ln(0.12) = -0.68t \t \approx \frac{ln(0.12)}{-0.68} \approx 3.57 \text{ s}.\]

Step 4: Calculate Mass of Nitrogen after 0.50s

Use the same formula with \( t = 0.50 \text{ s} \) to find \( N_t \):\[N_t = 2.0 e^{-0.68 \times 0.50} \approx 2.0 e^{-0.34} \approx 2.0 \times 0.7125 = 1.425 \text{ g}.\]The initial azomethane amount was 2.0 g; thus, \( 2.0 - 1.425 = 0.575 \text{ g} \) of azomethane has decomposed. Since 1 mole of azomethane produces 1 mole of nitrogen, \( 0.575 \text{ g} \) of azomethane decomposing equates to \( \frac{0.575}{58.07} \approx 0.0099 \text{ mol} \) of nitrogen. Mass of nitrogen:\[0.0099 \text{ mol} \times 28.02 \text{ g/mol} \approx 0.28 \text{ g}.\]

Key Concepts

Azomethane Decomposition

Azomethane is a chemical compound with the formula \( \mathrm{CH}_{3} \mathrm{~N} = \mathrm{NCH}_{3} \) that can undergo a decomposition reaction. In this particular reaction, azomethane breaks down into nitrogen gas \( \mathrm{N}_2 \) and ethane \( \mathrm{C}_2\mathrm{H}_6 \). This type of reaction is categorized as a decomposition reaction because a single compound breaks down into two or more simpler substances.
The chemical equation for azomethane decomposition is: \[ \mathrm{CH}_{3} \mathrm{~N} = \mathrm{NCH}_{3} (\mathrm{~g}) \rightarrow \mathrm{N}_{2}(\mathrm{~g}) + \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g}) \] When considering its practical applications, such reactions are often studied to understand reaction mechanisms and the conditions under which they occur. Azomethane decomposition is particularly interesting because it provides insights into breaking chemical bonds and the creation of new substances from pre-existing ones.

First-Order Kinetics

The decomposition of azomethane follows first-order kinetics. Kinetics refers to the rate at which a chemical reaction occurs, and a first-order reaction specifically means that the rate is directly proportional to the concentration of one reactant.
The mathematical expression for a first-order reaction is \[ N_t = N_0 e^{-kt} \] where:

• \( N_t \) is the remaining concentration or amount of the reactant at time \( t \),
• \( N_0 \) is the initial concentration or amount of the reactant,
• \( k \) is the rate constant, and
• \( t \) is the time elapsed since the start of the reaction.

This equation allows us to predict how much of a reactant will be left at any given time. In the case of azomethane decomposition, it helps understand how quickly azomethane is being converted into nitrogen and ethane.

Rate Constant

The rate constant \( k \) is a crucial parameter in chemical kinetics that defines the speed of a reaction. For azomethane decomposition at \( 425^\circ \mathrm{C} \), the rate constant is given as \( 0.68 \mathrm{~s}^{-1} \).
In a first-order reaction, the units for the rate constant are \( \mathrm{s}^{-1} \), which signifies that the reaction's rate is proportional to the concentration of the reactant. The value of \( k \) provides insight into how fast the reaction proceeds under certain conditions. A larger \( k \) value would indicate a faster reaction.
To find out how the reaction behaves over time, one can use the rate constant in the first-order kinetics equation. This helps to determine various factors, such as how much reactant remains after a certain period or how long it takes for the reactant to reach a specific concentration.

Chemical Kinetics

Chemical kinetics is the branch of chemistry that studies the rates of chemical processes. Understanding these rates allows chemists to infer about reaction mechanisms and determine optimal conditions for reactions.
In the context of azomethane decomposition, chemical kinetics involves studying how quickly azomethane is transformed into nitrogen and ethane and what factors influence this transformation.
Kinetics can involve:

• Temperature: Higher temperatures generally increase reaction rates.
• Reaction Medium: The solvent or medium can affect the speed of the reaction.
• Concentration: For a first-order reaction, the concentration of the reactant directly impacts the rate.
• Presence of Catalysts: Catalysts can drastically affect a reaction's rate without altering the overall chemical equilibrium.

The study of chemical kinetics helps in designing and optimizing industrial processes, understanding biological systems, and developing new materials and fuels.