Question

For the first order reaction, \(2 \mathrm{~N}_{2} \mathrm{O}(\mathrm{g}) \rightarrow 2 \mathrm{~N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\), what is the concentra- tion of \(\mathrm{N}_{2} \mathrm{O}\) after 3 half-lives if \(0.15 \mathrm{~mol}\) of \(\mathrm{N}_{2} \mathrm{O}\) is initially placed into in a \(1.00\) litre reaction vessel? a. \(9.1 \times 10^{-2} \mathrm{M}\) b. \(0.9 \times 10^{-2} \mathrm{M}\) c. \(2.68 \times 10^{-2} \mathrm{M}\) d. \(1.9 \times 10^{-2} \mathrm{M}\)

Short answer

d. \(1.9 \times 10^{-2} \text{ M}\)

Step-by-step solution

Understand the Half Life Concept for First-Order Reactions

In a first-order reaction, the half-life is the time required for the concentration of the reactant to decrease by half. If you start with concentration \([A_0]\), after one half-life, the concentration reduces to \([A_0]/2\). After two half-lives, it becomes \([A_0]/4\), and after three half-lives, it becomes \([A_0]/8\).

Calculate the Initial Concentration

The initial concentration \([A_0]\) is found by dividing the number of moles by the volume of the reaction vessel. Given that you start with 0.15 mol of \( \mathrm{N}_2\mathrm{O} \) in a 1.00 L vessel, \( [A_0] = \frac{0.15 \text{ moles}}{1.00 \text{ L}} = 0.15 \text{ M} \).

Apply the Formula for Three Half-Lives

After three half-lives, the concentration \([A]\) of \( \mathrm{N}_2\mathrm{O} \) is given by: \[ [A] = \frac{[A_0]}{2^n} \] where \( n \) is the number of half-lives. Here, \( n = 3 \), so: \[ [A] = \frac{0.15 \text{ M}}{2^3} = \frac{0.15 \text{ M}}{8} = 0.01875 \text{ M} \]

Match with Given Options

The calculated concentration \(0.01875 \text{ M}\) must be matched to the closest given option. Option "d. \(1.9 \times 10^{-2} \text{ M}\)" is approximately equal to \(0.01875 \text{ M}\).

Key concepts

Understanding Half-Life in First-Order Reactions

The concept of half-life is crucial in understanding how reactants diminish over time in first-order reactions. In a first-order reaction, the half-life is the time it takes for the concentration of a reactant to reduce to half its initial value. This characteristic makes the half-life for first-order reactions independent of the concentration.

Every half-life period, the concentration continues to halve. For instance:

• After one half-life, the initial concentration \([A_0]\) becomes \([A_0]/2\).
• After two half-lives, it reduces to \([A_0]/4\).
• After three half-lives, it becomes \([A_0]/8\).

With each passing half-life, the amount remains a fraction, allowing us to calculate the remaining concentration easily through a consistent formula. This is particularly useful when predicting how much of a substance will remain after several half-lives.

Basics of Reaction Kinetics

Reaction kinetics is the study of the speed of chemical reactions and the mechanisms behind them. In first-order reactions, kinetics helps us understand that the rate at which a reactant transforms into a product depends on the concentration of the reactant.

The rate law for a first-order reaction can be summarized as \( Rate = k[A] \), where \( k \) is the rate constant, and \([A]\) represents the concentration of the reactant. This equation highlights that the reaction speed is directly proportional to the concentration, meaning a higher concentration speeds up the reaction. However, as the reaction progresses, the reactant's concentration decreases, resulting in a slowing reaction rate.

This principle allows chemists to predict the dynamics of reactions over time and to envisage the changes occurring within the reaction vessel.

Performing Concentration Calculations

Calculating concentrations at various stages of a reaction is an essential part of reaction kinetics. For the exercise presented, determining the initial concentration is the first step. This is typically achieved by dividing the number of moles of the reactant by the volume of the container.

Given that the initial number of moles for \( \mathrm{N}_2\mathrm{O} \) was \( 0.15 \) mol in a 1.00 L vessel, the initial concentration \([A_0]\) is calculated as follows: \[ [A_0] = \frac{0.15 \text{ moles}}{1.00 \text{ L}} = 0.15 \text{ M} \]

To find the concentration after a given number of half-lives, the formula \([A] = \frac{[A_0]}{2^n}\) where \( n \) is the number of half-lives is applied. For three half-lives, the concentration is: \[ [A] = \frac{0.15 \text{ M}}{8} = 0.01875 \text{ M} \]

This calculation technique allows us to determine how much of a substance is left at any stage of a reaction and match it to specified options or scenarios as needed.