Problem 28 - Chemistry in 30 days

Question

The rate of a first order reaction is \(1.5 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{\circ}\) at \(0.5 \mathrm{x}\) reactant. The half-life of the reaction is (a) \(7.53 \mathrm{~min}\) (b) \(0.383 \mathrm{~min}\) (c) \(23.1 \mathrm{~min}\) (d) \(8.73\) min

Step-by-Step Solution

Answer

The half-life of the reaction is 23.1 min, option (c).

Step 1: Understand First Order Reaction

A first-order reaction has a rate that depends linearly on the concentration of one reactant. Its rate equation can be written as \( r = k[A] \), where \( r \) is the rate of the reaction, \( k \) is the rate constant, and \( [A] \) is the concentration of reactant A.

Step 2: Use Given Rate to Calculate Rate Constant

The given reaction rate is \( 1.5 \times 10^{-2} \ \mathrm{mol \ L^{-1} \ min^{-1}} \) and the concentration of the reactant is \( 0.5 \ \mathrm{mol \ L^{-1}} \). Substitute these values into the rate equation \( r = k[A] \) to find the rate constant \( k \). So, \( 1.5 \times 10^{-2} = k \times 0.5 \). Solve for \( k \):\[ k = \frac{1.5 \times 10^{-2}}{0.5} = 3 \times 10^{-2} \ \mathrm{min^{-1}} \].

Step 3: Calculate Half-Life for First Order Reaction

The half-life of a first-order reaction is given by the formula \( t_{1/2} = \frac{0.693}{k} \). Use the rate constant \( k = 3 \times 10^{-2} \ \mathrm{min^{-1}} \) calculated previously:\[ t_{1/2} = \frac{0.693}{3 \times 10^{-2}} = 23.1 \ \mathrm{min} \].

Step 4: Select Correct Answer Choice

Compare the calculated half-life to the provided options. The correct answer is the one that matches the calculation: (c) 23.1 min.

Key Concepts

Rate Equation

In a first-order reaction, one of the most fundamental concepts is the rate equation. This equation describes how the reaction rate is directly proportional to the concentration of a single reactant. It's typically represented as \( r = k[A] \). In this equation:

• \( r \) represents the rate of the reaction.
• \( k \) is the rate constant, a unique value for each reaction at a given temperature.
• \([A] \) denotes the concentration of the reactant "A".

The simplicity of this equation makes first-order reactions relatively easy to analyze and understand. By measuring how quickly the concentration of the reactant decreases over time, you can determine the rate constant, \( k \). Understanding this relationship is key to solving many problems in chemical kinetics.

Rate Constant

The rate constant, \( k \), is a crucial part of understanding how fast a reaction proceeds. In the context of a first-order reaction, \( k \) can be thought of as a proportionality factor linking the rate of reaction with the concentration of a single reactant.
One of the key aspects of \( k \) is that it remains constant for a given reaction at a particular temperature, even though reactant concentrations may change.

• To calculate \( k \), use the rate equation \( r = k[A] \).
• If you know the rate of the reaction and the concentration of the reactant, you can easily find \( k \) by rearranging the equation to \( k = \frac{r}{[A]} \).

In the given example, by knowing the rate \( 1.5 \times 10^{-2} \, \mathrm{mol \ L^{-1} \ min^{-1}} \) and a concentration of \( 0.5 \, \mathrm{mol \ L^{-1}} \), we can compute \( k \) to be \( 3 \times 10^{-2} \, \mathrm{min^{-1}} \). This value is essential for further calculations, such as determining the half-life.

Half-Life Calculation

The half-life of a reaction provides valuable insight into how long a reactant takes to reduce to half its initial concentration. For first-order reactions, calculating the half-life is straightforward due to the relationship: \[ t_{1/2} = \frac{0.693}{k} \] This formula highlights a crucial feature of first-order reactions: the half-life is independent of the initial concentration of the reactant. It depends solely on the rate constant \( k \).

• Simply plug in the value of \( k \) to find the half-life.
• For \( k = 3 \times 10^{-2} \, \mathrm{min^{-1}} \), substitute it into the formula to get:

\[ t_{1/2} = \frac{0.693}{3 \times 10^{-2}} = 23.1 \, \mathrm{min} \] This calculation indicates that regardless of how much reactant you start with, it will take 23.1 minutes for half of it to decompose. Understanding how to determine the half-life helps predict how the reaction progresses over time.