Question

Acetaldehyde decomposes at \(750 \mathrm{~K}\) : \(\mathrm{CH}_{3} \mathrm{CHO} \rightarrow \mathrm{CO}+\mathrm{CH}_{4^{-}}\)The reaction is first order in acetaldehyde and the half life of the reaction is found to be 530 seconds. What is the rate constant for the reaction at this temperature? a. \(3.7 \times 10^{-3} \mathrm{~s}^{-1}\) b. \(3.13 \times 10^{3} \mathrm{~s}^{-1}\) c. \(1.3 \times 10^{-3} \mathrm{~s}^{-1}\) d. \(2.3 \times 10^{-3} \mathrm{~s}^{-1}\)

Short answer

The rate constant is \(1.3 \times 10^{-3} \, \mathrm{s}^{-1}\) (option c).

Step-by-step solution

Understand the First-Order Reaction

For a first-order reaction, the half-life ( t_{1/2} ) is independent of the initial concentration. The formula for the half-life of a first-order reaction is: \[ t_{1/2} = \frac{0.693}{k} \] where \( k \) is the rate constant we need to find.

Identify Given Values

From the problem, we know the half-life of the reaction is \( t_{1/2} = 530 \) seconds. We need to use this information to find \( k \).

Solve for Rate Constant

Rearrange the formula for half-life to solve for the rate constant: \[ k = \frac{0.693}{t_{1/2}} \]Substitute the given value for the half-life: \[ k = \frac{0.693}{530} \]

Calculate Rate Constant

Perform the calculation: \[ k = \frac{0.693}{530} = 0.0013075 \approx 1.3 \times 10^{-3} \, \mathrm{s}^{-1} \]

Choose the Correct Answer

The calculated rate constant \( k = 1.3 \times 10^{-3} \, \mathrm{s}^{-1} \) matches with option c. Therefore, option c is the correct answer.

Key concepts

Rate Constant Calculation

When dealing with chemical reactions, particularly first-order reactions, understanding the concept of the rate constant is crucial. The rate constant, often denoted as \( k \), is a measure of how quickly a reaction proceeds. For first-order reactions, the relationship between the rate constant and the half-life of the reaction is especially simple and elegant.

The half-life \( (t_{1/2}) \) of a reaction is a useful indicator of the time it takes for the concentration of a reactant to decrease to half its initial amount. For a first-order reaction, the half-life is given by the equation:

• \( t_{1/2} = \frac{0.693}{k} \)

This equation reveals that the half-life is independent of the initial concentration, a unique feature of first-order reactions. Instead, it is directly related to the rate constant. To find \( k \) when the half-life is known, you simply rearrange the formula:

• \( k = \frac{0.693}{t_{1/2}} \)

By substituting in the known half-life value, you can easily determine the rate constant, which is integral in predicting how fast the reaction occurs.

Half-Life of Reactions

The concept of half-life in chemical kinetics offers a straightforward way to understand how long a reaction takes to reach a certain stage. Specifically, the half-life of a reaction is the period required for the quantity of a reactant to reduce to half of its initial value. This concept is particularly significant for first-order reactions.

For first-order reactions, the half-life equation \( t_{1/2} = \frac{0.693}{k} \) makes it easy to compute or predict the duration of half-life if the rate constant is already known. This is because:

• The half-life is constant and does not depend on the starting concentration of the reactant.
• It provides a useful measure when comparing the speed or longevity of different chemical reactions.

In essence, the half-life serves as a clock that helps chemists judge how fast or slow a chemical reaction is. It is a salient factor considered in the design and analysis of both experimental and industrial chemical processes.

Chemical Kinetics

Chemical kinetics is the branch of chemistry that concerns the rates of chemical reactions. It helps chemists understand how different conditions affect the speed at which reactions take place. At the heart of chemical kinetics are rate laws and equations that quantify the rate at which reactants transform into products over time.

Understanding how reactions proceed involves:

• Identifying the order of the reaction, which influences the rate law, such as zero-order, first-order, or second-order.
• Calculating rate constants, such as \( k \), which indicate the speed of the reaction under specific conditions.
• Determining activation energies, which is the energy barrier that must be overcome for a reaction to proceed.

Chemical kinetics doesn’t just provide insight into reaction speeds. It extends to mechanisms which are step-by-step descriptions of how reactants turn into products, giving a deeper understanding of the process. Learning these aspects is essential for anyone looking to delve into fields such as pharmaceuticals, where reaction rates can dictate drug effectiveness or industrial applications where process efficiency is king.