Question

Acetone is one of the most important solvents in organic chemistry, used to dissolve everything from fats and waxes to airplane glue and nail polish. At high temperatures, it decomposes in a first-order process to methane and ketene \(\left(\mathrm{CH}_{2}=\mathrm{C}=\mathrm{O}\right) .\) At \(600^{\circ} \mathrm{C},\) the rate constant is \(8.7 \times 10^{-3} \mathrm{~s}^{-1}\). (a) What is the half-life of the reaction? (b) How long does it take for \(40 . \%\) of a sample of acetone to decompose? (c) How long does it take for \(90 . \%\) of a sample of acetone to decompose?

Short answer

Half-life: 79.66 s; Time for 40% decomposition: 65.08 s; Time for 90% decomposition: 265.89 s.

Step-by-step solution

- Formula for half-life

The half-life for a first-order reaction is given by the formula: \[ t_{1/2} = \frac{0.693}{k} \]where \( k \) is the rate constant.

- Calculate half-life

Substitute the given rate constant into the formula: \[ t_{1/2} = \frac{0.693}{8.7 \times 10^{-3} \mathrm{s}^{-1}} \] Calculate the value: \[ t_{1/2} \approx 79.66 \; \text{s} \]

- Formula for time for decay

For a first-order reaction, the time to decay some fraction of the sample can be calculated using the integrated rate law: \[ t = \frac{\ln \left( \frac{[A]_0}{[A]} \right)}{k} \] where \( [A]_0 \) is the initial concentration and \( [A] \) is the remaining concentration after time \( t \).

- Determine time for 40% decomposition

For 40% decomposition, 60% of the sample remains. Therefore, the fraction remaining is 0.60. Substitute into the formula: \[ t = \frac{\ln \left( \frac{1.0}{0.60} \right)}{8.7 \times 10^{-3} \mathrm{s}^{-1}} \] Calculate the value: \[ t \approx 65.08 \; \text{s} \]

- Determine time for 90% decomposition

For 90% decomposition, 10% of the sample remains. Therefore, the fraction remaining is 0.10. Substitute into the formula: \[ t = \frac{\ln \left( \frac{1.0}{0.10} \right)}{8.7 \times 10^{-3} \mathrm{s}^{-1}} \] Calculate the value: \[ t \approx 265.89 \; \text{s} \]

Key concepts

Half-Life Formula

The half-life of a reaction is the time it takes for half the amount of a reactant to be consumed. This is a crucial concept in kinetics. For first-order reactions, the relationship simplifies through a specific formula. The half-life formula for first-order reactions is derived from the rate law and is given by:
\[ t_{1/2} = \frac{0.693}{k} \]

Here, \( k \) is the rate constant. This clear relationship signifies that the half-life is inversely proportional to the rate constant. So, if the rate constant is high, the half-life is short, meaning the reaction happens faster. Understanding this formula helps in calculating how quickly a reactant will reduce by half, which is often needed in practical scenarios.

Rate Constant

The rate constant, represented by \( k \), is a proportionality constant in the rate law that connects the reaction rate to the concentrations of reactants. For any chemical reaction, the rate constant gives an idea of how fast the reaction proceeds.

In first-order reactions, the rate constant has units of \[ \text{s}^{-1} \]

This indicates how frequently the reaction occurs on a per-second basis. For the decomposition of acetone at \( 600^{\circ}C \), the given rate constant is \( 8.7 \times 10^{-3}\ \text{s}^{-1} \). Higher rate constants mean reactions occur more quickly. By knowing \( k \), one can also use it to calculate half-life and determine the required amount of time for a reactant's concentration to reach a certain level.

Integrated Rate Law

The integrated rate law for first-order reactions is essential for calculating the time it takes for a reactant to reach a specific concentration. This law is given by the equation:
\[ t = \frac{\ln \left( \frac{[A]_0}{[A]} \right)}{k} \]

In this formula:

• \( [A]_0 \) is the initial concentration of the substance.
• \( [A] \) is the remaining concentration after time \( t \).
• \( k \) is the rate constant.

This equation shows a logarithmic relationship between the initial and current concentrations over time. It helps to solve how long it takes to reduce a given concentration by a certain fraction, such as decomposing acetone. Parameters such as the given rate constant, initial and remaining concentrations, are key inputs in using the integrated rate law.

Decomposition Reaction

A decomposition reaction involves the breaking down of a single reactant into two or more products. In the context of acetone, the decomposition at high temperatures into methane and ketene is a classic first-order reaction.

This type of reaction follows a predictable kinetic pattern, where the rate depends solely on the concentration of the decomposing reactant. In a first-order process, the composition of the mixture changes exponentially over time.

Understanding decomposition reactions helps in diverse fields like organic chemistry, environmental science, and even industrial processes. It explains how substances break down and form new products. For instance, knowing how acetone breaks down at \( 600^{\circ}C \) can be practically applied in chemical manufacturing and processing.