Question

When heated to \(600^{\circ} \mathrm{C}\), acetone \(\left(\mathrm{CH}_{3} \mathrm{COCH}_{3}\right)\) decomposes to give . CO and various hydrocarbons. The reaction is found to be first order in acetone concentration with a half- life of 81 sec. Given at \(600^{\circ} \mathrm{C}\) a 1-liter container into which acetone is injected at \(0.48 \mathrm{~atm}\), approximate how long would it take for the acetone pressure to drop to \(0.45 \mathrm{~atm}\) ?

Short answer

The time it takes for the acetone pressure to drop to 0.45 atm at \(600^{\circ} \mathrm{C}\) in the 1-liter container is approximately \(15.39 \, seconds\).

Step-by-step solution

Write the first-order reaction equation

For a first-order reaction, the rate law can be written as: \(rate = k[A]\) where \(rate\) is the rate of reaction, \(k\) is the rate constant, and \([A]\) is the concentration of the reacting substance (acetone in this case). Next, we can use the half-life to relate the rate constant.

Evaluate the rate constant using half-life

For a first-order reaction, the half-life (\(t_{1/2}\)) and the rate constant (\(k\)) are related by the equation: \[t_{1/2} = \frac{0.693}{k}\] We are given the half-life as 81 seconds, so we can solve for \(k\): \[k = \frac{0.693}{81}\]

Write the integrated rate law equation for first-order reaction

For a first-order reaction with respect to concentration, the integrated rate law equation can be written as follows: \[ln\frac{[A]_0}{[A]} = kt\] Where \([A]_0\) is the initial concentration of substance A, \([A]\) is the final concentration, \(k\) is the rate constant, and \(t\) is the time. Since we are given the intial and final pressures of acetone, in this case, we can write the equation using pressures instead of concentration with the same reasoning: \[ln\frac{P_0}{P} = kt\] Where \(P_0\) is the initial pressure of acetone and \(P\) is the final pressure. Now, we are ready to plug in the given values and find the time.

Calculate the time it takes for the acetone pressure to drop

The given values are: initial pressure of acetone (\(P_0\)) = 0.48 atm, final pressure (\(P\)) = 0.45 atm, and rate constant (\(k\)) = \(\frac{0.693}{81}\). Plug these values into the integrated rate law equation: \[ln\frac{0.48}{0.45} = (\frac{0.693}{81})t\] Now, solve for time (\(t\)): \[t = \frac{81 \times ln(\frac{0.48}{0.45})}{0.693}\] After calculating the expression, we find that the time it takes for the acetone pressure to drop to 0.45 atm is approximately: \[t \approx 15.39 \, seconds\] So, it would take approximately 15.39 seconds for the acetone pressure to drop to the desired value of 0.45 atm at 600ºC in the 1-liter container.

Key concepts

Acetone Decomposition

Acetone, or \(\text{CH}_3\text{COCH}_3\), is a common industrial solvent that decomposes at high temperatures such as \(600^{\circ} \text{C}\). The decomposition leads to the formation of carbon monoxide (CO) and various hydrocarbons, indicative of a breakdown of the acetone molecule.
This decomposition is significant in understanding reaction kinetics because it serves as an example of a first-order reaction.
In first-order reactions, the rate of the reaction is directly proportional to the concentration of one reactant.
Studying acetone decomposition under specified conditions allows us to analyze its kinetics, like how changes in temperature affect reaction rates and how quickly reactants transform into products. Learning how molecules like acetone break down gives insights into larger chemical processes, helping in designing industrial processes and controlling environmental emissions.

Rate Constant

The rate constant (\(k\)) is a crucial factor in reaction kinetics determining the speed of a chemical reaction and is unique for each reaction.
In a first-order reaction equation, it appears as \(rate = k[A]\), with \([A]\) representing the concentration of acetone in our scenario.For acetone decomposition at \(600^{\circ} \text{C}\), we relate the given half-life (\(t_{1/2}\)) to the rate constant using the formula:

• \[t_{1/2} = \frac{0.693}{k}\]
• Solving for \(k\) using the given half-life of 81 seconds results in \(k = \frac{0.693}{81}\).

This mathematical approach allows chemists to predict how quickly a reaction progresses, and the rate constant remains unchanged despite varying concentrations, highlighting the stability of reaction kinetics.

Integrated Rate Law

The integrated rate law for first-order reactions is an essential formula that links concentration changes over time. For these reactions, it is expressed as:

• \[ln\frac{[A]_0}{[A]} = kt\]

Where:

• \([A]_0\) is the initial concentration.
• \([A]\) is the concentration at time \(t\).
• \(k\) is the rate constant.
• \(t\) is the time elapsed.

In our exercise, pressure is employed in place of concentration because they are directly proportional, thus the equation becomes:

• \[ln\frac{P_0}{P} = kt\]

Understanding this equation is pivotal as it reveals the time-dependent nature of first-order reactions, providing insights into how quickly a reaction reaches completion under specific conditions. By manipulating these values, students can calculate how variables like initial pressure and time relate, demonstrating the strength of first-order kinetics analysis.

Half-Life of Reactions

Half-life, denoted as \(t_{1/2}\), describes the period it takes for half of a substance to undergo reaction, which is especially straightforward in first-order reactions since it remains constant throughout the reaction process.
It is calculated using the relationship between half-life and the rate constant:

• \[t_{1/2} = \frac{0.693}{k}\]

For the decomposition of acetone, the half-life is 81 seconds. This suggests that every 81 seconds, the concentration of acetone halves regardless of the initial amount.
Grasping this concept is critical for predicting the time required for a significant portion of a reaction to complete, aiding in effective planning and monitoring of industrial and experimental chemical reactions, and also applicable in calculating decay in radioactive substances.