Question

Consider the gas phase thermal decomposition of \(1.0 \operatorname{atm}\) of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COOC}\left(\mathrm{CH}_{3}\right)_{3}(g)\) to acetone \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}(g)\) and ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)(\mathrm{g}),\) which occurs with a rate constant of \(0.0019 \mathrm{s}^{-1} .\) After initiation of the reaction, at what time would you expect the pressure to be 1.8 atm?

Short answer

Based on the given initial conditions and rate constant for the thermal decomposition of the compound, it would take approximately 1687.41 seconds for the total pressure to reach 1.8 atm.

Step-by-step solution

Set up a reaction table

Let's set up a table for pressure, with the initial pressure of the reactant and the pressure change throughout the reaction. Here, let \(x\) represent the pressure change (in atm) of the reactant. | Substance | Initial Pressure | Change in Pressure | Pressure at time t | |----------------|------------------|--------------------|--------------------| |\((CH_3)_3COOC(CH_3)_3\)| 1.0 atm | -x | 1.0 - x | |\((CH_3)_2CO\) | 0 atm | x | x | |\(C_2H_6\) | 0 atm | x | x | At initiation of the reaction, the total pressure \(P_{total}\) can be written as: $$P_{total} = 1.0 - x + x + x = 1.0 + x$$

Write the integrated rate law

The first-order integrated rate law equation is given by: $$\ln\frac{[A]_0}{[A]_t} = kt$$ For an ideal gas, the pressure is directly proportional to the concentration. Therefore, we can rewrite the equation in terms of pressures: $$\ln\frac{P_0}{P_t} = kt$$ Where \(P_0\) is the initial pressure, \(P_t\) is the pressure at time t, and \(k\) is the rate constant.

Solve for time and calculate the time required

We are given the initial pressure of reactant \(P_0 = 1.0\) atm, the rate constant \(k = 0.0019\) s\(^{-1}\) and we need to find the time when the total pressure \(P_{total} = 1.8\) atm. Using the total pressure equation from step 1, we can find the pressure of the reactant at this time: $$1.0 + x = 1.8$$ $$x = 0.8$$ The pressure of the reactant at this time, \(P_t = 1.0-x = 0.2\) atm. Now, we can plug these values into the integrated rate law equation: $$\ln\frac{1.0}{0.2} = (0.0019 s^{-1})t$$ Calculate the time, t: $$t = \frac{\ln\frac{1.0}{0.2}}{0.0019 s^{-1}} \approx 1687.41 s$$ Therefore, it would take approximately 1687.41 seconds for the pressure to reach 1.8 atm.

Key concepts

Chemical Kinetics

Chemical kinetics is a fascinating area of chemistry that studies the rates of chemical processes and the factors influencing them. This branch of physical chemistry is crucial for understanding how chemical reactions occur and how to control them.

By analyzing the kinetics of a chemical reaction, scientists can determine the speed at which reactants turn into products. This rate can depend on various elements, such as temperature, concentration of reactants, presence of a catalyst, and physical state of the reactants. Kinetic studies also help chemists to elucidate reaction mechanisms—step-by-step sequences showing the path taken by molecules during a reaction.

In the context of the provided exercise, kinetic principles enabled us to predict the time required for a certain change in pressure during the thermal decomposition reaction, reflecting the progress of the chemical reaction over time.

Integrated Rate Law

To delve deeper into reaction kinetics, we encounter the integrated rate law. This mathematical equation relates concentrations of reactants or products to the time elapsed during the reaction.

For different orders of reactions—first-order, second-order, and zero-order—the form of the integrated rate law will differ. The integrated rate law helps in calculating the concentration of a reactant or product at any time during a chemical reaction. Essentially, it integrates the rate of the reaction over time to give us a snapshot of the reaction's progress at a specific moment.

In our specific exercise, we used the first-order integrated rate law to connect the initial pressure to the pressure at a later time. This approach allows for precise calculations of reaction times, which is particularly useful in industrial processes where timing can be critical.

First-Order Reaction

In first-order reactions, the rate of reaction is directly proportional to the concentration of a single reactant. Mathematically, this means that if you double the concentration of the reactant, the reaction rate also doubles.

The integrated rate law for a first-order reaction is an exponential decay function, where the natural logarithm of the concentration of the reactant decreases linearly over time. The beauty of first-order reactions is their simplicity, as they are described by a straightforward formula that allows for easy computation of variables like reaction time and concentration.

Our exercise is based on a first-order reaction, which simplified the calculation process and led to finding the time required for the pressure in the system to increase to 1.8 atm due to the gas phase reaction progression.

Gas Phase Reaction

A gas phase reaction is one that occurs between substances in the gaseous state. These reactions can be intriguing due to the unique properties of gases, such as their expansibility, compressibility, and low density. The behavior of gases is often described by the ideal gas law, which relates the pressure, volume, and temperature of a gas with the number of moles present.

It's vital to understand that in a gas phase reaction, changes in pressure can indicate changes in concentration, making pressure measurements a valuable tool when studying reaction kinetics. This ties back to our exercise, where the pressure change helped us track the progress of the thermal decomposition reaction.

Reaction Rate Constant

The reaction rate constant, often represented by the symbol 'k', is a proportionality constant in the rate law of a chemical reaction. This constant is unique for each reaction and is influenced by factors such as temperature and catalysts. It provides the relationship between the reactant concentrations and the rate of reaction.

In a first-order reaction, as we saw in the exercise, the rate constant relates the natural logarithm of the ratio of initial to remaining reactant concentration (or pressure, for gases) and the elapsed time. The reaction rate constant allows us to predict how quickly a reaction will proceed under given conditions. In the exercise, knowing the value of 'k' was essential to calculate when the pressure would rise to a specific value.