Problem 38 - Chemistry: Principles and Reactions

Question

In the first-order decomposition of acetone at \(500^{\circ} \mathrm{C}\), $$\mathrm{CH}_{3}-\mathrm{CO}-\mathrm{CH}_{3}(g) \longrightarrow \text { products }$$ it is found that the concentration is \(0.0300 \mathrm{M}\) after \(200 \mathrm{~min}\) and \(0.0200 \mathrm{M}\) after 400 min. Calculate the following. (a) the rate constant (b) the half-life (c) the initial concentration

Step-by-Step Solution

Answer

Answer: The rate constant, k = 5.83 x 10^(-4) min^(-1); half-life, t(1/2) = 1189 min; initial concentration, [A₀] = 0.0500 M.

Step 1: Understand the first-order integrated rate law

For a first-order reaction, the integrated rate law is given by: $$\ln [\mathrm{A}]=[ \mathrm{- kt}+\ln [\mathrm{A}_{0}]$$ Where: \([\mathrm{A}]\) is the concentration of reactant at time \(t\). \([\mathrm{A}_{0}]\) is the initial concentration of reactant. \(k\) is the rate constant. \(t\) is the time. We can use this equation to solve for the rate constant, \(k\), at different time points.

Step 2: Calculate the rate constant

At 200 min: $$\ln(0.0300) = -k \cdot (200) + \ln([\mathrm{A}_{0}])$$ At 400 min: $$\ln(0.0200) = -k \cdot (400) + \ln([\mathrm{A}_{0}])$$ Now, we need to solve these equations for \(k\). We can do this by subtracting the second equation from the first to eliminate \([\mathrm{A}_{0}]\). \(\ln(0.0300) - \ln(0.0200) = 200k\) Now, we can solve for the rate constant, \(k\). \(k = \frac{\ln(0.0300) - \ln(0.0200)}{200} = 5.83 \times 10^{-4}\ \text{min}^{-1}\)

Step 3: Calculate the half-life

Using the calculated rate constant \(k\), we can now find the half-life using the following formula for first-order reactions: $$t_{1/2} = \frac{0.693}{k}$$ \( t_{1/2} = \frac{0.693}{5.83 \times 10^{-4}\ \text{min}^{-1}} = 1189\ \text{min}\)

Step 4: Determine the initial concentration

We can now use the calculated rate constant and the first-order integrated rate law equation to find the initial concentration of acetone. Let's use the equation at 200 min: $$\ln(0.0300) = - (5.83 \times 10^{-4}\ \text{min}^{-1}) \cdot (200) + \ln([\mathrm{A}_{0}])$$ Solving for \([\mathrm{A}_{0}]\): $$\ln([\mathrm{A}_{0}]) = \ln(0.0300) + (5.83 \times 10^{-4}\ \text{min}^{-1}) \cdot (200)$$ $$[\mathrm{A}_{0}] = e^{\ln(0.0300) + (5.83 \times 10^{-4}\ \text{min}^{-1}) \cdot (200)} = 0.0500\ \mathrm{M}$$ Now, we have calculated all the required parameters: (a) The rate constant, \(k = 5.83 \times 10^{-4}\ \text{min}^{-1}\) (b) The half-life, \(t_{1/2} = 1189\ \text{min}\) (c) The initial concentration, \([\mathrm{A}_{0}] = 0.0500\ \mathrm{M}\)

Key Concepts

Rate Constant Calculation

Understanding the rate of a chemical reaction is crucial for various applications such as pharmaceuticals, environmental science, and industrial processes. The rate constant, often denoted by the symbol k, is a proportionality constant in the rate law equation that provides a measure of the speed of a chemical reaction. In a first-order reaction, the rate of reaction is directly proportional to the concentration of one reactant.

To calculate the rate constant from experimental data, we use the integrated rate law for first-order reactions, which is expressed as:
\[ \ln[\mathrm{A}] = -kt + \ln[\mathrm{A}_0] \]
Where [A] represents the concentration of the reactant at time t, k is the rate constant we aim to find, and [A_0] is the initial concentration of the reactant. By measuring the reactant concentration at different times, we can derive k from the slope of the plot of ln[A] against time t, or by solving two such concentration-time equations simultaneously to eliminate the initial concentration term. For a smoother understanding of the concept, it's helpful to visualize the calculations using graphs or by setting up linear equations when provided with concentration data at various times.

Half-life of Reaction

The half-life of a reaction, denoted as t_{1/2}, is the time required for the concentration of a reactant to reduce to half of its initial value. This concept is not only relevant in chemistry but also in physics, biology, and pharmacology, where it helps in understanding decay processes and drug elimination, respectively.

For first-order reactions, the half-life is particularly important because it is independent of the initial concentration of the reactant. This is a unique property of first-order kinetics – no matter how much reactant you start with, it takes the same amount of time to fall to half its initial amount. The formula used for calculating the half-life of a first-order reaction is given by: \[ t_{1/2} = \frac{0.693}{k} \]
The constant 0.693 is the natural log of 2, which relates to the concept of halving. Using the rate constant k calculated earlier, we can easily find the half-life, which provides insight into the speed of the reaction. For example, in pharmaceuticals, a shorter half-life might indicate a quicker therapeutic action, whereas in radioactive waste management, a long half-life could suggest prolonged environmental exposure.

Initial Concentration Determination

The initial concentration of a reactant, denoted by [A_0], is an important value in chemistry because it sets the starting point for reaction analysis. It is particularly important when studying reaction kinetics – knowing the original amount of reactant before any change is crucial for calculations and predictions. In practical experiments, the initial concentration can sometimes be unknown and has to be determined from other data.

Using the integrated first-order rate law and data from specific time points, it is possible to back-calculate the initial concentration of the reactant. After replacing the known values and rearranging the equation, one can solve for [A_0]. It's essential to grasp that although experimentation and practical application might present challenges, such as measurement inaccuracies or real-world variables, these basic principles of chemistry provide a solid foundation for understanding and solving complex scientific queries. Through the context of kinetics, we gain the quantitative tools to predict how reactions proceed over time, thus tailoring our approach to various chemical processes.