Problem 61 - Chemistry and Chemical Reactivity

Question

The decomposition of \(\mathrm{CO}_{2}\) is first order with respect to the concentration of \(\mathrm{CO}_{2}.\) $$2 \mathrm{CO}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$$ Data on this reaction are provided in the table below. $$\begin{array}{lc}\hline\left[\mathrm{CO}_{2}\right](\mathrm{mol} / \mathrm{L}) & \text { Time }(\mathrm{s}) \\\\\hline 0.38 & 0 \\\0.27 & 12 \\\\\hline\end{array}$$ (a) Write the rate equation for this reaction. (b) Use the data to determine the value of \(k\) (c) What is the half-life of \(\mathrm{CO}_{2}\) under these conditions?

Step-by-Step Solution

Answer

(a) Rate = \( k[\mathrm{CO}_2] \); (b) \( k \approx 0.0324 \, \text{s}^{-1} \); (c) Half-life \( \approx 21.4 \, \text{s} \)."}

Step 1: Define the Rate Equation

Since the decomposition of \( \mathrm{CO}_2 \) is first order, the rate equation is given by \( - \frac{d[\mathrm{CO}_2]}{dt} = k[\mathrm{CO}_2] \), where \( k \) is the rate constant.

Step 2: Use the Integrated Rate Law (First Order)

For a first order reaction, the integrated rate law is \( [\mathrm{CO}_2] = [\mathrm{CO}_2]_0 e^{-kt} \), or in logarithmic form: \[ \ln [\mathrm{CO}_2] = \ln [\mathrm{CO}_2]_0 - kt \] where \([\mathrm{CO}_2]_0\) is the initial concentration and \([\mathrm{CO}_2]\) is the concentration at time \(t\).

Step 3: Calculate the Rate Constant \( k \)

Starting with the logarithmic form of the first order integrated rate equation:\[ \ln [\mathrm{CO}_2] = \ln 0.38 - kt \]Substitute the given values:\[ \ln 0.27 = \ln 0.38 - 12k \]Calculate the difference:\[ \ln 0.27 - \ln 0.38 = -12k \]Use a calculator to solve for \(k\): \[ k = \frac{\ln 0.38 - \ln 0.27}{12} \approx 0.0324 \, \text{s}^{-1} \]

Step 4: Calculate the Half-Life

The half-life \( t_{1/2} \) for a first order reaction is given by \[ t_{1/2} = \frac{0.693}{k} \]Substitute the value of \(k\):\[ t_{1/2} = \frac{0.693}{0.0324} \approx 21.4 \, \text{s} \].

Key Concepts

Rate Equation

In a first-order reaction, the rate at which the reaction occurs is directly proportional to the concentration of a single reactant. For the decomposition of \( \mathrm{CO}_2 \), this concentration-dependent relationship can be represented as a rate equation:

• \( -\frac{d[\mathrm{CO}_2]}{dt} = k[\mathrm{CO}_2] \)

Here, \( [\mathrm{CO}_2] \) is the concentration of carbon dioxide, and \( k \) is the rate constant.
This equation tells us that as \( [\mathrm{CO}_2] \) decreases, the rate at which the reaction proceeds also decreases. Understanding the rate equation is crucial, as it forms the foundation for analyzing how fast a reaction is happening at any given moment.

Integrated Rate Law

The integrated rate law for a first-order reaction helps us understand how the concentration of reactants changes over time. For our reaction, it can take either an exponential or logarithmic form:

• Exponential: \( [\mathrm{CO}_2] = [\mathrm{CO}_2]_0 e^{-kt} \)
• Logarithmic: \( \ln [\mathrm{CO}_2] = \ln [\mathrm{CO}_2]_0 - kt \)

\([\mathrm{CO}_2]_0\) is the initial concentration, and \([\mathrm{CO}_2]\) is the concentration at time \( t \).
This formula allows us to predict how long it will take for the concentration to reach a certain amount or how much will be left after a given time. It's a versatile tool for chemists, providing insights into reaction kinetics effectively.

Half-Life Calculation

Half-life is the time it takes for half of the reactant to be consumed in a reaction. For a first-order process, it is expressed as:

• \( t_{1/2} = \frac{0.693}{k} \)

This equation tells us that the half-life is constant and independent of the initial concentration. That means whether you start with a little or a lot of \( \mathrm{CO}_2 \), the time it takes to deplete half will be the same.
In this exercise, substituting \( k = 0.0324 \, \text{s}^{-1} \) gives us a half-life of approximately 21.4 seconds. This consistent half-life is a hallmark of first-order kinetics.

Rate Constant Determination

Determining the rate constant \( k \) is essential to fully understand the kinetics of a reaction. Using the integrated rate law in its logarithmic form provides a pathway to find \( k \):

• By rearranging \( \ln 0.27 = \ln 0.38 - 12k \)
• Solve: \( k = \frac{\ln 0.38 - \ln 0.27}{12} \approx 0.0324 \, \text{s}^{-1} \)

The calculations involve natural logarithms of the initial and remaining concentrations over a fixed time interval, giving a numerical value for \( k \).
This rate constant explains the speed at which \( \mathrm{CO}_2 \) decomposes and is a key parameter for characterizing and predicting the behavior of the reaction.