Problem 46 - Chemistry The Central Science In Si Units

Question

From the following data for the second-order gas-phase decomposition of HI at \(430^{\circ} \mathrm{C},\) calculate the second-order rate constant and half- life for the reaction: \begin{tabular}{rl} \hline Time (s) & [HIYmol dm \(^{-3}\) \\ \hline 0 & 1 \\ 100 & 0.89 \\ 200 & 0.8 \\ 300 & 0.72 \\ 400 & 0.66 \end{tabular}

Step-by-Step Solution

Answer

The rate constant is \(0.00126\, \text{dm}^3/\text{mol/s}\) and the half-life is 793.65 seconds.

Step 1: Understanding Second-Order Kinetics

For a second-order reaction, the rate law is given by \( rate = k[HI]^2 \), where \( k \) is the rate constant, and the differential rate law integrates to the form \( \frac{1}{[HI]} - \frac{1}{[HI]_0} = kt \).

Step 2: Setting Up the Integrated Rate Law

Use the integrated rate law \( \frac{1}{[HI]} - \frac{1}{[HI]_0} = kt \) with initial concentration \([HI]_0 = 1\, \text{mol/dm}^3\) and the concentration at various times from the data provided.

Step 3: Calculating the Rate Constant k

Choose two data points to calculate \( k \). Let's use the data at 0 seconds (\([HI]_0 = 1\, \text{mol/dm}^3\)) and at 100 seconds (\([HI] = 0.89\, \text{mol/dm}^3\)). Substitute into the formula: \( \frac{1}{0.89} - \frac{1}{1} = k \times 100 \). Solve for \( k \): \[ \frac{1}{0.89} - 1 = 100k \Rightarrow k = \frac{0.126}{100} = 0.00126\, \text{dm}^3/\text{mol/s} \]

Step 4: Verifying the Rate Constant k

Repeat the calculation of \( k \) using another set of times, say 0 and 200 seconds, to verify the consistency of \( k \). For 200 sec, \( \frac{1}{0.8} - 1 = k \times 200 \Rightarrow k = \frac{0.25}{200} = 0.00125\, \text{dm}^3/\text{mol/s} \). The values are consistent.

Step 5: Calculating the Half-Life for a Second-Order Reaction

The half-life for a second-order reaction is given by \( t_{1/2} = \frac{1}{k[HI]_0} \). With \( k = 0.00126\, \text{dm}^3/\text{mol/s} \) and \([HI]_0 = 1\, \text{mol/dm}^3\), the half-life \( t_{1/2} \) is: \[ t_{1/2} = \frac{1}{0.00126 \times 1} = 793.65\, \text{s} \]

Step 6: Concluding Result

The second-order rate constant for the decomposition of HI at 430°C is approximately \( 0.00126\, \text{dm}^3/\text{mol/s} \), and the half-life is approximately 793.65 seconds.

Key Concepts

Rate Constant Calculation

In second-order reaction kinetics, the rate constant, denoted as \( k \), is a crucial part of understanding how quickly a reaction occurs. Specifically for a reaction of type \( 2A \rightarrow B \), the rate expression is given as \( rate = k[HI]^2 \). In practical terms, we use the integrated rate law formula \( \frac{1}{[HI]} - \frac{1}{[HI]_0} = kt \) to determine \( k \) from experimental data.
To calculate \( k \), we select data points from the experiment. The initial concentration \([HI]_0\) and concentration at a time \( t \) are used. For instance, with initial concentration at \( t = 0 \) seconds and data at \( t = 100 \) seconds, you substitute these values into the formula, solving for \( k \).
It's important to cross-verify by using different sets of data points. Consistency in the calculated value of \( k \) helps confirm its accuracy, as demonstrated by repeating the calculation with data from \( t = 0 \) and \( t = 200 \) seconds. Consistent values indicate reliability in the experimental setup and calculations.

Integrated Rate Law

The integrated rate law for second-order reactions is powerful because it links concentrations over time. This specific form of the rate law transforms a differential rate equation into one that is more useable with data, namely \( \frac{1}{[HI]} - \frac{1}{[HI]_0} = kt \), where \([HI]_0\) is the initial concentration and \( [HI] \) is the concentration at time \( t \).
This equation tells us how the concentration of reactants decreases with time. It's an effective tool for determining the rate constant from observable data, which otherwise would be hard to deduce from the primary rate law expression.
Additionally, by plotting \( \frac{1}{[HI]} \) against time \( t \), a straight line with slope \( k \) should be observable, if the reaction indeed follows second-order kinetics. This not only confirms the order of the reaction but also helps in accurately calculating the rate constant.

Half-Life Calculation

Calculating the half-life of a reaction helps us understand how long it takes for half of the reactant to be consumed. For a second-order reaction, the half-life formula is given by \( t_{1/2} = \frac{1}{k[HI]_0} \).
This formula shows that the half-life is not constant and depends on the initial concentration. As such, as the reaction proceeds and concentration decreases, the half-life becomes longer – unlike a first-order reaction where the half-life remains constant throughout the reaction.
Using the calculated rate constant \( k = 0.00126 \, \text{dm}^3/\text{mol/s} \) and \([HI]_0 = 1 \, \text{mol/dm}^3\), we determined the half-life to be approximately 793.65 seconds. This value provides a snapshot of the reaction's speed at its starting concentration, illustrating how second-order reactions evolve over time.