91Ó°ÊÓ

In chemical kinetics, understanding the order of a reaction is crucial in predicting how it progresses over time. A second-order reaction is a specific classification where the rate depends on the concentration of one reactant squared, or on the product of concentrations of two reactants. For the reaction \(\mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C}\), it's second-order if a plot of \(1/[\mathrm{A}]\) versus time gives a straight line. This linearity comes from the integrated rate law formulated for second-order reactions, which states \(1/[\mathrm{A}] = kt + 1/[\mathrm{A}_0]\). Here:

• \([\mathrm{A}]\) is the concentration of the reactant at time \(t\).
• \([\mathrm{A}_0]\) is the initial concentration.
• \(k\) represents the rate constant.

Understanding this linear relationship helps in identifying second-order reactions in experimental data. The plot's straightness confirms that the change in concentration is proportional to time, specific to second-order reactions.

The integrated rate law is a mathematical expression that relates the concentration of a reactant over time. For second-order reactions, like our example \(\mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C}\), it takes the form \(1/[\mathrm{A}] = kt + 1/[\mathrm{A}_0]\). This equation allows chemists to calculate how long it takes for a reactant concentration to drop to a certain level, or to find out the concentration at any given time.Key components include:

• \([\mathrm{A}_0]\) is the initial concentration, the starting point of the reaction, crucial for solving the rate equation.
• \([\mathrm{A}]\) is the concentration remaining at time \(t\).
• \(k\) is the rate constant unique to that reaction, influencing how quickly it proceeds.

This relationship, especially the inverse concentration plot versus time, is a fundamental tool in analyzing reaction kinetics, providing clear data on the progression of second-order reactions.

The rate constant \(k\) is a vital factor in determining the speed of a chemical reaction. Its calculation requires applying the integrated rate law, especially in second-order reactions. Here’s how the process unfolds:1. **Determine Concentration Changes:** First, note the initial concentration \([\mathrm{A}_0]\) and how it changes over time; in our case, it decreased from \(0.50\, M\) to \(0.30\, M\) after 40% reacted.2. **Use the Integrated Rate Equation:** Insert known values into the equation \(\frac{1}{[\mathrm{A}]} = kt + \frac{1}{[\mathrm{A}_0]}\), with \(t\) as time elapsed.3. **Rearrange to Solve for \(k\):** Rearrange to form \( k = \frac{1/[\mathrm{A}] - 1/[\mathrm{A}_0]}{t} \). This transformation isolates \(k\), making it solvable.In practical terms, we took the final concentrations and solved:

• Calculated \(1/0.30 = 3.33\, M^{-1}\).
• Calculated \(1/0.50 = 2\, M^{-1}\).
• Dividing the difference by time \((57\, s)\) yielded \( k \approx 0.0233\, M^{-1} s^{-1}\).

The methodical approach ensures precise calculation, reflecting the inherent rate of a reaction under given conditions.