Problem 51 - Chemistry and Chemical Reactivity

Question

To determine the concentration dependence of the rate of the reaction $$\mathrm{H}_{2} \mathrm{PO}_{3}^{-}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{HPO}_{3}^{2-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)$$ you might measure \(\left[\mathrm{OH}^{-}\right]\) as a function of time using a pH meter. (To do so, you would set up conditions under which \(\left[\mathrm{H}_{2} \mathrm{PO}_{3}^{-}\right]\) remains constant by using a large excess of this reactant.) How would you prove a second-order rate dependence for \(\left[\mathrm{OH}^{-}\right] ?\)

Step-by-Step Solution

Answer

Plot \( \frac{1}{[\mathrm{OH}^{-}]_t} \) versus time; a straight line confirms second-order reaction.

Step 1: Understanding Reaction Order

In a reaction, the rate law expresses the rate as a function of reactant concentrations. For a second-order reaction involving a single reactant, the rate law is \( rate = k [A]^2 \), where \([A]\) is the concentration of the reactant and \(k\) is the rate constant.

Step 2: Identifying Known Variables

You are given a reaction \( \mathrm{H}_{2} \mathrm{PO}_{3}^{-} + \mathrm{OH}^{-} \rightarrow \mathrm{HPO}_{3}^{2-} + \mathrm{H}_{2} \mathrm{O} \) and a concentration \( \left[\mathrm{OH}^{-}\right] \), with \( \left[\mathrm{H}_{2} \mathrm{PO}_{3}^{-}\right] \) in large excess. This means \( \left[\mathrm{H}_{2} \mathrm{PO}_{3}^{-}\right] \) effectively remains constant, allowing you to focus on \( \left[\mathrm{OH}^{-}\right] \).

Step 3: Recording Concentration Over Time

Using a pH meter, measure \( \left[\mathrm{OH}^{-}\right] \) at various time intervals to obtain \([\mathrm{OH}^{-}]_t\) data. This lets you observe how \( \left[\mathrm{OH}^{-}\right] \) changes over time.

Step 4: Modeling the Concentration Decrease

For a second-order reaction with respect to one reactant, the integrated rate law is \( \frac{1}{[A]_t} = \frac{1}{[A]_0} + kt \). Here, \([A]_0\) is the initial concentration of OH⁻, and \([A]_t\) is the concentration at time \( t \).

Step 5: Plotting and Analyzing Data

Plot \( \frac{1}{[\mathrm{OH}^{-}_t]} \) versus time \( t \). If the plot is a straight line, it indicates the reaction is second-order with respect to \( \mathrm{OH}^{-} \). The slope of the line will equal the rate constant \( k \).

Step 6: Conclusion from Data Analysis

If your plot of \( \frac{1}{[\mathrm{OH}^{-}_t]} \) versus \( t \) is linear, this confirms a second-order dependence in \( \left[\mathrm{OH}^{-}\right] \) for the reaction. If not, further analysis might be needed for higher or mixed-order dependencies.

Key Concepts

Second-Order Reaction

In reaction kinetics, a second-order reaction is identified when the rate of the reaction is proportional to the square of the concentration of one reactant. This type of reaction can be symbolically represented as \( rate = k [A]^2 \), where \([A]\) is the concentration of the reactant and \(k\) is the rate constant. This means as the concentration of the reactant doubles, the reaction rate increases fourfold. Understanding the order of a reaction is crucial because it influences the reaction's speed and half-life. In second-order reactions, the half-life varies dependent on the initial concentration, different from first-order reactions where it remains constant.

Rate Law

The rate law of a chemical reaction is an equation that links the rate of reaction to the concentration of reactants. For a second-order reaction, the rate law could be expressed as \( rate = k [A]^2 \) or \( rate = k [A][B] \) if two different reactants influence the rate. Determining the correct rate law is essential for predicting how changes in concentration affect the reaction rate. This information is vital in both experimental and industrial settings to control reaction conditions effectively. The rate constant \(k\) can also provide information about the reaction's speed and how it varies with different conditions like temperature.

Integrated Rate Law

In kinetics, the integrated rate law for a second-order reaction helps understand how the concentration of a reactant changes over time. The integrated form for a second-order reaction that involves a single reactant \([A]\) is given by \( \frac{1}{[A]_t} = \frac{1}{[A]_0} + kt \), where \([A]_0\) is the initial concentration and \([A]_t\) represents the concentration at time \(t\). This equation is derived from the basic rate law and helps in predicting the concentration at any given time. This mathematical form allows for straightforward plotting — if you plot \( \frac{1}{[A]_t} \) versus time, a second-order reaction will produce a linear graph. The slope of this line is equal to the rate constant \(k\), making it easy to determine the kinetics of the reaction from experimental data.

Concentration Measurement

Accurate measurement of reactant concentrations is critical to studying chemical kinetics. In the context of the given exercise, using a pH meter to gauge \( [\mathrm{OH}^-] \) concentration over time is a practical approach. This is because the concentration of hydroxide ions influences the pH, and changes in concentration can be monitored as changes in pH. Ensuring that all other variables are controlled, such as maintaining a constant concentration of \([\mathrm{H}_2\mathrm{PO}_3^-]\), allows for precise measurement of \( [\mathrm{OH}^-] \). By recording how \( [\mathrm{OH}^-] \) changes over time, you gather necessary data to use in calculating the rate constant and verifying the reaction order through integrated rate law plotting.