Problem 38 - Chemistry The Central Science In Si Units

Question

Consider the reaction of peroxydisulfate ion \(\left(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\right)\) with iodide ion ( \(1^{-}\) ) in aqueous solution: $$ \mathrm{S}_{2} \mathrm{O}_{8}^{2-}(a q)+3 \mathrm{I}^{-}(a q) \longrightarrow 2 \mathrm{SO}_{4}^{2-}(a q)+\mathrm{I}_{3}^{-}(a q) $$ At a particular temperature, the initial rate of disappearance of \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\) varies with reactant concentrations in the following manner: \begin{tabular}{lccc} \hline Experiment & {\(\left[\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\right](M)\)} & {[1]\((M)\)} & Initial Rate \((M / s)\) \\ \hline 1 & 0.018 & 0.036 & \(2.6 \times 10^{-6}\) \\ 2 & 0.027 & 0.036 & \(3.9 \times 10^{-6}\) \\ 3 & 0.036 & 0.054 & \(7.8 \times 10^{-6}\) \\ 4 & 0.050 & 0.072 & \(1.4 \times 10^{-5}\) \\ \hline \end{tabular} (a) Determine the rate law for the reaction and state the units of the rate constant. (b) What is the average value of the rate constant for the disappearance of \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\) based on the four sets of data? (c) How is the rate of disappearance of \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\) related to the rate of disappearance of \(\mathrm{I}^{-}\) ? (d) What is the rate of disappearance of \(1^{-}\) when \(\left[\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\right]=0.025 \mathrm{M}\) and \(\left[I^{-}\right]=0.050 M ?\)

Step-by-Step Solution

Answer

(a) Rate law: \( \text{Rate} = k[S_2O_8^{2-}][I^-] \), \( k \approx 4.0 \times 10^{-3} \text{M}^{-1} \text{s}^{-1} \). (b) Average \( k \approx 4.02 \times 10^{-3} \text{M}^{-1} \text{s}^{-1} \). (c) Rate of \( I^- \) is 3 times that of \( S_2O_8^{2-} \). (d) \( 1.51 \times 10^{-5} \text{M} \text{s}^{-1} \).

Step 1: Establish the rate law form

The rate law can be expressed as \( \text{Rate} = k \left[ S_2O_8^{2-} \right]^m \left[ I^- \right]^n \), where \( k \) is the rate constant, \( m \) is the order of reaction with respect to \( S_2O_8^{2-} \), and \( n \) is the order of reaction with respect to \( I^- \). We need to determine the values of \( m \), \( n \), and \( k \).

Step 2: Determine the reaction order with respect to \\ S_2O_8^{2-}

Comparing experiments 1 and 2, the concentration of \( S_2O_8^{2-} \) doubles while \( I^- \) remains constant, and the rate also doubles \( \left( \frac{3.9 \times 10^{-6}}{2.6 \times 10^{-6}} = 1.5 \approx 2 \right) \). This indicates the reaction is first order with respect to \( S_2O_8^{2-} \), i.e., \( m = 1 \).

Step 3: Determine the reaction order with respect to \\( I^- \\)

Comparing experiments 2 and 3, doubling \( I^- \) with constant \( S_2O_8^{2-} \) results in the rate doubling \( \left( \frac{7.8 \times 10^{-6}}{3.9 \times 10^{-6}} = 2 \right) \). This also suggests a first order with respect to \( I^- \), i.e., \( n = 1 \).

Step 4: Write the rate law and calculate the rate constant

The rate law is \( \text{Rate} = k \left[ S_2O_8^{2-} \right]^1 \left[ I^- \right]^1 \). Use any experiment to solve for \( k \): \( 2.6 \times 10^{-6} = k \times 0.018 \times 0.036 \) gives \( k \approx 4.0 \times 10^{-3} \, \text{M}^{-1} \text{s}^{-1} \).

Step 5: Calculate the average rate constant

Calculate \( k \) for each experiment and average them: Experiment 1: \( k = \frac{2.6 \times 10^{-6}}{0.018 \times 0.036} \)Experiment 2: \( k = \frac{3.9 \times 10^{-6}}{0.027 \times 0.036} \)Experiment 3: \( k = \frac{7.8 \times 10^{-6}}{0.036 \times 0.054} \)Experiment 4: \( k = \frac{1.4 \times 10^{-5}}{0.050 \times 0.072} \)Average \( k = 4.02 \times 10^{-3} \, \text{M}^{-1} \text{s}^{-1} \).

Step 6: Relate rates of disappearance

According to the stoichiometry of the reaction, 1 mole of \( S_2O_8^{2-} \) produces 3 moles of \( I^- \). Therefore, the rate of disappearance of \( I^- \) is 3 times the rate of \( S_2O_8^{2-} \): \( \text{Rate of } I^- = 3 \times \text{Rate of } S_2O_8^{2-} \).

Step 7: Calculate the rate of disappearance of \\( I^- \\)

Using the rate law, calculate the rate of disappearance of \( S_2O_8^{2-} \) first:\( \text{Rate} = k \times 0.025 \times 0.050 = 4.02 \times 10^{-3} \times 0.025 \times 0.050 \approx 5.03 \times 10^{-6} \, \text{M} \text{s}^{-1} \).The rate of disappearance of \( I^- \) is 3 times this rate: \( 3 \times 5.03 \times 10^{-6} \approx 1.51 \times 10^{-5} \, \text{M} \text{s}^{-1} \).

Key Concepts

Rate Laws

Understanding the rate law of a reaction is like having the roadmap to how fast reactants turn into products. In the world of chemical kinetics, the rate law expresses the relationship between the concentration of reactants and the rate at which they vanish or transform. The general form is:

• Rate = \( k [A]^m [B]^n \)
• \( k \) is the rate constant, \( [A] \) and \( [B] \) are concentrations of reactants.
• \( m \) and \( n \) denote the orders of the reaction concerning reactants \( A \) and \( B \).

The rate law is crucial because it enables you to predict how changing the concentration impacts the speed of the reaction. It's empirically determined, meaning you need experimental data to establish it, much like the data from experiments described in the exercise.

Reaction Order

Reaction order provides insights into how the concentration of each reactant affects the rate of a reaction. In simpler terms, if you want to know how much faster or slower a reaction becomes when you tweak the concentration of a reactant, you're looking at reaction orders. The order can be:

• First Order: The rate changes proportionally with the concentration change. For instance, if concentration doubles, so does the rate.
• Second Order: The effect on rate is squared relative to concentration change.
• Zero Order: The rate remains unchanged regardless of changes in concentration.

In the given exercise, both \( S_2O_8^{2-} \) and \( I^- \) show first-order behavior. Recognizing the order of a reaction helps chemists control reactions more efficiently.

Rate Constants

In chemical kinetics, the rate constant \( k \) is a fundamental factor. It connects the concentration of reactants with the rate of reaction through the rate law. Imagine it as a bridge between how much reactant you have and how quickly they turn into products. What is fascinating about \( k \) is:

• Its value is constant at a given temperature but can change if the temperature shifts.
• It also has units, which are dependent on the reaction order. For instance, it is \( \text{M}^{-1}\text{s}^{-1} \) when reaction is of overall second order, as in this exercise.
• Knowing \( k \) allows us to calculate how fast a reaction occurs, providing a powerful tool for prediction and analysis.

To find \( k \) in practice, just plug in known concentrations and rates from experimental data. A fantastically simple yet informative constant!

Stoichiometry of Reactions

The term stoichiometry might seem intimidating, but it's essentially chemistry's way of balancing the checkbook of reactions. It dictates the quantitative relationships among reactants and products in a chemical equation. Here's how it plays out:

• Balanced equations tell us how many moles of each reactant turn into products.
• In the provided reaction, \( 1 \text{ mole of } S_2O_8^{2-} \) produces \( 3 \text{ moles of } I^- \), showing a precise relation.
• The stoichiometry is directly reflected when calculating rates of disappearance: if one reactant disappears at a certain rate, another related one might disappear at a multiple of that rate.

Understanding stoichiometry allows chemists to scale reactions up or down, ensuring they can produce the right amount of products efficiently and effectively.