Consider the reaction of peroxydisulfate ion $\left(\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}\right)\( with iodide ion \)\left(\mathrm{I}^{-}\right)$ 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: $$ \begin{array}{lccc} \hline \text { Experiment } & {\left[\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}\right](M)} & {\left[\mathrm{I}^{-}\right](M)} & \text { Initial Rate }(\mathrm{M} / \mathrm{s}) \\ \hline 1 & 0.018 & 0.036 & 2.6 \times 10^{-6} \\ 2 & 0.027 & 0.036 & 3.9 \times 10^{-6} \\ 3 & 0.036 & 0.054 & 7.8 \times 10^{-6} \\ 4 & 0.050 & 0.072 & 1.4 \times 10^{-5} \\ \hline \end{array} $$ (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}^{-} ?(\mathbf{d})\( What is the rate of disappearance of \)\mathrm{I}^{-}$ when \(\left[\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}\right]=0.025 \mathrm{M}\) and \(\left[\mathrm{I}^{-}\right]=0.050 \mathrm{M} ?\)

Step by step solution

01

Determine the reaction order with respect to each reactant and establish the rate law.

To determine the reaction order with respect to each reactant, we can compare the change in initial concentrations and their effects on initial rates. We can use the general rate law expression for this reaction: Rate = k[\(\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}\)]^m[\(\mathrm{I}^{-}\)]^n where k is the rate constant, m and n are the reaction orders with respect to peroxydisulfate ion and iodide ion, respectively. Compare the data in Experiment 1 and 2, where the concentration of iodide ion is the same: Initial rate (Experiment 2) / Initial rate (Experiment 1) = (3.9 × 10^{-6}) / (2.6 × 10^{-6}) = 1.5 [\(\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}\)] (Experiment 2) / [\(\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}\)] (Experiment 1) = 0.027 / 0.018 = 1.5 Since the ratio of concentration equals the ratio of rates, we can conclude that the reaction is first-order with respect to peroxydisulfate ion. Now let's compare the data in Experiment 1 and 3, where the concentration of peroxydisulfate ion is doubled: Initial rate (Experiment 3) / Initial rate (Experiment 1) = (7.8 × 10^{-6}) / (2.6 × 10^{-6}) = 3 [\(\mathrm{I}^{-}\)] (Experiment 3) / [\(\mathrm{I}^{-}\)] (Experiment 1) = 0.054 / 0.036 = 1.5 Taking the square root of the ratio of rates (3) gives 3^(1/2) ≈ 1.73 ≈ 1.5, which indicates the reaction is second-order with respect to iodide ion. Now we can establish the rate law: Rate = k[\(\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}\)]^1[\(\mathrm{I}^{-}\)]^2

02

Determine the units of the rate constant.

The units of the rate constant depend on the rate law. For our rate law, we can find the units of k as follows: M/s (units of rate) = k × M (units of [\(\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}\)]) × M² (units of [\(\mathrm{I}^{-}\)]^2) Therefore, units of k = (M/s) / (M × M²) = M⁻¹s⁻¹

03

Calculate the average value of the rate constant based on the four sets of data.

To find the average value of the rate constant, we will use our rate law to calculate the value of k for each experiment, and then average those values together. Using: Rate = k[\(\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}\)]^1[\(\mathrm{I}^{-}\)]^2 For each experiment, we solve for k: k (Experiment 1) = (2.6 × 10^{-6}) / (0.018 × 0.036²) ≈ 1.60 M⁻¹s⁻¹ k (Experiment 2) = (3.9 × 10^{-6}) / (0.027 × 0.036²) ≈ 1.60 M⁻¹s⁻¹ k (Experiment 3) = (7.8 × 10^{-6}) / (0.036 × 0.054²) ≈ 1.60 M⁻¹s⁻¹ k (Experiment 4) = (1.4 × 10^{-5}) / (0.050 × 0.072²) ≈ 1.60 M⁻¹s⁻¹ Now we find the average value of k: k_avg = (1.60 + 1.60 + 1.60 + 1.60) / 4 = 1.60 M⁻¹s⁻¹

04

Relate the rate of disappearance of \(\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}\) to the rate of disappearance of \(\mathrm{I}^{-}\).

Looking at the balanced equation, we can see the reaction stoichiometry between peroxydisulfate ion and iodide ion (3 moles of I⁻ per mole of S2O8²⁻). Using this stoichiometry, we can write the relationship: Rate of disappearance of \(\mathrm{I}^{-}\) = 3 × Rate of disappearance of \(\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}\)

05

Calculate the rate of disappearance of \(\mathrm{I}^{-}\) for the given concentrations.

We are given the concentrations of both reactants for this step: [\(\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}\)] = 0.025 M and [\(\mathrm{I}^{-}\)] = 0.050 M We can use our rate law to calculate the rate of disappearance of peroxydisulfate ion: Rate = k[\(\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}\)]^1[\(\mathrm{I}^{-}\)]^2 Rate = (1.60 M⁻¹s⁻¹)(0.025 M)(0.050 M)² ≈ 2.0 × 10^{-6} M/s Now, we can find the rate of disappearance of iodide ion by relating the rates using the stoichiometry: Rate of disappearance of \(\mathrm{I}^{-}\) = 3 × Rate of disappearance of \(\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}\) Rate of disappearance of \(\mathrm{I}^{-}\) = 3 × (2.0 × 10^{-6} M/s) ≈ 6.0 × 10^{-6} M/s

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!