Question

The iodate ion is reduced by sulfite according to the following reaction: $$\mathrm{IO}_{3}^{-}(a q)+3 \mathrm{SO}_{3}^{2-}(a q) \longrightarrow \mathrm{I}^{-}(a q)+3 \mathrm{SO}_{4}^{2-}(a q)$$ The rate of this reaction is found to be first order in \(\mathrm{IO}_{3}^{-}\) , first order in \(\mathrm{SO}_{3}^{2-}\) , and first order in \(\mathrm{H}^{+}\) . (a) Write the rate law for the reaction. (b) By what factor will the rate of the reaction change if the pH is lowered from 5.00 to 3.50\(?\) Does the reaction proceed more quickly or more slowly at the lower pH? (c) By using the concepts discussed in Section 14.6, ex-plain how the reaction can be pH-dependent even though H' does not appear in the overall reaction.

Short answer

The rate law for the reaction is: Rate = k[IO₃⁻][SO₃²⁻][H⁺]. The rate of the reaction changes by a factor of 31.6 when the pH is lowered from 5.00 to 3.50, and the reaction proceeds more quickly at lower pH. The pH dependency results from H⁺ ions being involved in one or more elementary steps of the reaction mechanism, even though they don't appear in the overall balanced equation.

Step-by-step solution

(a) Write the rate law for the reaction.

The rate law expresses the reaction rate as a product of the concentrations of the reactants raised to their respective orders. We are given that the reaction is first order in IO₃⁻, first order in SO₃²⁻, and first order in H⁺. Therefore, the rate law for the reaction can be written as: Rate = k[IO₃⁻][SO₃²⁻][H⁺] where k is the rate constant.

(b) Determine the factor by which the rate changes when pH is changed and if the reaction is faster or slower at lower pH.

We know that pH is the negative logarithm of the hydrogen ion concentration: pH = -log[H⁺], so [H⁺] = 10^(-pH). Now, let's find the hydrogen ion concentrations at pH 5.00 and 3.50: [H⁺]_5.00 = 10^(-5.00) [H⁺]_3.50 = 10^(-3.50) Now let's find the ratio of the reaction rates at these two pH: Rate ratio = (Rate at pH 3.50) / (Rate at pH 5.00) Rate ratio = (k[IO₃⁻][SO₃²⁻][H⁺]_3.50) / (k[IO₃⁻][SO₃²⁻][H⁺]_5.00) Since [IO₃⁻] and [SO₃²⁻] remain unchanged, they cancel out in the ratio: Rate ratio = [H⁺]_3.50 / [H⁺]_5.00 Now substitute the concentrations of H⁺: Rate ratio = (10^(-3.50)) / (10^(-5.00)) Rate ratio = 10^(1.5) Calculating the exact value: Rate ratio ≈ 31.6 Thus, the rate of the reaction changes by a factor of 31.6 when the pH is lowered from 5.00 to 3.50. Since the reaction is first-order in H⁺, the reaction rate is directly proportional to the H⁺ concentration, which means the reaction proceeds more quickly at lower pH.

(c) Explain the pH dependency of the reaction although H⁺ doesn't appear in the overall reaction.

Although H⁺ ions are not involved in the overall reaction, they might be involved in one or more elementary steps of the reaction mechanism. For example, a possible elementary step could be the transfer of a proton (H⁺) between reactants, leading to the formation of intermediate species that participate in subsequent steps. The lower the pH, the higher the concentration of H⁺ ions, which increases the rate of proton transfer in the elementary steps. The pH dependency of the given reaction indicates that H⁺ ions play a role in the reaction mechanism, even if they do not appear in the overall balanced equation.

Key concepts

Rate Law

In chemical kinetics, the rate law is an equation that relates the reaction rate with the concentration of reactants. For the reaction between iodate (\(\mathrm{IO}_3^-\)) and sulfite (\(\mathrm{SO}_3^{2-}\)), the rate law is determined based on the order of reaction with respect to each reactant. Given that it is first order in \(\mathrm{IO}_3^-\), first order in \(\mathrm{SO}_3^{2-}\), and first order in \(\mathrm{H}^+\), the rate law can be expressed as:

• Rate = \(k[\mathrm{IO}_3^-][\mathrm{SO}_3^{2-}][\mathrm{H}^+]\)

Here, \(k\) is the rate constant. Each concentration is raised to the power of its respective order, which in this case is one. This means the rate of the reaction depends linearly on the concentration of each reactant. Understanding the rate law is essential to determining how changes in concentration affect the reaction rate directly, offering insights into reaction dynamics.

Reaction Order

The reaction order tells us how the concentration of a reactant influences the overall reaction rate. With reaction orders typically being integers (such as zero, first, or second), they define the power to which the concentration term of a reactant is raised in the rate law equation:

• For the iodate reduction by sulfite, the reaction is first order in each reactant: \(\mathrm{IO}_3^-\), \(\mathrm{SO}_3^{2-}\), and \(\mathrm{H}^+\).

Therefore, each doubling of a reactant's concentration will result in the doubling of the rate if all other conditions remain the same. Total reaction order, in this case, is the sum of orders of all contributing reactants, which is three (sum of the individual orders). Reaction order is pivotal for predicting how the reaction rate responds to concentration changes, crucial for controlling industrial chemical processes.

pH Impact on Reaction Rate

The pH of a solution directly influences its hydrogen ion concentration (\([\mathrm{H}^+]\)), which can have a significant impact on the reaction rate. In our iodine and sulfite reaction:

• The rate law shows that the reaction is first order with respect to \([\mathrm{H}^+]\).
• This means the rate is directly proportional to the concentration of \(\mathrm{H}^+\) ions.

As the pH decreases from 5.00 to 3.50, \([\mathrm{H}^+]\) increases, thus increasing the reaction rate by a factor of approximately 31.6. In acidic solutions (lower pH), more H⁺ ions are available, hence the reaction speeds up. This dependence on pH is fundamental in biochemical systems, where enzyme activities and reaction speeds are often pH-sensitive.

Reaction Mechanism

The reaction mechanism outlines the step-by-step sequence of elementary reactions by which a chemical change occurs. Though \(\mathrm{H}^+\) does not appear in the overall balanced equation for the reaction between iodate and sulfite, it still influences the mechanism:

• \(\mathrm{H}^+\) might participate in intermediate steps where proton transfer occurs.
• These steps involve the formation of transient intermediate species crucial for reaction progress.
• The role of \(\mathrm{H}^+\) as a catalyst or facilitator in these steps is what makes the reaction pH-dependent.

Understanding the reaction mechanism is important as it provides a more complex view of how the reaction proceeds at the molecular level. It explains why changes in pH, despite not printing changes in stoichiometry, can significantly alter the rate by modifying these intermediate stages.