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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 \(\mathrm{pH}\) is lowered from \(5.00\) to 3.50? Does the reaction proceed faster or slower at the lower pH? (c) By using the concepts discussed in Section 14.6, explain how the reaction can be pH- dependent even though \(\mathrm{H}^{+}\) does not appear in the overall reaction.

Short Answer

Expert verified
The rate law for the given reaction is: \(Rate = k[\mathrm{IO}_{3}^{-}][\mathrm{SO}_{3}^{2-}][\mathrm{H}^{+}]\). When the pH is lowered from 5 to 3.5, the rate change factor is calculated as \(10^{1.5}\), indicating that the reaction is faster at the lower pH value. The pH-dependency of the reaction arises from the involvement of \(\mathrm{H}^{+}\) in intermediate steps, affecting the equilibrium of these intermediate species and consequently the overall reaction rate.

Step by step solution

01

Write the Rate Law

Since the iodate ion (\(\mathrm{IO}_{3}^{-}\)) is first order, sulfite ion (\(\mathrm{SO}_{3}^{2-}\)) is first order, and hydrogen ion (\(\mathrm{H}^{+}\)) is first order, we can write the rate law as follows: \[Rate = k[\mathrm{IO}_{3}^{-}][\mathrm{SO}_{3}^{2-}][\mathrm{H}^{+}]\]
02

Calculate the change in \(\mathrm{H}^{+}\) concentration

We are given the initial pH value of 5 and the new pH value of 3.5, we can calculate the change in the concentration of \(\mathrm{H}^{+}\) using: \[ \mathrm{H^+} = 10^{-pH} \] Calculate the \(\mathrm{H}^{+}\) for both situations: Initial pH: \[ [\mathrm{H}^{+}_{1}] = 10^{-5} \] New pH: \[ [\mathrm{H}^{+}_{2}] = 10^{-3.5} \]
03

Calculate the rate change factor

Next, we divide the new concentration by the initial concentration to find the factor by which the rate will change: Rate Change Factor: \( \frac{[\mathrm{H}^{+}_{2}]}{[\mathrm{H}^{+}_{1}]} \) \[ Rate\:Change\:Factor = \frac{10^{-3.5}}{10^{-5}} \] Now simplify: \[ Rate\:Change\:Factor = 10^{(5-3.5)} = 10^{1.5} \]
04

Determine if the reaction is faster or slower

Since the rate change factor is greater than 1, the rate of the reaction will increase when lowering the pH from 5 to 3.5, and thus, the reaction proceeds faster at the lower pH.
05

Explain pH-dependency

Even though \(\mathrm{H}^{+}\) does not appear in the overall reaction, it is involved in intermediate steps, influencing the reaction pathway. The pH changes the \(\mathrm{H}^{+}\) concentration, which modifies the equilibrium of these intermediate species. This can, in turn, affect the rate of the overall reaction. In this case, a lower pH leads to a higher concentration of \(\mathrm{H}^{+}\). Since the reaction rate is first-order in \(\mathrm{H}^{+}\), a higher concentration of \(\mathrm{H}^{+}\) will result in a faster reaction.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rate Law
In reaction kinetics, the rate law offers a mathematical expression that defines the relationship between the concentration of reactants and the rate of a chemical reaction. For the given reaction involving iodate and sulfite ions, the rate law is a direct outcome of the reaction being first order with respect to each reactant: iodate (\(\mathrm{IO}_{3}^{-}\)), sulfite (\(\mathrm{SO}_{3}^{2-}\)), and hydrogen ion (\(\mathrm{H}^{+}\)). This implies that the reaction rate is directly proportional to the concentration of each of these species. Therefore, the rate law can be expressed as:
  • \[Rate = k[\mathrm{IO}_{3}^{-}][\mathrm{SO}_{3}^{2-}][\mathrm{H}^{+}]\] where \(k\) is the rate constant.
The rate constant, \(k\), is a proportionality factor that depends on the specific reaction and the conditions under which it occurs. Understanding the rate law enables chemists to predict how changes in concentration affect the reaction rate.
pH Dependence
The pH of a solution can significantly influence the rate of a reaction, especially when it involves hydrogen ions (\(\mathrm{H}^{+}\)) as a reactant. In this reaction, the rate of the reaction is first order in \(\mathrm{H}^{+}\), meaning it depends on the concentration of these ions. Lowering the pH from 5.00 to 3.50 increases the concentration of \(\mathrm{H}^{+}\) ions (calculated by converting the pH to hydrogen ion concentration using the formula: \(\mathrm{H^+} = 10^{-pH}\)).
  • At pH 5.00: \([\mathrm{H}^{+}_{1}] = 10^{-5}\)
  • At pH 3.50: \([\mathrm{H}^{+}_{2}] = 10^{-3.5}\)
Lowering the pH increases the \(\mathrm{H}^{+}\) concentration, leading to an increased reaction rate. When comparing concentrations at these pH levels, the rate increases by a factor of 10 to the power of the difference in pH:
  • Rate Change Factor: \(10^{(5-3.5)} = 10^{1.5}\)
This factor shows the reaction proceeds faster at a lower pH.
Reaction Order
Reaction order indicates the power to which the concentration of a reactant is raised in the rate law. It is an empirical value and does not necessarily correlate with the stoichiometric coefficients in the balanced equation. For this reaction, each component: iodate (\(\mathrm{IO}_{3}^{-}\)), sulfite (\(\mathrm{SO}_{3}^{2-}\)), and hydrogen (\(\mathrm{H}^{+}\)) contributes equally with an order of one, meaning the overall reaction is third-order.A reaction that is first order in a particular reactant means that if you double the concentration of that reactant, the rate of reaction also doubles. Similarly, if you halve the concentration, the reaction rate is halved.
  • First-order concentration effect: Linear relationship with rate.
  • Overall reaction order: Sum of individual orders, which in this case equals 3.
Understanding reaction order helps in making predictions about how changes in conditions can modify the reaction rate.
Concentration Effect
The concentration effect in chemical kinetics refers to how changing the concentration of reactants affects the rate of a reaction. As expressed in the rate law, the rate is proportional to the concentration of reactants to their respective orders.When examining the influence of reactants like iodate and sulfite ions, as well as hydrogen ions, we see how these concentrations directly influence the reaction rate:
  • Increasing concentration of \(\mathrm{IO}_{3}^{-}\) results in a proportional increase in rate.
  • Similarly, increasing \(\mathrm{SO}_{3}^{2-}\) concentration results in a corresponding increase in rate.
  • The increase of \(\mathrm{H}^{+}\) due to lower pH increases the rate by a factor dictated by its order.
The reaction's sensitivity to each reactant's concentration allows for targeted adjustments to reaction conditions to achieve desired rates effectively. By understanding these effects, chemists can better control reactions in both industrial and laboratory settings.

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Most popular questions from this chapter

Consider two solutions, solution \(\mathrm{A}\) and solution B. \(\left[\mathrm{H}^{+}\right]\) in solution \(\mathrm{A}\) is 500 times greater than that in solution \(\mathrm{B}\). What is the difference in the \(\mathrm{pH}\) values of the two solutions?

A \(0.100 M\) solution of chloroacetic acid \(\left(\mathrm{ClCH}_{2} \mathrm{COOH}\right)\) is \(11.0 \%\) ionized. Using this information, calculate \(\left.\left[\mathrm{ClCH}_{2} \mathrm{COO}^{-}\right],\left[\mathrm{H}^{+}\right],\left[\mathrm{ClCH}_{2} \mathrm{COOH}\right)\right]\), and \(K_{a}\) for chloroacetic acid.

Lactic acid, \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\), received its name because it is present in sour milk as a product of bacterial action. It is also responsible for the soreness in muscles after vigorous exercise. (a) The \(\mathrm{pK}_{a}\) of lactic acid is \(3.85\). Compare this with the value for propionic acid \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}, \mathrm{p} K_{a}=4.89\right)\), and explain the differ- ence. (b) Calculate the lactate ion concentration in a \(0.050 \mathrm{M}\) solution of lactic acid. (c) When a solution of sodium lactate, \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COONa}\), is mixed with an aqueous copper(II) solution, it is possible to obtain a solid salt of copper(II) lactate as a blue-green hydrate, \(\left(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COO}\right)_{2} \mathrm{Cu} \cdot x \mathrm{H}_{2} \mathrm{O} .\) Elemental analysis of the solid tells us that the solid is \(22.9 \% \mathrm{Cu}\) and \(26.0 \% \mathrm{C}\) by mass. What is the value for \(x\) in the formula for the hydrate? (d) The acid-dissociation constant for the \(\mathrm{Cu}^{2+}(a q)\) ion is \(1.0 \times 10^{-8}\). Based on this value and the acid-dissociation constant of lactic acid, predict whether a solution of copper(II) lactate will be acidic, basic, or neutral. Explain your answer.

Phenylacetic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{COOH}\right)\) is one of the substances that accumulates in the blood of people with phenylketonuria, an inherited disorder that can cause mental retardation or even death. A \(0.085 \mathrm{M}\) solution of \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{COOH}\) has a pH of \(2.68\). Calculate the \(\mathrm{K}_{a}\) value for this acid.

Based on their compositions and structures and on conjugate acid-base relationships, select the stronger base in each of the following pairs: (a) \(\mathrm{NO}_{3}^{-}\) or \(\mathrm{NO}_{2}^{-}\), (c) \(\mathrm{HCO}_{3}^{-}\) or \(\mathrm{CO}_{3}^{2-}\). (b) \(\mathrm{PO}_{4}{ }^{3-}\) or \(\mathrm{AsO}_{4}{ }^{3-}\)

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