/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 113 A solid sample of benzoic acid, ... [FREE SOLUTION] | 91影视

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A solid sample of benzoic acid, a carboxylic acid, is in equilibrium with an aqueous solution of benzoic acid. A tiny quantity of \(\mathrm{D}_{2} \mathrm{O},\) water containing the isotope \({ }^{2} \mathrm{H}\), deuterium, is added to the solution. The solution is allowed to stand at constant temperature for several hours, after which some of the solid benzoic acid is removed and analyzed. The benzoic acid is found to contain a tiny quantity of deuterium, D, and the formula of the deuterium-containing molecules is \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOD}\). Explain how this can happen.

Short Answer

Expert verified
Deuterium from \(\text{D}_2\text{O}\) replaces hydrogen in benzoic acid through equilibrium reactions, forming deuterated acid.

Step by step solution

01

Understanding the Concept of Equilibrium

When benzoic acid, a carboxylic acid, is in equilibrium with its aqueous solution, the acid can transfer protons (H+) between the solid and liquid phases. This is due to the nature of acid-base equilibrium, where an acid can donate protons to a base, in this case, primarily water.
02

Introduction of Deuterium and its Role

When a small amount of \(\text{D}_2\text{O}\), which contains deuterium, a heavier isotope of hydrogen with one proton and one neutron, is added, it participates in chemical equilibrium similarly to normal water (\(\text{H}_2\text{O}\)). Deuterium can replace hydrogen atoms in molecules forming deuterium analogs.
03

Proton (or Deuterium) Exchange Mechanism

In the aqueous solution, a series of proton exchange reactions can occur due to the dynamic nature of equilibrium. The regular hydrogen in benzoic acid can be replaced by deuterium from the added \(\text{D}_2\text{O}\), resulting in the formation of \(\text{C}_6\text{H}_5\text{COOD}\), a deuterated benzoic acid molecule.
04

Reversibility and Time Allowance at Constant Temperature

By allowing the solution to stand for several hours at constant temperature, the equilibrium reaches a state where exchange reactions have distributed deuterium into the solid form of benzoic acid. This is why, upon removal and analysis, solid benzoic acid is found with a small quantity of deuterium.

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Key Concepts

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

Acid-Base Equilibrium
Acid-base equilibrium is a fundamental concept in chemistry involving the transfer of protons (H鈦) between acids and bases. In our example, benzoic acid is in equilibrium with its aqueous solution. What's happening here is that benzoic acid can give away protons to water molecules. This interaction allows the protons to move between the solid benzoic acid and the dissolved molecules in the water, maintaining a balance.

A key aspect of equilibrium is that both forward and backward reactions occur at the same rate, meaning the concentrations of reactants and products remain constant over time. When deuterated water ( D鈧侽) is introduced, it behaves like any regular water in this equilibrium but adds an interesting twist due to the presence of deuterium. Adding deuterated water to the mix doesn't disturb the equilibrium significantly, it just introduces deuterium atoms into the environment, which can participate in exchanges like their hydrogen counterparts.
Deuterium
Deuterium is an isotope of hydrogen, and it is heavier because it contains one neutron along with the one proton typical of hydrogen atoms. So, on its own, deuterium forms a molecule like water called deuterium oxide ( D鈧侽). Although similar to regular water, it's distinguishable by its physical properties like being heavier.
  • Deuterium can replace hydrogen in many compounds.
  • Its chemical behavior is very close to regular hydrogen.
  • It serves as a useful tracer in experimental chemistry because of these properties.
In the context of our benzoic acid scenario, deuterium from D鈧侽 can substitute a regular hydrogen atom in the acid, forming a compound called C鈧咹鈧匔OOD. This substitution highlights deuterium's role in the proton exchange reactions that occur during equilibrium with hydrogen.
Proton Exchange Reaction
Proton exchange reactions are key in understanding isotope exchanges like those seen with deuterium. These reactions involve swapping protons between molecules. In the case of benzoic acid, it involves trading places between hydrogen and deuterium atoms due to their similar chemical properties.
  • The aqueous environment supports dynamic interactions where hydrogen can be swapped with deuterium from D鈧侽.
  • This exchange doesn't require added energy due to the equilibrium established.
  • Such reactions can occur repeatedly, given the right conditions.
When you let the solution sit for hours, the benzoic acid can incrementally substitute its hydrogen atoms for deuterium across its molecules. Over time, some of these switches become evident in the solid form of benzoic acid. This incremental exchange results in a measurable amount of deuterium integrated into the compound, observable during analysis.

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Most popular questions from this chapter

Hydrogen, bromine, and \(\mathrm{HBr}\) in the gas phase are in equilibrium in a container of fixed volume. \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Br}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HBr}(\mathrm{g}) \quad \Delta_{r} H^{\circ}=-103.7 \mathrm{~kJ} / \mathrm{mol}\) How will each of these changes affect the indicated quantities? Write "increase," "decrease," or "no change." \begin{tabular}{l} \hline Change & {\(\left[\mathrm{Br}_{2}\right]\)} & {\([\mathrm{HBr}]\)} & \(K_{c}\) & \(K_{\mathrm{p}}\) \\ \hline Some \(\mathrm{H}_{2}\) is added to the \\ container. \\ The temperature of the gases \\ in the container is increased. \\ The pressure of \(\mathrm{HBr}\) is \\ increased. \end{tabular}

Think of an experiment you could do to demonstrate that the equilibrium $$ 2 \mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) $$ is a dynamic process in which the forward and reverse reactions continue to occur after equilibrium has been achieved. Describe how such an experiment might be carried out.

Nitrosyl chloride, NOCl, decomposes to \(\mathrm{NO}\) and \(\mathrm{Cl}_{2}\) at high temperatures. $$ 2 \mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) $$ Suppose you place \(2.00 \mathrm{~mol} \mathrm{NOCl}\) in a \(1.00-\mathrm{L}\) flask, seal it, and raise the temperature to \(462^{\circ} \mathrm{C}\). When equilibrium has been established, \(0.66 \mathrm{~mol} \mathrm{NO}\) is present. Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the decomposition reaction from these data.

Carbon dioxide reacts with carbon to give carbon monoxide according to the equation $$ \mathrm{C}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g}) $$ At \(700 .{ }^{\circ} \mathrm{C},\) a \(2.0-\mathrm{L}\) sealed flask at equilibrium contains $$ 0.10 \mathrm{~mol} \mathrm{CO}, 0.20 \mathrm{~mol} \mathrm{CO}_{2}, \text { and } 0.40 \mathrm{~mol} \mathrm{C} . \text { Calculate } $$ the equilibrium constant \(K_{\mathrm{P}}\) for this reaction at the specified temperature.

For each of these chemical reactions, predict whether the equilibrium constant at \(25^{\circ} \mathrm{C}\) is greater than 1 or less than \(1,\) or state that insufficient information is available. Also indicate whether each reaction is product-favored or reactant-favored. (a) \(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}) \quad \Delta_{\mathrm{r}} H^{\circ}=-115 \mathrm{~kJ} / \mathrm{mol}\) (b) \(2 \mathrm{O}_{3}(\mathrm{~g}) \rightleftharpoons 3 \mathrm{O}_{2}(\mathrm{~g})\) \(\Delta_{\mathrm{r}} H^{\circ}=-285 \mathrm{~kJ} / \mathrm{mol}\) (c) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NCl}_{3}(\mathrm{~g})\) \(\Delta_{1} H^{\circ}=460 \mathrm{~kJ} / \mathrm{mol}\)

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