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Chapter 18: Problem 93

For the decomposition of formic acid, $$\begin{aligned} \mathrm{HCOOH}(l) \longrightarrow & \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}(g) \\ \Delta H^{\circ}=+29 \mathrm{~kJ} / \mathrm{mol} \text { at } 25^{\circ} \mathrm{C} \end{aligned}$$ a. Does the tendency of this reaction to proceed to a state of minimum energy favor the formation of water and carbon monoxide or formic acid? Explain. b. Does the tendency of this reaction to proceed to a state of maximum entropy favor the formation of products or reactants? Explain.

Short Answer

Expert verified
a. Minimum energy favors reactants (formic acid). b. Maximum entropy favors products (H2O and CO).

Step by step solution

01

Understanding Enthalpy Change

The given reaction has an enthalpy change of \( \Delta H^{\circ} = +29 \) kJ/mol. A positive \( \Delta H \) indicates that the reaction is endothermic, meaning it absorbs energy from the surroundings. Typically, reactions tend to proceed towards a state of minimum energy, which favors exothermic reactions (negative \( \Delta H \)). Thus, the tendency to minimize energy favors the reactants, formic acid, since breaking it down into products requires energy input.
02

Assessing Entropy Change

Consider the change in entropy, which is a tendency towards disorder. In this reaction, formic acid decomposes into water and carbon monoxide. Initially, there is one molecule of liquid reactant, and upon decomposing, there are a liquid and a gas molecule. Gases generally have higher entropy than liquids, and the number of molecules increases, which tends to increase entropy. Thus, the reaction proceeds toward maximum entropy, favoring the products.

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Key Concepts

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

Enthalpy Change
Enthalpy change is an essential concept in thermodynamics, which measures the heat absorbed or released during a chemical reaction at constant pressure.
For the decomposition of formic acid, the enthalpy change is given as \( \Delta H^{\circ} = +29 \) kJ/mol.
This positive value indicates that the reaction is endothermic, meaning it requires energy input from the surrounding environment. In simpler terms, the decomposition absorbs heat rather than releasing it.

In chemical reactions, systems often prefer to move towards a state of minimum energy, which makes exothermic reactions (negative \( \Delta H \)) more favorable. In this context, since the decomposition of formic acid into water and carbon monoxide requires heat, it is not spontaneously favored if energy conservation is the primary goal.

The reaction, therefore, leans towards maintaining the formic acid rather than breaking it into products due to the external energy requirement.
Entropy Change
Entropy is a measure of randomness or disorder within a system. An increase in entropy typically indicates a move towards a more disordered state.
During the decomposition of formic acid, a single molecule (formic acid) transforms into one water molecule and one carbon monoxide molecule.

This reaction results in an increase in the number of particles and introduces a gaseous product, carbon monoxide, which has higher entropy compared to liquids due to the greater freedom of motion of gas molecules.

As a general tendency, reactions proceed towards maximum entropy or randomness. For the decomposition of formic acid, the increase in entropy favors the formation of the products, water and carbon monoxide. This is because breaking apart the formic acid into separate molecules results in a higher level of disorder.
Endothermic Reaction
An endothermic reaction is characterized by the absorption of energy from its surroundings.
This occurs because the products have higher potential energy compared to the reactants.
In the case of formic acid decomposition, the reaction absorbs \( +29 \) kJ/mol of heat, indicating it's endothermic.

In practical scenarios, an endothermic reaction might feel cold to the touch because it draws heat from the environment to proceed.
This requirement for external energy input is a crucial factor why such reactions are less likely to occur spontaneously without a continuous energy supply.

In simpler terms, if you were conducting this reaction in a lab, you would need to supply heat continuously to keep it going, as the products (water and carbon monoxide) are formed by taking in energy rather than giving off heat as they form.

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Most popular questions from this chapter

The entropy change \(\Delta S\) for a phase transition equals \(\Delta H / T,\) where \(\Delta H\) is the enthalpy change. Why is it that the entropy change for a system in which a chemical reaction occurs spontaneously does not equal \(\Delta H / T\) ?

Which of the following are spontaneous processes? a.A cube of sugar dissolves in a cup of hot tea. b.A rusty crowbar turns shiny. c.Butane from a lighter burns in air. d.A clock pendulum, initially stopped, begins swinging. e.Hydrogen and oxygen gases bubble out from a glass of pure water.

Define the free energy \(G\). How is \(\Delta G\) related to \(\Delta H\) and \(\Delta S ?\)

On the basis of \(\Delta G^{\circ}\) for each of the following reactions, decide whether the reaction is spontaneous or nonspontaneous as written. Or, if you expect an equilibrium mixture with significant amounts of both reactants and products, say so. a.\(\mathrm{SO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \longrightarrow 3 \mathrm{~S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) ; \Delta G^{\circ}=-91 \mathrm{~kJ}\) b.\(2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{O}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) ; \Delta G^{\circ}=-211 \mathrm{~kJ}\) c. \(\mathrm{HCOOH}(l) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) ; \Delta G^{\circ}=119 \mathrm{~kJ}\) d.\(\mathrm{I}_{2}(s)+\mathrm{Br}_{2}(l) \longrightarrow 2 \operatorname{IBr}(g) ; \Delta G^{\circ}=7.5 \mathrm{~kJ}\) e.\(\mathrm{NH}_{4} \mathrm{Cl}(s) \longrightarrow \mathrm{NH}_{3}(g)+\mathrm{HCl}(g) ; \Delta G^{\circ}=92 \mathrm{~kJ}\)

The heat of vaporization of carbon disulfide, \(\mathrm{CS}_{2}\), at \(25^{\circ} \mathrm{C}\) is \(27.2 \mathrm{~kJ} / \mathrm{mol}\). What is the entropy change when \(1.00 \mathrm{~mol}\) of vapor in equilibrium with liquid condenses to liquid at \(25^{\circ} \mathrm{C}\) ? The entropy of this vapor at \(25^{\circ} \mathrm{C}\) is \(243 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})\). What is the entropy of the liquid at this temperature?
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