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(a) Why are tables of standard enthalpies of formation so useful? (b) What is the value of the standard enthalpy of formation of an element in its most stable form? (c) Write the chemical equation for the reaction whose enthalpy change is the standard enthalpy of formation of sucrose (table sugar), \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s), \Delta H_{f}^{\circ}\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right]\)

Short Answer

Expert verified
(a) Tables of standard enthalpies of formation are useful because they provide information about enthalpy changes during the formation of compounds from their constituent elements, enabling chemists to calculate enthalpy changes without experimental measurements. They help in predicting stability, synthesis energy, and energy released during decomposition. (b) The standard enthalpy of formation of an element in its most stable form is always zero, as the formation of an element from itself does not require any energy changes. (c) The chemical equation for the formation of sucrose is: \(12 \: \mathrm{C}(s)+11 \: \mathrm{O}_{2}(g)+22 \: \mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)\)

Step by step solution

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Part (a): Importance of standard enthalpies of formation tables

Tables of standard enthalpies of formation (∆Hf°) are useful because they provide valuable information about the enthalpy changes that occur during the formation of compounds from their constituent elements. This enables chemists to easily calculate the enthalpy changes of different reactions without having to measure them experimentally. They can be used along with Hess's Law to determine the enthalpy change for any chemical reaction, by subtracting the sum of the standard enthalpies of formation of the reactants from those of the products. These tables help in predicting the stability of a compound, the energy needed to synthesize it, and the energy released during its decomposition.
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Part (b): Enthalpy of formation for an element

The standard enthalpy of formation of an element in its most stable form is always zero. This is by definition because the formation of an element from itself does not require any energy changes or cause any enthalpy change.
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Part (c): Chemical equation for sucrose formation

The standard enthalpy of formation of sucrose is the enthalpy change associated with the formation of one mole of sucrose from its constituent elements in their standard states. Since sucrose has the molecular formula \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)\), the chemical equation for its formation is as follows: \(12 \: \mathrm{C}(s)+11 \: \mathrm{O}_{2}(g)+22 \: \mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)\) This is the chemical equation for the reaction whose enthalpy change is the standard enthalpy of formation of sucrose.

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Key Concepts

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

Enthalpy Change Calculation
Understanding enthalpy change calculations is crucial for students who want to grasp the energetic impact of chemical reactions. Enthalpy, represented by the symbol 'H', is the measure of the total heat content of a system at constant pressure. Essentially, when a compound forms from its elements, or reacts with another compound, there is an enthalpy change, \( \Delta H \), which can be either exothermic (releasing heat) or endothermic (absorbing heat).

Calculating this change in enthalpy allows us to predict whether a reaction will emit or require energy. The standard enthalpy of formation, \( \Delta H_f^\circ \), refers to the enthalpy change when one mole of a compound is formed from its elements in their standard states. For sucrose, this would involve the combination of carbon, hydrogen, and oxygen in the right proportions to produce a mole of sucrose, as shown in the exercise solution.

To calculate enthalpy changes for complex reactions, we can simply refer to tables of standard enthalpies of formation. By summing the enthalpies of formation for the reactants and subtracting the sum for the products, we get the enthalpy change for the overall reaction, per the equations:
\[ \Delta H_{reaction} = \biggl( \sum \Delta H_f^\circ \text{(products)} \biggr) - \biggl( \sum \Delta H_f^\circ \text{(reactants)} \biggr) \]
It is this robust method that allows students to solve a variety of problems related to chemical energetics without the need for complex experiments.
Hess's Law
Hess's Law is a powerful tool in the arsenal of chemistry and a best friend to students battling through thermodynamics. It states that the total enthalpy change of a chemical reaction is the same, regardless of the number of steps the reaction is carried out in. This means that the enthalpy of a reaction is a state function—it only depends on the initial and final states, regardless of the path taken.

In practice, Hess's Law enables us to piece together known reactions like a puzzle to determine the enthalpy change of an overall reaction, even if it has never been measured directly. We are in essence accounting clerks, tallying up the energy changes of individual steps. For example, if we were to determine the enthalpy change in the formation of sucrose, we could use known enthalpies of formation for carbon in graphite form, oxygen gas, and hydrogen gas to find the sum of the energy needed or released.

Applying Hess's Law can seem like traveling through a maze - we sometimes need to reverse reactions (changing the sign of their enthalpies) or scale them to match the moles needed in the overall equation. However, by systematically arranging individual reactions with known enthalpies, we can solve for the unknown, much like completing the final piece of a jigsaw puzzle.
Chemical Reaction Energetics
At the heart of chemistry lies the study of how substances interact with one another—chemical reaction energetics are an essential piece of this puzzle. This field examines the energy changes that accompany chemical reactions, often visualized in the form of potential energy diagrams. These diagrams represent the energy of reactants and products and show the energy absorbed or released (the enthalpy change) during a reaction.

Energetics encompass not only the amount of energy but its quality—the difference between heat (often as a result of bond breaking) and work (such as expansion against pressure). When studying the energetics of a reaction, like the formation of sucrose from carbon, hydrogen, and oxygen, it's vital to consider the conservation of energy. The fundamental principle at play is the first law of thermodynamics, which reminds us that energy cannot be created nor destroyed but transformed from one form to another.

Therefore, by tracking the energy flow during a reaction, we can make critical decisions about the feasibility of industrial processes, the efficiency of fuel reactions, and much more. The energetics of chemical reactions are not just numbers on a page; they represent the practical and sometimes breathtaking interplay of matter and energy that shapes our world.

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Most popular questions from this chapter

The gas-phase reaction shown, between \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\), was run in an apparatus designed to maintain a constant pressure. (a) Write a balanced chemical equation for the reaction depicted and predict whether \(w\) is positive, negative, or zero. (b) Using data from Appendix \(\mathrm{C}\), determine \(\Delta H\) for the formation of one mole of the product. Why is this enthalpy change called the enthalpy of formation of the involved product? [Sections 5.3 and 5.7\(]\)

(a) Calculate the kinetic energy in joules of a \(1200-\mathrm{kg}\) automobile moving at \(18 \mathrm{~m} / \mathrm{s}\). (b) Convert this energy to calories. (c) What happens to this energy when the automobile brakes to a stop?

(a) When a 4.25 -g sample of solid ammonium nitrate dissolves in \(60.0 \mathrm{~g}\) of water in a coffee-cup calorimeter (Figure 5.18), the temperature drops from \(22.0^{\circ} \mathrm{C}\) to \(16.9^{\circ} \mathrm{C}\). Calculate \(\Delta H\left(\right.\) in \(\left.\mathrm{kJ} / \mathrm{mol} \mathrm{NH}_{4} \mathrm{NO}_{3}\right)\) for the solution process $$ \mathrm{NH}_{4} \mathrm{NO}_{3}(s) \longrightarrow \mathrm{NH}_{4}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) $$ Assume that the specific heat of the solution is the same as that of pure water. (b) Is this process endothermic or exothermic?

At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat \(\mathrm{KClO}_{3}\) : \(2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) \quad \Delta H=-89.4 \mathrm{~kJ}\) For this reaction, calculate \(\Delta H\) for the formation of (a) 1.36 mol of \(\mathrm{O}_{2}\) and (b) \(10.4 \mathrm{~g}\) of \(\mathrm{KCl}\) (c) The decomposition of \(\mathrm{KClO}_{3}\) proceeds spontaneously when it is heated. Do you think that the reverse reaction, the formation of \(\mathrm{KClO}_{3}\) from \(\mathrm{KCl}\) and \(\mathrm{O}_{2},\) is likely to be feasible under ordinary conditions? Explain your answer.

Calcium carbide \(\left(\mathrm{CaC}_{2}\right)\) reacts with water to form acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and \(\mathrm{Ca}(\mathrm{OH})_{2}\). From the following enthalpy of reaction data and data in Appendix C, calculate \(\Delta H_{f}^{\circ}\) for \(\mathrm{CaC}_{2}(s):\) $$ \begin{aligned} \mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{C}_{2} \mathrm{H}_{2}(g) \\ \Delta H^{\circ}=-127.2 \mathrm{~kJ} \end{aligned} $$

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