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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 allow for convenient calculation of enthalpy changes in chemical reactions using Hess's Law, providing valuable thermodynamic information for various applications. (b) The standard enthalpy of formation for an element in its most stable form is defined as zero. (c) The chemical equation for the formation of sucrose is: 12 \(\mathrm{C}(s, \text{graphite})\) + 11 \(\mathrm{O}_{2}(g)\) + 11 \(\mathrm{H}_{2}(g)\) → \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)\).

Step by step solution

01

Answer (a): Usefulness of tables of standard enthalpies of formation

Tables of standard enthalpies of formation are useful because they provide a convenient and standardized way of calculating the enthalpy change for various chemical reactions. By knowing the enthalpy of formation values for the reactants and the products in a reaction, we can determine the overall enthalpy change for the reaction using Hess's Law. This information is crucial in understanding the thermodynamic properties of a reaction, which can be helpful in various industrial, environmental, and research applications.
02

Answer (b): Enthalpy of Formation for an Element in its Most Stable Form

The standard enthalpy of formation for an element in its most stable form is always defined as zero (0). This is because the formation of an element in its standard state from itself requires no change in energy, as no chemical transformation or bonds breaking/forming occurs. As standard enthalpies of formation are relative values based on the formation of a compound from its constituent elements, a reference point needs to be established. By convention, this reference point is set as zero for elements in their most stable forms.
03

Answer (c): Chemical Equation for Sucrose Formation

To write the chemical equation for the formation of sucrose from its constituent elements, we must consider the fact that it has a chemical formula of \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)\). To find its enthalpy of formation, we need to form it from its constituent elements in their most stable states: Carbon in the form of graphite (\(\mathrm{C}(s, \text{graphite})\)), Hydrogen as H2 gas (\(\mathrm{H}_{2}(g)\)), and Oxygen as O2 gas (\(\mathrm{O}_{2}(g)\)). The chemical equation for the formation of sucrose is: 12 \(\mathrm{C}(s, \text{graphite})\) + 11 \(\mathrm{O}_{2}(g)\) + 11 \(\mathrm{H}_{2}(g)\) → \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)\)

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Key Concepts

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

Enthalpy Change
Enthalpy change, denoted by \( \Delta H \), represents the heat absorbed or released during a chemical reaction at constant pressure. It is a thermodynamic quantity that gives insights into the energy changes occurring in a reaction. The sign of \( \Delta H \), positive or negative, indicates whether the process is endothermic (heat absorbed) or exothermic (heat released), respectively. In educational contexts, mastering the concept of enthalpy change is crucial for predicting the feasibility and spontaneity of reactions. Understanding \( \Delta H \), students can infer much about a reaction's behavior without performing it, which is pivotal in planning and analyzing chemical processes.

In the case of formation reactions, the \( \Delta H_{f}^{\circ} \) of a compound is particularly significant as it reveals the energy required to form one mole of that compound from its elements in their standard states. For sucrose, the enthalpy change of formation would inform us about the energy dynamics involved in its creation from carbon (graphite), hydrogen (gas), and oxygen (gas).
Hess's Law
Hess's Law is a foundational concept in thermodynamics stating that the total enthalpy change for a chemical reaction is the same, regardless of the route by which the chemical reaction occurs, assuming that the initial and final conditions are the same. This principle essentially means that if a reaction can be expressed as a series of steps, the sum of the enthalpy changes for each step will equal the overall enthalpy change for the complete reaction.

This law allows chemists to calculate the enthalpy change of a complex reaction by breaking it down into a series of simpler reactions for which standard enthalpies of formation are available. It is especially useful when direct measurement of the enthalpy change for a reaction is not feasible. For students struggling to grasp this concept, envisioning energy levels as steps that must always add up to the same overall change, regardless of the path taken, can facilitate comprehension.
Thermodynamic Properties

Understanding Thermodynamic Properties


Thermodynamic properties of substances, such as enthalpy, internal energy, entropy, and Gibbs free energy, serve as critical parameters for understanding and predicting the behavior of chemical reactions. These properties can help determine whether a reaction will occur spontaneously and the extent to which it will proceed. For students, it can be helpful to think of thermodynamic properties as the 'personality traits' of a chemical process, defining how it will 'act' under various conditions.

For example, the standard enthalpy of formation is a thermodynamic property that can be utilized to calculate the energy changes in reactions through Hess's Law. In educational settings, emphasizing the interrelation between thermodynamic properties and real-world applications, such as in energy generation, material synthesis, and environmental sciences, can help students appreciate their practical significance.
Chemical Reaction Equation

Dissecting the Chemical Reaction Equation


The chemical reaction equation serves as a symbolic representation of a chemical reaction, where reactants are transformed into products. It conveys essential information such as the nature and quantity of the involved substances and the stoichiometry of the reaction. Grasping this concept is foundational in chemistry education, as it forms the basis for understanding how different elements combine and transform into new compounds.

For the formation of sucrose, the reaction equation outlines the exact elements and their specific states needed to synthesize a mole of sucrose. The coefficients in the reaction equation indicate the proportion of moles required for each reactant. Proper interpretation and balancing of these equations are crucial skills in chemistry education to predict the outcomes of reactions and to comprehend the changes in enthalpy associated with the process.

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Most popular questions from this chapter

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 \(\mathrm{CaC}_{2}(s)\) : $$ \begin{array}{r} \mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{C}_{2} \mathrm{H}_{2}(g) \\ \Delta H^{\circ}=-127.2 \mathrm{~kJ} \end{array} $$

(a) Under what condition will the enthalpy change of a process equal the amount of heat transferred into or out of the system? (b) During a constant- pressure process, the system releases heat to the surroundings. Does the enthalpy of the system increase or decrease during the process? (c) In a constant-pressure process, \(\Delta H=0\). What can you conclude about \(\Delta E, q\), and \(w\) ?

Consider the following reaction: $$ 2 \mathrm{CH}_{3} \mathrm{OH}(g) \longrightarrow 2 \mathrm{CH}_{4}(g)+\mathrm{O}_{2}(g) \quad \Delta H=+252.8 \mathrm{~kJ} $$ (a) Is this reaction exothermic or endothermic? (b) Calculate the amount of heat transferred when \(24.0 \mathrm{~g}\) of \(\mathrm{CH}_{3} \mathrm{OH}(g)\) is decomposed by this reaction at constant pressure. (c) For a given sample of \(\mathrm{CH}_{3} \mathrm{OH}\), the enthalpy change during the reaction is \(82.1 \mathrm{~kJ}\). How many grams of methane gas are produced? (d) How many kilojoules of heat are released when \(38.5 \mathrm{~g}\) of \(\mathrm{CH}_{4}(g)\) reacts completely with \(\mathrm{O}_{2}(g)\) to form \(\mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})\) at constant pressure? 5.45 When solutions containing silver ions and chloride ions are mixed, silver chloride precipitates: $$ \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \operatorname{AgCl}(s) \quad \Delta H=-65.5 \mathrm{~kJ} $$ (a) Calculate \(\Delta H\) for the production of \(0.450 \mathrm{~mol}\) of \(\mathrm{AgCl}\) by this reaction. (b) Calculate \(\Delta H\) for the production of \(9.00 \mathrm{~g}\) of \(\mathrm{AgCl}\). (c) Calculate \(\Delta H\) when \(9.25 \times 10^{-4} \mathrm{~mol}\) of \(\mathrm{AgCl}\) dissolves in water.

You may have noticed that when you compress the air in a bicycle pump, the body of the pump gets warmer. (a) Assuming the pump and the air in it comprise the system, what is the sign of \(w\) when you compress the air? (b) What is the sign of \(q\) for this process? (c) Based on your answers to parts (a) and (b), can you determine the sign of \(\Delta E\) for compressing the air in the pump? If not, what would you expect for the sign of \(\Delta E\) ? What is your reasoning? [Section 5.2]

Using values from Appendix \(C\), calculate the standard enthalpy change for each of the following reactions: (a) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (b) \(\mathrm{Mg}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{MgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)\) (c) \(\mathrm{N}_{2} \mathrm{O}_{4}(g)+4 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{SiCl}_{4}(l)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{SiO}_{2}(s)+4 \mathrm{HCl}(g)\)

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