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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 glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{~s}), \Delta \mathrm{H}_{f}^{\circ}\left[\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right]\).

Short Answer

Expert verified
(a) Tables of standard enthalpies of formation are useful because they allow us to calculate the enthalpy change for a given reaction using Hess's law. (b) The standard enthalpy of formation of an element in its most stable form is always zero. (c) The chemical equation for the standard enthalpy of formation of glucose is: \[ 6 \mathrm{C(graphite)} + 6 \mathrm{O_{2}(g)} + 12 \mathrm{H_{2}(g)} \rightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{s}) \]

Step by step solution

01

(a) Importance of Tables of Standard Enthalpies of Formation

Tables of standard enthalpies of formation are useful because they allow us to calculate the enthalpy change for a given reaction. The enthalpy change for a reaction, \(\Delta H_{\text{rxn}}^{\circ}\), can be determined by subtracting the sum of the standard enthalpies of formation of the reactants from the sum of the standard enthalpies of formation of the products. This is known as Hess's law, which states that the overall enthalpy change of a reaction is independent of the route by which the reaction occurs.
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(b) Standard Enthalpy of Formation of an Element in its Most Stable Form

The standard enthalpy of formation of an element in its most stable form is always zero. This is because the formation of an element in its standard state (under standard conditions of temperature and pressure) does not involve any changes in enthalpy. The element is already in its most stable state, so there's no energy change involved in its formation.
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(c) Chemical Equation for the Standard Enthalpy of Formation of Glucose

To write the chemical equation for the reaction whose enthalpy change is the standard enthalpy of formation of glucose, \(\Delta \mathrm{H}_{f}^{\circ}\left[\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right]\), we need to consider the reactants and products that are involved in the formation of glucose from its constituent elements in their standard states. The constituent elements of glucose are carbon (C), hydrogen (H), and oxygen (O). Under standard conditions, the most stable form of carbon is graphite, hydrogen is diatomic gas (H\(_2\)), and oxygen is diatomic gas (O\(_2\)). The stoichiometric ratios in glucose are 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms. The chemical equation for the reaction whose enthalpy change is the standard enthalpy of formation of glucose can be written as: \[ 6 \mathrm{C(graphite)} + 6 \mathrm{O_{2}(g)} + 12 \mathrm{H_{2}(g)} \rightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{s}) \] In this reaction, 6 moles of graphite carbon, 6 moles of diatomic oxygen gas, and 12 moles of diatomic hydrogen gas react to form 1 mole of solid glucose.

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Key Concepts

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

Hess's Law
Hess's Law is a fundamental principle in chemistry that provides a simplified approach to calculate the enthalpy change of reactions. It states that the overall enthalpy change for a chemical reaction is consistent, regardless of the path taken. This means:
  • You can break down complicated reactions into a series of simpler reactions and still obtain the same total enthalpy change.
  • You can use known enthalpy changes of these simpler reactions to calculate the unknown enthalpy change of a more complex reaction.
This principle is especially useful when looking directly at the reactions might be challenging or not possible under normal laboratory conditions. Thanks to Hess's Law, by using standard tables of enthalpies of formation, we can piece together the puzzle of reaction enthalpies and provide a reliable measure of energy changes in chemical processes.
Chemical Reaction Equations
Chemical reaction equations are symbolic representations that describe what happens in a chemical reaction. They follow a particular format that includes:
  • Reactants: The starting materials in a reaction, found on the left side of the equation.
  • Products: The substances formed by the reaction, shown on the right side of the equation.
  • Arrow (\( ightarrow\)): Indicates the direction of the reaction from reactants to products.
Writing these equations requires balancing the number of atoms for each element to ensure they are the same on both sides, maintaining the law of conservation of mass. For example, when writing the equation for the formation of glucose, we consider the most stable forms of its constituent elements (carbon as graphite, hydrogen as \(\text{H}_2\) gas, and oxygen as \(\text{O}_2\) gas) to create a balanced equation. Such equations not only help in understanding reaction mechanics but are critical in calculating energy changes using enthalpy values.
Energy Change in Reactions
Energy change in reactions is essential for understanding how substances transform during a chemical process. It represents the difference in energy between reactants and products. This energy can be absorbed or released, and reactions can be categorized as either:
  • Exothermic: Reactions that release energy, usually in the form of heat, causing the surroundings to warm up.
  • Endothermic: Reactions that absorb energy, leading to a cooling effect on the surrounding environment.
The enthalpy change (\(\Delta H\)) for a reaction can be determined using the standard enthalpies of formation from tables. By applying Hess’s Law, we can calculate the energy change even if the direct measurement is not feasible. Recognizing the energy change in reactions is crucial for understanding reaction feasibility, safety, and the design of chemical processes.

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Most popular questions from this chapter

When a 9.55-g sample of solid sodium hydroxide dissolves in \(100.0 \mathrm{~g}\) of water in a coffee-cup calorimeter (Figure 5.17), the temperature rises from \(23.6^{\circ} \mathrm{C}\) to \(47.4^{\circ} \mathrm{C}\). Calculate \(\Delta H\) (in \(\mathrm{kJ} / \mathrm{mol} \mathrm{NaOH}\) ) for the solution process $$ \mathrm{NaOH}(s) \stackrel{-\cdots}{\mathrm{Na}}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ Assume that the specific heat of the solution is the same as that of pure water.

Gasoline is composed primarily of hydrocarbons, including many with eight carbon atoms, called octanes. One of the cleanest-burning octanes is a compound called \(2,3,4\) -trimethylpentane, which has the following structural formula: The complete combustion of one mole of this compound to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\) leads to \(\Delta H^{\circ}=-5064.9 \mathrm{~kJ} / \mathrm{mol}\). (a) Write a balanced equation for the combustion of 1 mol of \(\mathrm{C}_{8} \mathrm{H}_{18}(l) .(\mathrm{b})\) Write a balanced equation for the formation of \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\) from its elements. (c) By using the information in this problem and data in Table \(5.3\), calculate \(\Delta H_{f}^{\circ}\) for \(2,3,4\) -trimethylpentane.

Limestone stalactites and stalagmites are formed in caves by the following reaction: \(\mathrm{Ca}^{2+}(a q)+2 \mathrm{HCO}_{3}^{-}(a q) \longrightarrow\) \(\mathrm{CaCO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(I)\) If \(1 \mathrm{~mol}\) of \(\mathrm{CaCO}_{3}\) forms at \(298 \mathrm{~K}\) under \(1 \mathrm{~atm}\) pressure, the reaction performs \(2.47 \mathrm{~kJ}\) of \(P-V\) work, pushing back the atmosphere as the gaseous \(\mathrm{CO}_{2}\) forms. At the same time, \(38.95 \mathrm{~kJ}\) of heat is absorbed from the environment. What are the values of \(\Delta H\) and of \(\Delta E\) for this reaction?

(a) What is the kinetic energy in joules of an \(850-\mathrm{lb}\) motorcycle moving at \(66 \mathrm{mph} ?\) (b) By what factor will the kinetic energy change if the speed of the motorcycle is decreased to \(33 \mathrm{mph} ?\) (c) Where does the kinetic energy of the motorcycle go when the rider brakes to a stop?

(a) What is meant by the term standard conditions, with reference to enthalpy changes? (b) What is meant by the term enthalpy of formation? (c) What is meant by the term standard enthalpy of formation?

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