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(a) Why does the standard enthalpy of formation of both the very reactive fluorine \(\left(\mathrm{F}_{2}\right)\) and the almost inert gas nitrogen \(\left(\mathrm{N}_{2}\right)\) both read zero? (b) Write the chemical equation for the reaction whose enthalpy change is the standard enthalpy of formation of naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\).

Short Answer

Expert verified
(a) The standard enthalpy of formation for both fluorine (F₂) and nitrogen (N₂) gas is zero because they are already in their most stable forms under standard conditions (pressure of 1 atmosphere and temperature of 298 K), and no heat is absorbed or released in their formation. (b) The chemical equation for the reaction representing the standard enthalpy of formation of naphthalene (C₁₀H₈) is: \[ 10 \mathrm{C_{(graphite)}} + 4 \mathrm{H_{2(g)}} \rightarrow \mathrm{C_{10}H_{8(s)}}\]

Step by step solution

01

Define Standard Enthalpy of Formation

Standard enthalpy of formation refers to the heat absorbed or released when one mole of a compound is formed from its constituent elements in their most stable forms, under standard conditions. Standard conditions are defined as a pressure of 1 atmosphere and a temperature of 298 K.
02

Apply the Definition to F₂ and N₂

In the given problem, we are considering the standard enthalpy of formation for fluorine (F₂) and nitrogen (N₂) gas. Both of these are diatomic molecules and represent the most stable forms of their respective elements under standard conditions. Since they are already in their most stable forms, there is no actual formation reaction occurring for these species. This means, there is no heat absorbed or released in their formation (as they are already formed). Therefore, the standard enthalpy of formation for both F₂ and N₂ is zero.
03

Write the Chemical Equation for the Formation of Naphthalene

The chemical formula of naphthalene is C₁₀H₈. To write the equation for its formation from its constituent elements, we need to consider the most stable forms of carbon and hydrogen under standard conditions. Carbon is most stable as solid graphite and hydrogen is most stable as diatomic gas (H₂). The formation reaction of naphthalene is: \[ 10 \mathrm{C_{(graphite)}} + 4 \mathrm{H_{2(g)}} \rightarrow \mathrm{C_{10}H_{8(s)}}\] This is the chemical equation for the reaction whose enthalpy change is the standard enthalpy of formation of naphthalene.

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Key Concepts

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

Diatomic Molecules
Diatomic molecules are those composed of only two atoms. These molecules can either consist of two atoms of the same element or different elements. A common example of diatomic molecules involving the same elements are gases like oxygen ( 2), hydrogen ( 2), nitrogen ( 2), and fluorine ( 2). These diatomic molecules are prevalent in nature and play essential roles in various chemical reactions.

The intriguing point with diatomic molecules like 2 and 2 is their stability. Under standard conditions (1 atm pressure and 298 K temperature), these gases exist naturally in their most stable form. This is significant because when an element is already in its most stable form as a diatomic molecule, there's no need for an additional reaction to form it from other forms. This intrinsic stability is why their standard enthalpy of formation is zero, indicating no energy change is required to form them from their elemental states.
Chemical Equations
Chemical equations are symbolic representations of chemical reactions. They show the reactants and their amounts on the left side of the equation and the products on the right. Balancing chemical equations ensures that the same number of each type of atom appears on both sides, illustrating the conservation of mass principle.

In the context of naphthalene (2) formation, we use chemical equations to describe its creation from stable, elemental forms of carbon and hydrogen. In a balanced chemical equation:
  • Reactants: Represent the starting elements, such as carbon in its most stable form (") and hydrogen as diatomic gas (").
  • Products: Include the desired compound, here naphthalene (").
The balanced equation for the formation of naphthalene is:\[ 10 \ \mathrm{C_{(graphite)}} + 4 \ \mathrm{H_{2(g)}} \rightarrow \mathrm{C_{10}H_{8(s)}} \]This equation represents the enthalpy change, or the energy required or released when forming naphthalene from its elements under standard conditions.
Stable Forms of Elements
Elements have various forms they can exist in, but each element has a form that is considered most stable under standard conditions (1 atm, 298 K). These stable forms are critical when determining standard enthalpies of formation, as they set the baseline for calculating energy changes.

For many elements, the stable form is the diatomic molecule, like nitrogen ( 2) and hydrogen ( 2). Other elements have different stable forms; carbon, for example, is most stable as graphite.
  • The importance of knowing an element's stable form lies in computations like enthalpy of formation. The standard enthalpy of formation is defined for compounds created from these stable forms.
  • Understanding the stable form is crucial as reactions typically start from these forms and transform into new molecular compounds.
Thus, when calculating energy changes, you always start from these stable reference points, leading to the understanding of new compound formations and their corresponding energy dynamics.

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Most popular questions from this chapter

Consider the combustion of liquid methanol, \(\mathrm{CH}_{3} \mathrm{OH}(l):\) $$ \begin{aligned} \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) & \\ \Delta H=&-726.5 \mathrm{~kJ} \end{aligned} $$ (a) What is the enthalpy change for the reverse reaction? (b) Balance the forward reaction with whole-number coefficients. What is \(\Delta H\) for the reaction represented by this equation? (c) Which is more likely to be thermodynamically favored, the forward reaction or the reverse reaction? (d) If the reaction were written to produce \(\mathrm{H}_{2} \mathrm{O}(g)\) instead of \(\mathrm{H}_{2} \mathrm{O}(l),\) would you expect the magnitude of \(\Delta H\) to increase, decrease, or stay the same? Explain.

It is interesting to compare the "fuel value" of a hydrocarbon in a hypothetical world where oxygen is not the combustion agent. The enthalpy of formation of \(\mathrm{CF}_{4}(g)\) is \(-679.9 \mathrm{~kJ} / \mathrm{mol}\). Which of the following two reactions is the more exothermic? $$ \begin{aligned} \mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) & \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \\ \mathrm{CH}_{4}(g)+4 \mathrm{~F}_{2}(g) & \longrightarrow \mathrm{CF}_{4}(g)+4 \mathrm{HF}(g) \end{aligned} $$

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 \(\mathrm{CH}_{3} \mathrm{OH}(g)\) at constant pressure?

A sample of a hydrocarbon is combusted completely in \(\mathrm{O}_{2}(g)\) to produce \(21.83 \mathrm{~g} \mathrm{CO}_{2}(g), 4.47 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}(g),\) and \(311 \mathrm{~kJ}\) of heat. (a) What is the mass of the hydrocarbon sample that was combusted? (b) What is the empirical formula of the hydrocarbon? (c) Calculate the value of \(\Delta H_{f}^{\circ}\) per empiricalformula unit of the hydrocarbon. (d) Do you think that the hydrocarbon is one of those listed in Appendix C? Explain your answer.

(a) What is the electrostatic potential energy (in joules) between two electrons that are separated by \(460 \mathrm{pm} ?\) (b) What is the change in potential energy if the distance separating the two electrons is increased to \(1.0 \mathrm{nm}\) ? (c) Does the potential energy of the two particles increase or decrease when the distance is increased to \(1.0 \mathrm{nm}\) ?

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