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Chapter 5: Problem 70

The standard enthalpy change for the following reaction is \(436.4 \mathrm{~kJ} / \mathrm{mol}\) : $$\mathrm{H}_{2}(g) \longrightarrow \mathrm{H}(g)+\mathrm{H}(g)$$ Calculate the standard enthalpy of formation of atomic hydrogen (H).

Short Answer

Expert verified
The standard enthalpy of formation of atomic hydrogen is \(218.2 \, \mathrm{kJ/mol}\).

Step by step solution

01

Understand the Problem

The problem requires us to find the standard enthalpy of formation of atomic hydrogen from the given standard enthalpy change of the reaction: \(\mathrm{H}_{2}(g) \longrightarrow \mathrm{H}(g)+\mathrm{H}(g)\). This means we need to determine how much energy is required to form atomic hydrogen from diatomic hydrogen gas.
02

Interpret the Given Reaction

The reaction \(\mathrm{H}_{2}(g) \longrightarrow \mathrm{H}(g)+\mathrm{H}(g)\) represents the dissociation of one mole of diatomic hydrogen into two moles of atomic hydrogen. The provided standard enthalpy change of the reaction is \(436.4\, \mathrm{kJ/mol}\). This is the energy required to break the bond in \(1\, \text{mole}\) of \(\mathrm{H}_2\) to form \(2\, \text{moles}\) of atomic \(\mathrm{H}\).
03

Define Enthalpy of Formation

The standard enthalpy of formation for a substance is the energy change when one mole of a substance is formed from its elements in their standard states. We want to find the enthalpy change to form 1 mole of atomic hydrogen \(\mathrm{H}(g)\) from its standard state, which is \(\mathrm{H}_{2}(g)\).
04

Calculate the Enthalpy of Formation for Atomic Hydrogen

Since the enthalpy change given is for \(1\, \text{mole}\) of \(\mathrm{H}_2\) forming \(2\, \text{moles}\) of \(\mathrm{H}(g)\), the enthalpy of formation for one mole of atomic \(\mathrm{H}(g)\), is half of this value. Therefore, the enthalpy of formation of \(\mathrm{H}(g)\) is:\[\Delta H_f^{\circ}(\mathrm{H}) = \frac{436.4\, \mathrm{kJ/mol}}{2} = 218.2\, \mathrm{kJ/mol}.\]
05

Verify Calculation

Review the calculations to ensure they align with the definition and setup provided. One mole of \(\mathrm{H}_2\) creates two moles of \(\mathrm{H}(g)\), so dividing the enthalpy change equally is consistent with the problem statement.

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Key Concepts

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

Standard Enthalpy Change
The standard enthalpy change refers to the heat exchange that accompanies a chemical reaction conducted under standard conditions. These conditions typically include a temperature of 298 K and a pressure of 1 atm. In the context of the given exercise, the standard enthalpy change of 436.4 kJ/mol corresponds to the dissociation of molecular hydrogen (\(\mathrm{H}_2\)) into atomic hydrogen (\(\mathrm{H}\)).
  • Standard conditions ensure that results are comparable and consistent across different experiments.
  • The symbol for standard enthalpy change is often \(\Delta H^\circ\), where the degree symbol represents the standard states of the reactants and products.
Understanding this concept helps clarify why we use the given values and how they relate to real-world chemical processes.
Dissociation Energy
Dissociation energy is the amount of energy required to break a chemical bond in a molecule and separate its atoms. Specifically, in our problem, it refers to the energy needed to break the bond between two hydrogen atoms in a diatomic hydrogen molecule (\(\mathrm{H}_2\)) to form two individual hydrogen atoms (\(\mathrm{H}(g)\)).
  • This energy is vital in reactions because it dictates how much energy is needed to initiate a reaction.
  • The dissociation energy of a bond influences the stability of the molecule; stronger bonds require more energy to break.
Therefore, the value of 436.4 kJ/mol signifies the energy required for this specific bond dissociation.
Atomic Hydrogen
Atomic hydrogen is a single hydrogen atom in the gaseous state, denoted as (\(\mathrm{H}(g)\)). It is highly reactive and tends to form bonds with other atoms, converting back to its more stable diatomic form (\(\mathrm{H}_2\)).
  • Atomic hydrogen is essential in various industrial processes, including welding and the Haber process for ammonia production.
  • Its reactivity arises from its unpaired electron, which makes it eager to bond with other atoms to achieve stability.
In the enthalpy of formation exercise, understanding atomic hydrogen's nature helps explain why energy is involved in breaking down molecular hydrogen.
Molecular Hydrogen
Molecular hydrogen, represented as \(\mathrm{H}_2\), is the simplest known molecule and consists of two hydrogen atoms bonded together. It is the most abundant form of hydrogen due to its stability compared to atomic hydrogen.
  • It naturally occurs in the atmosphere and is a critical component in numerous chemical reactions.
  • The bond between the two hydrogen atoms in \(\mathrm{H}_2\) is strong, which explains why it requires significant energy to dissociate into atomic hydrogen.
Understanding this molecule's properties is crucial in determining the energy requirements in chemical processes like the one depicted in the exercise.

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Most popular questions from this chapter

An excess of zinc metal is added to \(50.0 \mathrm{~mL}\) of a \(0.100 \mathrm{M} \mathrm{AgNO}_{3}\) solution in a constant-pressure calorimeter like the one pictured in Figure 5.8 . As a result of the reaction \(\mathrm{Zn}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+2 \mathrm{Ag}(s)\) the temperature rises from \(19.25^{\circ} \mathrm{C}\) to \(22.17^{\circ} \mathrm{C}\). If the heat capacity of the calorimeter is \(98.6 \mathrm{~J} /{ }^{\circ} \mathrm{C},\) calculate the enthalpy change for the given reaction on a molar basis. Assume that the density and specific heat of the solution are the same as those for water, and ignore the specific heats of the metals.

At \(25^{\circ} \mathrm{C}\), the standard enthalpy of formation of \(\mathrm{HF}(a q)\) is \(-320.1 \mathrm{~kJ} / \mathrm{mol} ;\) of \(\mathrm{OH}^{-}(a q),\) it is \(-229.6 \mathrm{~kJ} / \mathrm{mol} ;\) of \(\mathrm{F}^{-}(a q)\) it is \(-329.1 \mathrm{~kJ} / \mathrm{mol} ;\) and of \(\mathrm{H}_{2} \mathrm{O}(l),\) it is \(-285.8 \mathrm{~kJ} / \mathrm{mol}\). (a) Calculate the standard enthalpy of neutralization of \(\mathrm{HF}(a q)\) \(\mathrm{HF}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{F}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (b) Using the value of \(-56.2 \mathrm{~kJ}\) as the standard enthalpy change for the reaction \(\mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) calculate the standard enthalpy change for the reaction \(\mathrm{HF}(a q) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q)\)

The enthalpy of combustion of benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) is commonly used as the standard for calibrating constant-volume bomb calorimeters; its value has been accurately determined to be \(-3226.7 \mathrm{~kJ} / \mathrm{mol}\). When \(1.9862 \mathrm{~g}\) of benzoic acid are burned in a calorimeter, the temperature rises from \(21.84^{\circ} \mathrm{C}\) to \(25.67^{\circ} \mathrm{C}\). What is the heat capacity of the bomb? (Assume that the quantity of water surrounding the bomb is exactly \(2000 \mathrm{~g} .\) )

A quantity of \(2.00 \times 10^{2} \mathrm{~mL}\) of \(0.862 \mathrm{M} \mathrm{HCl}\) is mixed with \(2.00 \times 10^{2} \mathrm{~mL}\) of \(0.431 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of the \(\mathrm{HCl}\) and \(\mathrm{Ba}(\mathrm{OH})_{2}\) solutions is the same at \(20.48^{\circ} \mathrm{C}\). For the process $$\mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)$$ the heat of neutralization is \(-56.2 \mathrm{~kJ} / \mathrm{mol}\). What is the final temperature of the mixed solution? Assume the specific heat of the solution is the same as that for pure water.

Given the thermochemical data, \(\mathrm{A}+\mathrm{B} \longrightarrow 2 \mathrm{C} \quad \Delta H_{1}=600 \mathrm{~kJ} / \mathrm{mol}\) \(\begin{array}{ll}2 \mathrm{C}+\mathrm{D} \longrightarrow 2 \mathrm{E} & \Delta H_{1}=210 \mathrm{~kJ} / \mathrm{mol}\end{array}\) Determine the enthalpy change for each of the following: a) \(4 \mathrm{E} \longrightarrow 4 \mathrm{C}+2 \mathrm{D}\) d) \(2 C+2 E \longrightarrow 2 A+2 B+D\) b) \(\mathrm{A}+\mathrm{B}+\mathrm{D} \longrightarrow 2 \mathrm{E}\) e) \(\mathrm{E} \longrightarrow \frac{1}{2} \mathrm{~A}+\frac{1}{2} \mathrm{~B}+\frac{1}{2} \mathrm{D}\) c) \(\mathrm{C} \longrightarrow \frac{1}{2} \mathrm{~A}+\frac{1}{2} \mathrm{~B}\)
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