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The convention of arbitrarily assigning a zero enthalpy value for the most stable form of each element in the standard state at \(25^{\circ} \mathrm{C}\) is a convenient way of dealing with enthalpies of reactions. Explain why this convention cannot be applied to nuclear reactions.

Short Answer

Expert verified
Nuclear reactions involve much larger energy changes due to nuclear binding energies, making the zero enthalpy convention unsuitable.

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01

Definition of Standard Enthalpy

In chemistry, the standard enthalpy of formation is defined as the change in enthalpy when one mole of a substance is formed from its elements in their most stable forms under standard conditions (usually 25°C and 1 atm). The convention is to set the enthalpy of formation of these elements to zero.
02

Nature of Chemical versus Nuclear Reactions

Chemical reactions involve the rearrangement of electrons in atoms and molecules while the nuclei of atoms remain unchanged. In contrast, nuclear reactions involve changes in the energy and composition of atomic nuclei, leading to the release of binding energy.
03

Energy Scales: Chemical vs. Nuclear

The energy scale of nuclear reactions is vastly different from chemical reactions. Nuclear binding energies are on the order of millions of electron volts (MeV) per nucleus, whereas chemical bond energies are a few electron volts (eV) per molecule.
04

Application of Zero Enthalpy Convention

In chemical reactions, the zero enthalpy convention is useful because the relative energy changes due to forming or breaking chemical bonds are what determine the overall energy changes. However, the massive energy changes due to nuclear binding energies do not align with this convention.
05

Conclusion: Limitation with Nuclear Reactions

The zero enthalpy convention is inappropriate for nuclear reactions because it ignores nuclear binding energies, which are significantly larger and of different nature than chemical bond energies. Therefore, assigning zero enthalpy to nuclear reactions does not give a true representation of energy changes.

Key Concepts

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

Understanding Chemical Reactions
Chemical reactions are all about the rearrangement of atoms. They happen when atoms and molecules interact to form new products. In these processes, the nuclei of atoms remain unchanged. Instead, it's the electrons that get shuffled around, which means bonds between atoms are broken, formed, or reshuffled. This is why chemical reactions often involve changes in energy.
  • Breaking Bonds: It requires energy to break bonds, often absorbed from heat.
  • Forming Bonds: Making new bonds releases energy, which can lead to heat or light.
  • Balance of Energy: If more energy is released than absorbed, the reaction can be exothermic (releasing heat).
Reactions are typically expressed with balanced chemical equations showing reactants and products involved. The concept of enthalpy in these reactions is crucial as it helps us understand how energy is transferred during the reaction. Standard enthalpy of formation is a term here, used to denote this energy change when compounds form from their elements under standard conditions.
The Intricacies of Nuclear Reactions
Unlike chemical reactions, nuclear reactions occur within the nucleus of an atom. These reactions involve the alteration in the composition of an atomic nucleus, causing significant energy changes. During nuclear reactions, the components or the nucleons (protons and neutrons) within a nucleus rearrange or are released to form new atoms.
  • Nuclear Stability: Changes or transformations can lead to either stable or new elements.
  • Types of Nuclear Reactions: Includes fission (splitting of heavy nucleus) and fusion (combination of lighter nuclei).
  • Energy Release: These reactions often release substantial amounts of energy, measured in million electron volts (MeV).
The nuclear binding energy, which is the measure of the nucleus being held together, is crucial here. Because of its high energy scale, the zero enthalpy convention used for chemical reactions does not apply to nuclear reactions, as it would not accurately reflect the energy shifts involved.
Differing Energy Scales: Why It Matters
Energy scales dramatically differ between chemical and nuclear reactions. While chemical reactions deal with energy changes on the order of a few electron volts (eV), nuclear reactions take energy changes to a whole new level, often involving millions of electron volts (MeV).
  • Comparison: Chemical reactions involve small-scale energy transitions compared to nuclear's major shifts.
  • Convention Limitation: The zero enthalpy convention is effective for chemical reactions under standard conditions, as it tracks relative energy changes efficiently.
  • Implications for Nuclear: Due to its sheer magnitude, applying a zero baseline for nuclear reactions would ignore the massive binding energy changes, providing an inaccurate energy landscape.
Understanding these differences is critical for anyone studying reaction dynamics. It not only underlies why enthalpy conventions differ between reaction types but also highlights the vast potentials nuclear reactions hold in energy production compared to chemical reactions.

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Most popular questions from this chapter

Determine the enthalpy change for the gaseous reaction of sulfur dioxide with ozone to form sulfur trioxide given the following thermochemical data: $$ \begin{aligned} 2 \mathrm{SO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{2}(g) & \Delta H^{\circ}=-602.8 \mathrm{~kJ} / \mathrm{mol} \\ 3 \mathrm{SO}(g)+2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{SO}_{3}(g) & \\\ \Delta H_{\mathrm{rxn}}^{\circ}=-1485.03 \mathrm{~kJ} / \mathrm{mol} \\ \frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{O}_{3}(g) & \Delta H_{\mathrm{rxn}}^{\circ}=142.2 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$

Define these terms: enthalpy and enthalpy of reaction. Under what condition is the heat of a reaction equal to the enthalpy change of the same reaction?

Give an example for each of the following situations: (a) adding heat to a system raises its temperature, (b) adding heat to a system does not change its temperature, and (c) a system's temperature changes despite no heat being added to it or removed from it.

From the following heats of combustion, \(\begin{aligned} \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H_{\mathrm{rxn}}^{\circ}=-726.4 \mathrm{~kJ} / \mathrm{mol} \\\ \mathrm{C}(\text { graphite })+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) & \\ \Delta H_{\mathrm{rxn}}^{\circ}=-393.5 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) & \\ \Delta H_{\mathrm{rxn}}^{\circ}=-285.8 \mathrm{~kJ} / \mathrm{mol} \end{aligned}\) calculate the enthalpy of formation of methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) from its elements: $$ \mathrm{C}(\text { graphite })+2 \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(l) $$

State Hess's law. Explain, with one example, the usefulness of Hess's law in thermochemistry.

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