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Chapter 19: Problem 83

In addition to the process described in the text, a second process called the carbon-nitrogen cycle occurs in the sun: $$ \begin{aligned} { }_{1}^{1} \mathrm{H}+{ }_{6}^{12} \mathrm{C} \longrightarrow{ }_{7}^{13} \mathrm{~N}+{ }_{0}^{0} \gamma \\ { }_{7}^{13} \mathrm{~N} & \longrightarrow{ }_{6}^{13} \mathrm{C}+{ }_{+1}^{0} \mathrm{e} \\ { }_{1}^{1} \mathrm{H}+{ }_{6}^{13} \mathrm{C} &{ }_{7}^{14} \mathrm{~N}+{ }_{0}^{0} \gamma \\ { }_{1}^{1} \mathrm{H}+{ }_{7}^{14} \mathrm{~N} \longrightarrow &{ }_{8}^{15} \mathrm{O}+{ }_{0}^{0} \gamma \\ { }_{8}^{15} \mathrm{O} \longrightarrow{ }_{7}^{15} \mathrm{~N}+{ }_{+1}^{0} \mathrm{e} \\ { }_{1}^{1} \mathrm{H}+{ }_{7}^{15} \mathrm{~N} \longrightarrow{ }_{6}^{12} \mathrm{C}+{ }_{2}^{4} \mathrm{He}+{ }_{0}^{0} \gamma \\ \hline \end{aligned} $$ reaction: \(\quad 4{ }_{1}^{1} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He}+2{ }_{+1}^{0} \mathrm{e}\) a. What is the catalyst in this process? b. What nucleons are intermediates? c. How much energy is released per mole of hydrogen nuclei in the overall reaction? (The atomic masses of \({ }_{1} \mathrm{H}\) and \({ }_{2}^{4} \mathrm{He}\) are \(1.00782 \mathrm{u}\) and \(4.00260 \mathrm{u}\), respectively. \()\)

Short Answer

Expert verified
The catalyst in the carbon-nitrogen cycle is \({}_{6}^{12}\mathrm{C}\). The intermediate nucleons are \({}_{7}^{13}\mathrm{N}\), \({}_{6}^{13}\mathrm{C}\), \({}_{7}^{14}\mathrm{N}\), \({}_{8}^{15}\mathrm{O}\), and \({}_{7}^{15}\mathrm{N}\). The energy released per mole of hydrogen nuclei in the overall reaction is approximately \(2.57\times10^{12}\,\text{J/mole}\).

Step by step solution

01

a. Identifying the catalyst

To find the catalyst in this process, we need to look for a species that appears in the beginning of a reaction and then is regenerated at the end of the process. From the given reactions, we can see that the catalyst is \({}_{6}^{12}\mathrm{C}\). It reacts with \({}_{1}^{1}\mathrm{H}\) in the first reaction and is regenerated in the last reaction.
02

b. Identifying the intermediate nucleons

Intermediate nucleons are the species that are produced and consumed within the process but are not present at the beginning or end of the overall reaction. In this case, the intermediate nucleons are: - \({}_{7}^{13}\mathrm{N}\) - \({}_{6}^{13}\mathrm{C}\) - \({}_{7}^{14}\mathrm{N}\) - \({}_{8}^{15}\mathrm{O}\) - \({}_{7}^{15}\mathrm{N}\) These nucleons are formed and consumed in the carbon-nitrogen cycle but do not appear in the overall reaction.
03

c. Calculating energy released per mole of hydrogen nuclei in the overall reaction

We are given the atomic masses of \({}_{1}\mathrm{H}\) and \({}_{2}^{4}\mathrm{He}\) as \(1.00782 \mathrm{u}\) and \(4.00260 \mathrm{u}\), respectively. The overall reaction is: \(4{ }_{1}^{1} \mathrm{H} \longrightarrow{ }_{2}^{4}\mathrm{He}+2{ }_{+1}^{0}\mathrm{e}\). To find the energy released per mole of hydrogen nuclei, we will first calculate the mass difference between the reactants and products and then convert that mass difference to energy. Step 1: Calculate the mass difference (in atomic mass units) Mass of reactants = \(4 \times 1.00782 \mathrm{u} = 4.03128 \mathrm{u}\) Mass of products = \(4.00260 \mathrm{u}\) (ignore the mass of electrons as it's negligible) Mass difference = \(4.03128 \mathrm{u} - 4.00260 \mathrm{u} = 0.02868 \mathrm{u}\) Step 2: Convert mass difference to energy Using the energy-mass equivalence formula, \(E=mc^2\), where \(E\) is energy, \(m\) is mass, and \(c\) is the speed of light (\(3.00\times10^{8}\,\text{m/s}\)). First, we need to convert the mass difference from atomic mass units to kg. \(1\,\mathrm{u} = 1.66054 \times 10^{-27}\,\text{kg}\). Mass difference in kg = \(0.02868\,\mathrm{u} \times 1.66054\times10^{-27}\,\text{kg/u} = 4.764\times10^{-29}\,\text{kg}\) Now, calculate the energy released per nucleus using \(E=mc^2\). \(E = (4.764\times10^{-29}\,\text{kg})(3.00\times10^{8}\,\text{m/s})^2 = 4.27\times10^{-12}\,\text{J}\) Step 3: Calculate the energy released per mole of hydrogen nuclei Given that \(1\,\text{mole} = 6.022\times10^{23}\) entities, the energy released per mole of hydrogen nuclei can be calculated as follows: Energy released per mole of hydrogen nuclei = \((4.27\times10^{-12}\,\text{J})\times(6.022\times10^{23}\,\text{mole}^{-1}) = 2.57\times10^{12}\,\text{J/mole}\) So, the energy released per mole of hydrogen nuclei in the overall reaction is approximately \(2.57\times10^{12}\,\text{J/mole}\).

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Key Concepts

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

Stellar Nucleosynthesis
Stellar nucleosynthesis is the process by which elements are created within the cores of stars through nuclear reactions. This cosmic forge operates under extreme temperatures and pressures, allowing atomic nuclei to collide and fuse, forming heavier elements.

For instance, the sun and other stars of similar size primarily fuse hydrogen atoms to form helium in the core through the proton-proton chain reaction. However, in more massive stars, a collection of processes including the carbon-nitrogen cycle, fusion of heavier elements, and supernova nucleosynthesis take place, leading to the creation of all naturally occurring elements in the universe. The elements generated by these processes are eventually dispersed into space during supernova explosions or planetary nebulae, contributing to the cosmic matter that will form new stars, planets, and possibly life.
Nuclear Reactions in the Sun
The nuclear furnace at the heart of our solar system, the sun, fuels its radiance through a sequence of nuclear reactions. Specifically, the carbon-nitrogen cycle, an alternative to the proton-proton chain, illustrates these complex nuclear interactions. It starts with a carbon-12 nucleus capturing a proton and proceeds through successive transformations involving nitrogen and oxygen isotopes.

As these reactions cycle, intermediate nuclei are created and then transformed further or decay, generating energy. The carbon-nitrogen cycle is heavily reliant on carbon-12 acting as a catalyst, since it emerges unaltered at the cycle's completion. This process is predominant in larger stars with hotter cores where higher temperatures enable the carbon-nitrogen cycle to proceed efficiently.
Energy Release in Nuclear Reactions
The energy emitted in nuclear reactions is astounding and happens due to the conversion of mass into energy, described by Einstein's famous equation, \(E=mc^2\). This principle is pronounced in the carbon-nitrogen cycle.

The mass of the reactants, usually heavier due to the strong nuclear forces that bind protons and neutrons, is more than the mass of the resultant nuclei post-fusion. Thus, the difference in mass, albeit tiny, when multiplied by the square of the speed of light, results in a substantial amount of energy released. This energy sustains the star's outward pressure to balance gravitational collapse and makes life on Earth possible. In detailed calculations for energy release, the mass of electrons is generally omitted as it is negligible compared to that of nucleons.

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Most popular questions from this chapter

A reported synthesis of the transuranium element bohrium (Bh) involved the bombardment of berkelium-249 with neon-22 to produce bohrium-267. Write a nuclear reaction for this synthesis. The half-life of bohrium-267 is \(15.0\) seconds. If 199 atoms of bohrium-267 could be synthesized, how much time would

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Which do you think would be the greater health hazard: the release of a radioactive nuclide of \(\mathrm{Sr}\) or a radioactive nuclide of Xe into the environment? Assume the amount of radioactivity is the same in each case. Explain your answer on the basis of the chemical properties of Sr and Xe. Why are the chemical properties of a radioactive substance important in assessing its potential health hazards?

Phosphorus-32 is a commonly used radioactive nuclide in biochemical research, particularly in studies of nucleic acids. The half-life of phosphorus-32 is \(14.3\) days. What mass of phosphorus-32 is left from an original sample of \(175 \mathrm{mg}\) \(\mathrm{Na}_{3}{ }^{32} \mathrm{PO}_{4}\) after \(35.0\) days? Assume the atomic mass of \({ }^{32} \mathrm{P}\) is \(32.0 \mathrm{u}\).

Breeder reactors are used to convert the nonfissionable nuclide \({ }_{92}^{238} \mathrm{U}\) to a fissionable product. Neutron capture of the \({ }_{92}^{238} \mathrm{U}\) is followed by two successive beta decays. What is the final fissionable product?
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