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In addition to the process described in the text, a second process called the carbon-nitrogen cycle occurs in the sun: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 \(i \mathrm{H}\) and 4 He are 1.00782 u and 4.00260 u, respectively.)

Short Answer

Expert verified
The catalyst in the carbon-nitrogen cycle is carbon-12 (C₃). The intermediate nucleons in the process are nitrogen-13 (N) and oxygen-16 (O). The energy released per mole of hydrogen nuclei in the overall reaction is approximately \(2.57 \times 10^{12}\) J/mol.

Step by step solution

01

Understand the Carbon-Nitrogen Cycle

The carbon-nitrogen cycle is an alternative process to the proton-proton chain that powers the Sun. This cycle involves nitrogen, carbon, and oxygen working as catalysts to convert hydrogen into helium, with a release of energy in the form of gamma rays and neutrinos. The carbon-nitrogen cycle is more prominent in stars that are larger and hotter than the Sun.
02

Identify the Catalyst in the Carbon-Nitrogen Cycle

The catalyst in the carbon-nitrogen cycle is carbon-12 (C₃). Carbon works with nitrogen and oxygen, effectively driving the fusion of hydrogen atoms into helium atoms. The carbon-12 nucleus returns to its initial state upon completing the cycle, enabling it to be reused indefinitely. Thus, the answer to part a is carbon-12 (C₃).
03

Find the Intermediate Nucleons

The intermediate nucleons in the carbon-nitrogen cycle are nitrogen and oxygen nuclei. In this process, carbon undergoes several reactions, transforming carbon-12 into nitrogen-13 and then oxygen-16, before returning to carbon-12 again. The nitrogen-13 and oxygen-16 nuclei serve as intermediates throughout the cycle. Therefore, the answer to part b is nitrogen-13 (N) and oxygen-16 (O).
04

Calculate the Energy Released per Mole of Hydrogen Nuclei

To calculate the energy released in the overall reaction, we can use the mass-energy equivalence equation: \[E = (\Delta m) c^2\] Where \(E\) is the energy released, \(\Delta m\) is the change in mass, and \(c\) is the speed of light. Before we do the calculation, we need to find the change in mass. For this, we can use the given atomic masses: 1 H: 1.00782 u 4 He: 4.00260 u The overall reaction involves the fusion of 4 hydrogen nuclei into 1 helium nucleus: 4(1 H) → 4 He The mass of 4 hydrogen nuclei is: Mass of 4 H = 4 × 1.00782 u = 4.03128 u The change in mass during the reaction is: \(\Delta m =\) Mass of 4 H - Mass of 4 He = 4.03128 u - 4.00260 u = 0.02868 u To convert the mass difference from atomic mass units (u) to kilograms (kg), we'll use the following conversion: 1 u = 1.66054 × 10⁻²⁷ kg Now, we'll finally calculate the energy released per mole of hydrogen nuclei: \(\Delta m_{kg} =\) 0.02868 u × 1.66054 × 10⁻²⁷ kg/u = 4.76296 × 10⁻²⁹ kg \(E = (\Delta m_{kg}) c^2\) \(E = (4.76296 × 10⁻²⁹ kg) (2.99792 × 10⁸ m/s)^2\) \(E = 4.27501 × 10⁻¹² J\) Now to convert the energy per hydrogen nuclei to energy per mole of hydrogen nuclei, we multiply by Avogadro's number: Energy released per mole of hydrogen nuclei = \(4.27501 × 10⁻¹² J × 6.022 × 10²³ mol⁻¹ ≈ 2.57 \times 10^{12}\) J/mol Thus, 2.57 × 10¹² J (joules) of energy is released per mole of hydrogen nuclei in the overall reaction.

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Key Concepts

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

Fusion Reactions
Fusion reactions are a type of nuclear reaction where two light atomic nuclei combine to form a heavier nucleus. This process is fundamental in stars, including our Sun, where hydrogen nuclei fuse to create helium, releasing immense energy.
The carbon-nitrogen cycle is one fusion reaction that mainly operates in stars larger and hotter than the Sun. It involves a series of reactions where carbon, nitrogen, and oxygen act as catalysts. In this cycle, hydrogen nuclei (protons) are gradually converted into helium through a sequence of fusion reactions. Each step in the cycle contributes to the build-up of helium and releases energy.
Fusion is key to understanding stellar life cycles. The energy released helps to counter gravity, stabilizing the star. These reactions highlight nature's ability to convert tiny mass changes into significant energy outputs, which can be further understood through Einstein's equation, \(E=mc^2\).
Catalysts in Nuclear Processes
In nuclear processes, catalysts are elements that facilitate reactions without being consumed in the process. A prime example of this is the carbon-nitrogen cycle occurring in stars.
Carbon-12 acts as a catalyst by aiding in the conversion of hydrogen to helium. It first reacts with protons to form nitrogen and then oxygen, but crucially, it reforms back into carbon-12 by the end of the cycle. This ability to return to its original form allows carbon to participate continuously in the reactions without being depleted.
This cyclic nature is reminiscent of catalysts in chemical reactions on Earth, where a catalyst remains unchanged after the reaction. Such cycles are efficient, allowing the star to sustain fusion over its lifetime.
Energy Calculation in Chemistry
Energy calculations are crucial in understanding chemical and nuclear reactions. These calculations often involve determining the energy released during reactions, using mass changes and Einstein’s mass-energy equivalence principle.
In the context of the carbon-nitrogen cycle, the fusion of hydrogen nuclei into helium is a perfect example. Initially, the masses of four hydrogen nuclei and one helium nucleus need to be considered. The mass difference between the reactants and the products is calculated, which reveals the mass defect.
This mass defect is then translated into energy using the equation \(E = \Delta m \, c^2\). Here, \(\Delta m\) is the change in mass and \(c\) is the speed of light. The resulting energy is initially calculated per reaction, but for practical chemistry and astronomy, it is typically converted to energy per mole using Avogadro’s number.
These calculations allow scientists to comprehend the immense energy produced in nuclear processes, highlighting the beauty of the tiny mass difference leading to massive energy outputs.

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Most popular questions from this chapter

Zirconium is one of the few metals that retains its structural integrity upon exposure to radiation. The fuel rods in most nuclear reactors therefore are often made of zirconium. Answer the following questions about the redox properties of zirconium based on the half-reaction $$\mathrm{ZrO}_{2} \cdot \mathrm{H}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}+4 \mathrm{e}^{-} \longrightarrow \mathrm{Zr}+4 \mathrm{OH}^{-} \quad \mathscr{C}^{\circ}=-2.36 \mathrm{V}$$a. Is zirconium metal capable of reducing water to form hydrogen gas at standard conditions? b. Write a balanced equation for the reduction of water by zirconium. c. Calculate \(\mathscr{E}^{\circ}, \Delta G^{\circ},\) and \(K\) for the reduction of water by zirconium metal. d. The reduction of water by zirconium occurred during the accidents at Three Mile Island in \(1979 .\) The hydrogen produced was successfully vented and no chemical explosion occurred. If \(1.00 \times 10^{3} \mathrm{kg}\) Zr reacts, what mass of \(\mathrm{H}_{2}\) is produced? What volume of \(\mathrm{H}_{2}\) at 1.0 atm and \(1000 .^{\circ} \mathrm{C}\) is produced? e. At Chernobyl in \(1986,\) hydrogen was produced by the reaction of superheated steam with the graphite reactor core:$$\mathbf{C}(s)+\mathbf{H}_{2} \mathbf{O}(g) \longrightarrow \mathbf{C O}(g)+\mathbf{H}_{2}(g)$$ It was not possible to prevent a chemical explosion at Chernobyl. In light of this, do you think it was a correct decision to vent the hydrogen and other radioactive gases into the atmosphere at Three Mile Island? Explain.

In each of the following radioactive decay processes, supply the missing particle. a. \(^{60} \mathrm{Co} \rightarrow^{60} \mathrm{Ni}+?\) b. \(^{97} \mathrm{Tc}+? \rightarrow^{97} \mathrm{Mo}\) c. \(^{99} \mathrm{Tc} \rightarrow^{99} \mathrm{Ru}+?\) d. \(^{239} \mathrm{Pu} \rightarrow^{235} \mathrm{U}+?\)

Radioactive copper- 64 decays with a half-life of 12.8 days. a. What is the value of \(k\) in \(s^{-1} ?\) b. A sample contains \(28.0 \space\mathrm{mg}\space^{64} \mathrm{Cu}\). How many decay events will be produced in the first second? Assume the atomic mass of \(^{64} \mathrm{Cu}\) is \(64.0 \mathrm{u}\). c. A chemist obtains a fresh sample of \(^{64} \mathrm{Cu}\) and measures its radioactivity. She then determines that to do an experiment, the radioactivity cannot fall below \(25 \%\) of the initial measured value. How long does she have to do the experiment?

Radioactive cobalt-60 is used to study defects in vitamin \(\mathbf{B}_{12}\) absorption because cobalt is the metallic atom at the center of the vitamin \(\mathrm{B}_{12}\) molecule. The nuclear synthesis of this cobalt isotope involves a three-step process. The overall reaction is iron-58 reacting with two neutrons to produce cobalt-60 along with the emission of another particle. What particle is emitted in this nuclear synthesis? What is the binding energy in J per nucleon for the cobalt-60 nucleus (atomic masses: \(^{60} \mathrm{Co}=\) \(\left.59.9338 \mathrm{u} ;^{1} \mathrm{H}=1.00782 \mathrm{u}\right) ?\) What is the de Broglie wave-length of the emitted particle if it has a velocity equal to \(0.90 c\) where \(c\) is the speed of light?

A rock contains \(0.688 \mathrm{mg}^{206} \mathrm{Pb}\) for every \(1.000 \mathrm{mg}\) \(^{238} \mathrm{U}\) present Assuming that no lead was originally present, that all the \(^{206} \mathrm{Pb}\) formed over the years has remained in the rock, and that the number of nuclides in intermediate stages of decay between \(^{238} \mathrm{U}\) and \(^{206} \mathrm{Pb}\) is negligible, calculate the age of the \(\text { rock. For }^{238} \mathbf{U}, t_{1 / 2}=4.5 \times 10^{9} \text { years. }\)

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