Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 49

Calculate the amount of energy released per gram of hydrogen nuclei reacted for the following reaction. The atomic masses are \(^{1}_{1}{H}, 1.00782 \mathrm{u} ; \frac{2}{1} \mathrm{H}, 2.01410 \mathrm{u} ;\) and an electron, \(5.4858 \times\) \(10^{-4}\) u. (Hint: Think carefully about how to account for the electron mass.)$$\mathrm{i} \mathrm{H}+\mathrm{i} \mathrm{H} \longrightarrow_{\mathrm{i}}^{2} \mathrm{H}+_{+\mathrm{i}}^{0}$$

Short Answer

Expert verified
The energy released per gram of hydrogen nuclei reacted for the given reaction can be calculated by first finding the mass difference between the initial reactants and final products, then converting the mass difference into energy using Einstein's equation, and finally converting the energy per reaction to energy per gram of hydrogen nuclei reacted. Following these steps, we find that the energy released per gram of hydrogen nuclei reacted is: Energy released per gram = Energy released per reaction × Number of reactions in 1 gram of hydrogen nuclei

Step by step solution

01

Write down the given reaction and identify the initial reactants and final products

The given reaction can be written as: \(^{1}_{1}\mathrm{H} + ^{1}_{1}\mathrm{H} \longrightarrow ^{2}_{1}\mathrm{H} + e^-\) In this reaction: - Initial reactants are: \(^{1}_{1}\mathrm{H}\) and \(^{1}_{1}\mathrm{H}\) - Final products are: \(_{1}^{2}\mathrm{H}\) and \(e^-\)
02

Find the mass difference between the initial reactants and final products

To calculate the mass difference, we need to subtract the total mass of the final products from the total mass of the initial reactants. Mass of initial reactants = \(2 \times 1.00782 \mathrm{u}\) (since two \(^{1}_{1}\mathrm{H}\) are reacted) Mass of final products = \(1 \times 2.01410 \mathrm{u} + 1 \times 5.4858 \times 10^{-4} \mathrm{u}\) (one \(_{1}^{2}\mathrm{H}\) and one \(e^-\)) Now, let's calculate the mass difference: Mass difference = Mass of initial reactants - Mass of final products Mass difference = \((2 \times 1.00782) - (2.01410 + 5.4858 \times 10^{-4}) \mathrm{u}\)
03

Convert the mass difference into energy

Now we will use Einstein's expression, \(E = mc^2\), to convert the mass difference into energy. Here, \(c\) is the speed of light, which is approximately \(3 \times 10^8 \mathrm{m.s^{-1}}\). First, let's convert the mass difference from unified atomic mass units (u) to kilograms (kg), using the conversion factor \(1.66054 \times 10^{-27}\mathrm{kg.u^{-1}}\): Mass difference = Mass difference in u × \(1.66054 \times 10^{-27}\mathrm{kg.u^{-1}}\) Now we can calculate the energy using \(E = mc^2\): Energy released = Mass difference × \((3 \times 10^8)^2 \mathrm{m^2.s^{-2}}\)
04

Calculate the energy released per gram of hydrogen nuclei reacted

We were asked to find the amount of energy released per gram of reacting hydrogen nuclei. Since we have calculated the energy released per reaction, we can now convert it into energy per gram. First, we need to find the mass of 1 gram of hydrogen nuclei in terms of the number of hydrogen nuclei: 1 gram of hydrogen nuclei = \(\frac{1 \mathrm{g}}{1.00782 \mathrm{u} \times 1.66054 \times 10^{-27} \mathrm{kg.u^{-1}}}\) Now, we multiply the energy released per reaction by the number of reactions in 1 gram of hydrogen nuclei: Energy released per gram = Energy released per reaction × Number of reactions in 1 gram of hydrogen nuclei

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Americium-241 is widely used in smoke detectors. The radiation released by this element ionizes particles that are then detected by a charged-particle collector. The half-life of \(^{241} \mathrm{Am}\) is 433 years, and it decays by emitting \(\alpha\) particles. How many \(\alpha\) particles are emitted each second by a 5.00 -g sample of \(^{241} \mathrm{Am} ?\)

The earth receives \(1.8 \times 10^{14} \mathrm{kJ} / \mathrm{s}\) of solar energy. What mass of solar material is converted to energy over a 24 -h period to provide the daily amount of solar energy to the earth? What mass of coal would have to be burned to provide the same amount of energy? (Coal releases \(32 \mathrm{kJ}\) of energy per gram when burned.)

Given the following information: Mass of proton \(=1.00728 \mathrm{u}\) Mass of neutron \(=1.00866 \mathrm{u}\) Mass of electron \(=5.486 \times 10^{-4} \mathrm{u}\) Speed of light \(=2.9979 \times 10^{8} \mathrm{m} / \mathrm{s}\) Calculate the nuclear binding energy of \(\frac{24}{12} \mathrm{Mg},\) which has an atomic mass of 23.9850 u.

Many elements have been synthesized by bombarding relatively heavy atoms with high-energy particles in particle accelerators. Complete the following nuclear equations, which have been used to synthesize elements. a. \(\quad+\frac{4}{2} H e \rightarrow 243 B k+\frac{1}{0} n\) b. \(^{238} \mathrm{U}+^{12}_{6} \mathrm{C} \rightarrow$$\quad$$+6_{0}^{1} n\) c. \(^{249} \mathrm{Cf}+$$\quad$$\rightarrow \frac{260}{105} D b+4 \frac{1}{6} n\) d. \(^{249} \mathrm{Cf}+^{10}_{5} \mathrm{B} \rightarrow \frac{257}{153} \mathrm{Lr}+\)__________

Which of the following statement(s) is(are) true? a. A radioactive nuclide that decays from \(1.00 \times 10^{10}\) atoms to \(2.5 \times 10^{9}\) atoms in 10 minutes has a half-life of 5.0 minutes.b. Nuclides with large \(Z\) values are observed to be \(\alpha\) -particle producers. c. As \(Z\) increases, nuclides need a greater proton-to-neutron ratio for stability. d. Those "light" nuclides that have twice as many neutrons as protons are expected to be stable.
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Inorganic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free