Question

How do the energies released by nuclear processes compare in magnitude with the energies of ordinary chemical processes?

Short answer

The energies released in nuclear processes, such as fission of uranium-235, are much larger than those of ordinary chemical processes, like the combustion of methanol. In our comparison, the energy per interaction for uranium fission was \(200 \times 10^6 \mbox{ eV}\), while for methanol combustion, it was only \(1.5 \mbox{ eV}\). The ratio of energy released in uranium fission to methanol combustion is around \(1.33 \times 10^8\). This shows that nuclear processes can release approximately \(10^8\) times more energy than chemical processes.

Step-by-step solution

Understand the differences between nuclear and chemical processes:

Nuclear and chemical processes involve different interactions within matter. Chemical processes occur when the electrons surrounding the atoms interact, leading to the formation and breaking of chemical bonds. Examples include combustion, precipitation, or oxidation reactions. On the other hand, nuclear processes involve interactions within atomic nuclei, such as nuclear fission, where heavy atomic nuclei break into smaller nuclei, and nuclear fusion, where light atomic nuclei combine to form heavier nuclei.

Calculate energy released in chemical reactions:

Let's take an example of a chemical reaction: combustion of methanol (CH3OH). The balanced equation for the combustion of methanol is: CH3OH + O2 -> CO2 + 2H2O The amount of energy released in this reaction per mole of methanol is approximately 1450 kJ/mol. To convert this into energy per interaction, we can divide this energy per mole by Avogadro's number (\(6.022 \times 10^{23}\) interactions per mole) to get the energy per interaction in joules, and then convert it to electronvolts (eV) using \(1 \mbox{ eV }= 1.6 \times 10^{-19}\) J. Energy per interaction for methanol combustion in eV: \(\frac{1450 \mbox{ kJ/mol}}{6.022 \times 10^{23}\mbox{ interactions/mol}} \cdot \frac{1000 \mbox{ J}}{\mbox{ kJ}} \cdot \frac{1 \mbox{ eV}}{1.6 \times 10^{-19}\mbox{ J}} \approx 1.5 \mbox{ eV}\)

Calculate energy released in nuclear reactions:

For nuclear processes, let's take an example of the fission of uranium-235 (U-235) by a neutron. One U-235 nucleus can generate about 200 MeV (million electronvolts) in this reaction. Converting this to eV, we have: Energy per interaction for uranium fission: 200 MeV \(=\) 200 * \(10^6\) eV

Compare the energy magnitudes:

Now we can compare the energy magnitudes between chemical and nuclear processes: Energy per interaction for methanol combustion: \(1.5 \mbox{ eV}\) Energy per interaction for uranium fission: \(200 \times 10^6 \mbox{ eV}\) The ratio of energy released in uranium fission to methanol combustion is: \(\frac{200 \times 10^6 \mbox{ eV}}{1.5 \mbox{ eV }} \approx 1.33 \times 10^8\) The energies released in nuclear processes are much larger than those released in ordinary chemical processes, in this case around \(10^8\) times greater. This result is representative of nuclear reactions in general compared to chemical reactions. It highlights the substantial energy that can be harnessed through nuclear reactions, such as fission and fusion, compared to those from ordinary chemical reactions.