Consider the reaction $$2{ }_{1}^{2} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He}$$ (a) Calculate \(\Delta E\) in kilojoules per gram of deuterium fused. (b) How much energy is potentially available from the fusion of all the deuterium in seawater? The percentage of deuterium in water is about \(0.0017 \%\). The total mass of water in the oceans is \(1.3 \times 10^{24} \mathrm{~g}\). (c) What fraction of the deuterium in the oceans would have to be consumed to supply the annual energy requirements of the world \(\left(2.3 \times 10^{17} \mathrm{~kJ}\right) ?\)

First, let's find the mass difference between the initial deuterium nuclei and the final helium nucleus. Note that 1 deuterium nucleus has a mass of approximately 2 atomic mass units (amu) and the helium nucleus has a mass of approximately 4 amu. The mass difference for this reaction is: $$Δm = 2(2\text{ amu}) - 4\text{ amu} = 0\text{ amu}$$ The mass difference is zero since the initial and final mass are the same. However, we need to consider the binding energy to accurately calculate the energy released in the reaction.

The binding energy of deuterium is about 2.224 MeV, and the binding energy of helium is about 28.3 MeV. One atomic mass unit (amu) can be converted to energy units (MeV) using: $$1\text{ amu} \approx 931.5\text{ MeV}$$ The total initial binding energy of two deuterium nuclei is: $$2\times2.224\text{ MeV} = 4.448\text{ MeV}$$

Now, we will calculate the difference in binding energy between the initial deuterium nuclei and the final helium nucleus: $$ΔE = 28.3\text{ MeV} - 4.448\text{ MeV} = 23.852\text{ MeV}$$ Next, we need to convert this energy difference to kilojoules per gram. Convert MeV to J: $$23.852\text{ MeV}\times\frac{1.602\times10^{-13}\text{ J}}{1\text{ MeV}}=3.821\times10^{-12}\text{ J}$$ Finally, convert energy per reaction to kilojoules per gram. The molar mass of deuterium is approximately 2 g/mol, so we have: $$\frac{3.821\times10^{-12}\text{ J}}{2\text{ amu}}\times\frac{1\text{ mol}}{1\times10^{24}\text{ amu}}\times\frac{1\text{ kg}}{10^3\text{ g}}\times\frac{10^3\text{ kJ}}{1\text{ J}} = 1.910\times10^{-11}\text{ kJ/g}$$ Answer for part (a): The energy released per gram of deuterium fused is \(1.910\times10^{-11}\mathrm{\ kJ/g}\).

First, we need to calculate the total mass of deuterium in the oceans, using the given percentage and total mass of water: $$1.3\times10^{24}\text{ g}\times\frac{0.0017}{100} = 2.21\times10^{22}\text{ g}$$ Now, we can use the energy released per gram of deuterium fused calculated in part (a) to find the total energy available from the fusion of all deuterium in the oceans: $$2.21\times10^{22}\text{ g}\times1.910\times10^{-11}\text{ kJ/g} = 4.221\times10^{11}\text{ kJ}$$ Answer for part (b): The total energy potentially available from the fusion of all deuterium in seawater is \(4.221\times10^{11}\mathrm{\ kJ}\).

Now, let's compare the total energy available from the fusion of deuterium to the given annual energy requirements of the world: $$\frac{2.3\times10^{17}\text{ kJ}}{4.221\times10^{11}\text{ kJ}} = 5.455\times10^{-5}$$ Answer for part (c): The fraction of deuterium in the oceans needed to be consumed to supply the annual energy requirements of the world is approximately \(5.455\times10^{-5}\).