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What is the standard enthalpy change for the fusion reaction of hydrogen atoms to form helium atoms? Give the equation for the reaction and show all calculations involved in determining the answer.

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The fusion reaction of hydrogen atoms to form helium atoms is known as nuclear fusion, which occurs in the core of the Sun and other stars. In this process, four hydrogen nuclei  protons  combine to form one helium nucleus  two protons and two neutrons  with the release of a large amount of energy. The overall reaction can be represented as:4 H-1  4 protons   He-4  2 protons + 2 neutrons  + 2 e+ + 2 _e + energyHere, e+ represents a positron, and _e represents an electron neutrino.To calculate the standard enthalpy change for this reaction, we need to consider the mass difference between the reactants and products, which is converted into energy according to Einstein's famous equation, E=mc.Mass of 4 H-1 atoms = 4  1.007825 amu = 4.0313 amuMass of 1 He-4 atom = 4.001506 amuMass difference  m  = 4.0313 amu - 4.001506 amu = 0.029794 amuTo convert the mass difference into energy, we need to convert amu  atomic mass units  to kg and then use the E=mc equation.1 amu = 1.66054  10^-27  kgm = 0.029794 amu  1.66054  10^-27  kg/amu = 4.9509  10^-29  kgNow, we can calculate the energy released using E=mc:E =  4.9509  10^-29  kg    3.00  10^8 m/s E = 4.4558  10^-12  JSince this is the energy released for one reaction, we need to find the energy per mole of reactions. There are 6.022  10^23 particles in one mole, so:Energy per mole = 4.4558  10^-12  J  6.022  10^23Energy per mole = 2.681  10^12 J/molThe standard enthalpy change for the fusion reaction of hydrogen atoms to form helium atoms is approximately +2.681  10^12 J/mol. The positive sign indicates that the reaction is exothermic, meaning it releases energy.

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