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Ch.20 - Nuclear Chemistry
Chapter 20, Problem 86

Thorium-232 decays by a 10-step series of nuclear reactions, ultimately yielding lead-208, along with 6 α particles and 4 β particles. How much energy (in kJ/mol) is released during the overall process? The relevant masses are 232Th = 232.038 054, 208Pb = 207.976 627, electron = 0.000 548 6, and 4He = 4.002 603.

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Step 1: Identify the nuclear reaction process. Thorium-232 decays to lead-208 through a series of nuclear reactions, emitting 6 alpha particles and 4 beta particles.
Step 2: Calculate the total mass of the reactants. The initial mass is the mass of Thorium-232, which is 232.038 054 u.
Step 3: Calculate the total mass of the products. This includes the mass of lead-208, 6 alpha particles, and 4 beta particles. Use the given masses: 208Pb = 207.976 627 u, 4He = 4.002 603 u, and electron = 0.000 548 6 u.
Step 4: Determine the mass defect. Subtract the total mass of the products from the total mass of the reactants to find the mass defect, which represents the mass converted to energy.
Step 5: Use Einstein's equation, E=mc^2, to calculate the energy released. Convert the mass defect from atomic mass units to kilograms, and then calculate the energy in joules. Finally, convert the energy to kJ/mol by considering Avogadro's number.