Determine how much energy is released when polonium-210 decays according to . [Atomic masses: polonium-210 = 209.982857 amu; helium-4 = 4.002603 amu; lead-206 = 205.974449 amu]

Question

Determine how much energy is released when polonium-210 decays according to  . [Atomic masses: polonium-210 = 209.982857 amu; helium-4 = 4.002603 amu; lead-206 = 205.974449 amu]

Quizplus · Accepted Answer

The nuclear equation for the decay of polonium-210 is:
^210Po → ^206Pb + ^4He
The difference in mass between the reactant and the products is:

209.982857 amu - (206.014469 amu + 4.002603 amu) = -0.034215 amu

This mass difference corresponds to the mass defect, which is converted to energy using the equation E = mc^2:

E = Δm c^2
E = -0.034215 amu x (1.66054 x 10^-27 kg/amu) x (2.99792 x 10^8 m/s)^2
E = -5.221 x 10^-13 J

Multiplying by Avogadro's number and converting to kilojoules per mole gives:

E = -5.221 x 10^-13 J x (6.022 x 10^23/mol) / (1000 J/kJ)
E = -3.14 kJ/mol
The negative value indicates that energy is released in the reaction. The closest answer choice is C, which is calculated as:

5.22 x 10^-3 kJ/mol (expressed in scientific notation)
Therefore, the best answer is C.

Determine How Much Energy Is Released When Polonium-210 Decays According