Based on the following atomic mass values - 1H, 1.00782 amu; 2H, 2.01410 amu; 3H

Question

Based on the following atomic mass values - 1H, 1.00782 amu; 2H, 2.01410 amu; 3H, 3.01605 amu; 3He, 3.01603 amu; 4He, 4.00260 amu-and the mass of the neutron given in the text, calculate the energy change per mole in each of the following nuclear reactions, all of which are possibilities for a controlled fusion process.

Part A

21H+31H42He+10n

Express your answer using four significant figures.

Part B

21H+21H32He+10n

Express your answer using three significant figures.

Part C

21H+32He42He+11H

Express your answer using four significant figures.

Explanation / Answer

During a nuclear reaction, the combined mass of the products is less than the combined mass of the reactants. The “difference in mass” is converted into energy and may be calculated using Einstein’s mass-energy equivalence, E=mc^2

Now 1 amu = 1.6605 x 10^-24 g = 1.6605 x 10^-27 kg

c = 299 792 458 m/s

E = (1.6605 x 10^-27)( 299 792 458)^2 = 1.492 x 10^-10 J              (1 kg m^2 /s^2 = 1 J)

Thus 1 amu = 1.492 x 10^-10 J = 931.5 MeV                        (1 MeV = 1.602 1x 10^- 13 J)

Given :

1H, 1.00782 amu; 2H, 2.01410 amu; 3H, 3.01605 amu; 3He, 3.01603 amu; 4He, 4.00260 amu-

Mass of neutron 10n= 1.6749 x 10-27 kg = 1.0086654 a.m.u.

Part A

21H+31H42He+10n

Express your answer using four significant figures.

Sum of the masses of reactants = 2.01410 + 3.01605= 5.03015 a.m.u.
Sum of the masses of products =4.00260 + 1.00866 = 5.0112 amu

Therefore, M = Sum of the masses of products – Sum of the masses of reactants

= 5.0112-5.03015= – 0.01895 a.m.u.

A decrease in mass indicates that energy is being released.

E = 0.01895 a.m.u..x 1.492 x 10^-10 J /a.m.u = 2.827 x 10^-12 J

Now convert energy from J/atom to kJ/ mole

E = (2.827 x 10^-12 J/atom) (1 kJ/1000 J)(6.023 x 1023 atoms/mol)

E = 1.702 x 10^9 kJ/mol

Part B

21H+21H32He+10n

Express your answer using three significant figures.

Sum of the masses of reactants = 2 (2.01410) = 4.0282 a.m.u.
Sum of the masses of products = 3.01603 + 1.00866 = 4.0247 amu

Therefore, M = Sum of the masses of products – Sum of the masses of reactants

= 4.0247 -4.0282 = – 0.0035 a.m.u.

A decrease in mass indicates that energy is being released.

E = 0.0035 a.m.u..x 1.492 x 10^-10 J /a.m.u = 5.236 x 10^-13 J

Now convert energy from J/atom to kJ/ mole

E = (5.236 x 10^-13 J/atom) (1 kJ/1000 J)(6.023 x 1023 atoms/mol)

E = 3.15 x 10^8 kJ/mol

Part C

21H+32He42He+11H

Express your answer using four significant figures

Sum of the masses of reactants = 2.01410 + 3.01603 = 5.03013 a.m.u.
Sum of the masses of products =4.00260 + 1.00782 = 5.01042 amu

Therefore, M = Sum of the masses of products – Sum of the masses of reactants

= 5.01042 -5.03013= – 0.0197 a.m.u.

A decrease in mass indicates that energy is being released.

E = 0.0197 a.m.u..x 1.492 x 10^-10 J /a.m.u = 2.941 x 10^-12 J

Now convert energy from J/atom to kJ/ mole

E = (2.941 x 10^-12 J/atom) (1 kJ/1000 J)(6.023 x 1023 atoms/mol)

E = 1.771 x 10^9 kJ/mol

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