Chapter 21 Nuclear Chemistry Lecture Presentation Chapter 21 Nuclear Chemistry James F. Kirby Quinnipiac University Hamden, CT © 2015 Pearson Education, Inc.
Chemical vs. Nuclear Reactions Comparison of Chemical Reactions and Nuclear Reactions Chemical reactions Nuclear reactions 1. Atoms are rearranged by the breaking and forming of bonds. 2. Only electrons are involved in breaking and forming of bonds. 3. Reactions are accompanied by absorption or release of relatively small amounts of energy. 4. Rates of reaction are influenced by temperature, pressure, concentration, and catalysts. 1. Elements or isotopes are converted from one to another. 2. Protons, neutrons, electrons, and other elementary particles may be involved. 3. Reactions are accompanied by absorption or relase of tremendous amounts of energy. 4. Rates of reaction normally are not affected by temperature, pressure, and catalysts.
21.1 Radioactivity and Nuclear Equations Some nuclei are instable. They fall apart spontaneously, and emit particles in doing so. The process is called radioactive decay or nuclear decay. All isotopes of the elements with atomic numbers higher than 84 are radioactive. Mass Number X A Z Element Symbol Atomic Number Nuclear terminology proton neutron b particle positron a particle 1 1 H or 1 1 p 0 1 n −1 0 e or −1 0 β +1 0 e or +1 0 β 2 4 He or 2 4 α A 1 1 4 Z 1 −1 +1 2
21.1 Radioactivity and Nuclear Equations Conserve mass number (A). U 235 92 + 1n Cs 138 55 + Rb 96 37 + 2 1n 235 + 1 = 138 + 96 + 2x1 The sum of protons plus neutrons in the products must equal the sum of protons plus neutrons in the reactants. Conserve atomic number (Z) or nuclear charge. U 235 92 + 1n Cs 138 55 + Rb 96 37 + 2 1n 92 + 0 = 55 + 37 + 2x0 The sum of nuclear charges in the products must equal the sum of nuclear charges in the reactants.
EXERCISE #1 212 Po 2 4 α + 82 208 Pb Pb A: 212 = 4 + x x = 208 212Po decays by alpha emission. Write the balanced nuclear equation for the decay of 212Po. 212 Po 2 4 α + 82 208 Pb Pb 84 82 A: 212 = 4 + x x = 208 Z: 84 = 2 + y y = 82 EXERCISE #2 Write the balanced equation for the nuclear reaction 26 56 Fe(d,α) 25 54 Mn where d represents the deuterium nucleus (that is, 1 2 H ). 26 56 Fe + 1 2 H 2 4 α + 25 54 Mn
21.1 Radioactivity and Nuclear Equations Types of Radioactive Decay Positron decay Beta decay 6 11 C 5 11 B + +1 0 β 6 14 C 7 14 N + −1 0 β 19 38 K 18 38 Ar + +1 0 β 19 40 K 20 40 Ca + −1 0 β Z: −1 A: unchanged Z: +1 A: unchanged (n +1) (n −1) Electron-capture decay Alpha decay 18 37 Ar + −1 0 e 17 37 Cl 84 212 Po 82 208 Pb + 2 4 α Inner-orbital electron Z: −2 A: −4 26 55 Fe + −1 0 e 25 55 Mn (n −2) Z: −1 A: unchanged (n +1)
EXERCISE #3 Which of the following produces a b particle? a) 31 68 Ga+ −1 0 e 30 68 Zn b) 29 62 Cu +1 0 β + 28 62 Ni c) 87 212 Fr 2 4 α + 85 208 At d) 51 129 Sb −1 0 β + 52 129 Te electron-capture decay positron decay a decay b Decay
21.4 Rates of Radioactive Decay Rate of Decay Half-life Time required for the number of nuclides to reach half the original value. The rate of decay is first-order and proportional to the number of nuclides. rate = − ∆N ∆t = kN t1/2 = 0.693 k ln N = ln N0 – kt N = amount of atoms at time t N0 = amount of atoms at time t = 0 k = decay constant
EXERCISE #4 The half-life of molybdenum-99 is 66.0 hrs. How much of a 1.00-mg sample of 42 99 Mo is left after 330 hrs? ln N = ln N0 – kt k t1/2 = 0.693 k 0.693 𝑡 1/2 = 0.693 66 hrs k = = 0.0105 hr−1 ln N = ln N0 – kt = ln(1.00) – (0.0105 hr−1)(330 hrs) = −3.465 N = e−3.465 = 0.031 mg
21.3 Nuclear Transmutations The change of one element into another, brought about by the collision of two particles. 13 27 Al + 2 4 He 15 30 P + 0 1 n 98 249 Cf + 8 18 O 106 263 Sg + 4 0 1 n 7 14 N + 2 4 He 8 17 O + 1 1 p 7 14 N (a, p) 8 17 O
21.7-21.8 Nuclear Power Nuclear fission A process in which a heavy nucleus (Z >200) splits to form smaller nuclei with intermediate mass and one or more neutrons. 0 1 n + 92 235 U 56 142 Ba + 36 91 Kr + 3 0 1 n Nuclear fusion A process in which two smaller nuclei combine to form a heavier, more stable nucleus. 2 3 He + 1 2 H 2 4 He + 1 0 e Copyright © Cengage Learning. All rights reserved
21.7 Nuclear Power: Fission Nuclear fission A self-sustaining fission process is called a chain reaction. Nuclear chain reaction Event Neutrons causing fission event Result subcritical < 1 Reaction stops critical = 1 Sustained reactions supercritical > 1 Violent explosion Atomic bomb The TNT was set off first. The explosion forces the section of fissionable material together to form an amount considerably larger than the critical mass.
21.8 Nuclear Power: Fusion Nuclear fusion Fusion Reaction Energy Released 1 2 H + 1 2 H 1 3 H + 1 1 H 3.0 x 108 kJ/mol 1 2 H + 1 3 H 2 4 He + 0 1 n 1.7 x 109 kJ/mol 3 6 Li + 1 2 H 2 2 4 He 2.2 x 109 kJ/mol
END OF CHAPTER 21