chapter20_part1

chapter20_part1 - Chapter 20 Nuclear Chemistry 908...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 20: Nuclear Chemistry 908 End-of-Chapter Solutions for Chapter 20 Summary Problem Answer: (a) 86 222 Rn 2 4 He + 84 218 Po (b) 92 238 U + 1 n 92 239 U , 92 239 U 93 239 Np + " 1 e , 93 239 Np 94 239 Pu + " 1 e (c) 92 235 U + 1 n 51 132 Sb + 41 101 Nb + 3 1 n (d) – 3.37 × 10 8 kJ/g reactants (e) (i) 1.0 × 10 4 y (ii) 8.1 × 10 5 y (f) provisions for natural events for 800,000 years (g) (i) 53 131 I 54 131 Xe + " 1 e (ii) 4.0 × 10 10 Bq Strategy and Explanation: Balance the atomic numbers and mass numbers. Use the periodic table to get atomic numbers or the symbols of elements. (a) Radon, Rn, has atomic number 86. Alpha emission produces 2 4 He . 86 = 2 + 84. 222 = 4 + 218. The product is an element with atomic number 84, Po. 86 222 Rn 2 4 He + 84 218 Po (b) Uranium, U, has atomic number 92. Neutron absorption means that 92 238 U reacts with 1 n . 238 + 1 = 239. 92 + 0 = 92. The product still has atomic number 92, U. 92 238 U + 1 n 92 239 U Beta decay results in the production of " 1 e . 92 = 93 – 1. 239 = 239 + 0. The product is an element with atomic number 93, Np. 92 239 U 93 239 Np + " 1 e Beta decay results in the production of " 1 e . 93 = 94 – 1. 239 = 239 + 0. The product is an element with atomic number 94, Pu. 93 239 Np 94 239 Pu + " 1 e (c) Fission occurs with the absorption of a single neutron, the production of two smaller atoms and the emission of one or more neutrons. Look up atomic numbers for U (92), Sb (51), and Nb (41). 92 + 0 = 51 + 41. 235 + 1 = 236 = 132 + 101 + 3. The subatomic particles emitted must be three neutrons. 92 235 U + 1 n 51 132 Sb + 41 101 Nb + 3 1 n (d) Use the method described in Section 20.3. Calculate the mass defect ( Δ m). Use Einstein’s equation: E = ( Δ m)c 2 to calculate the total energy released, then divide by the mass of the reactants. The stoichiometry of the nuclear equation indicates that one mol of each of the products and reactants are involved. Δ m = m products – m reactants Δ m = (m 4 He + m 1 n ) – (m 2 H + m 3 H ) Δ m = (4.00150 g + 1.00867 g) – (2.01355 g + 3.01550 g) = – 0.01888 g The nuclear binding energy = E b = ( Δ m)c 2 E b = ( Δ m)c 2 = " 0.01888 g # 1 kg 1000 g $ % & ’ ( ) # (2.99792 # 10 8 m/s) 2 # 1 J 1 kgm 2 s " 2 # 1 kJ 1000 J = – 1.70 × 10 9 kJ A total of 1.70 × 10 9 kJ released. To find the kJ/gram of reactants, divide the E b by the total mass of reactants: E b per gram reactants = 1.70 " 10 9 kJ 3.01550 g 1 3 H + 2.01355 g 1 2 H = 3.37 × 10 8 kJ/g reactants Chapter 20: Nuclear Chemistry 909 (e) Determine the value of the rate constant, k, from the half-life: k = ln2 t 1/2 = ln2 2.44 " 10 4 y = 2.84 × 10 –5 y –1 The activity at initial time is 100%, so A = 100%. Use Equation 20.2 to determine the time required to get to A at 75% and 10%: ln A A " # $ % & ’ = – kt (i) For A = 75% ln 75% 100% " # $ % & ’ = – (2.84 × 10 –5 y –1 )t ln(0.75) = – 0.288 = – (2.84 × 10 –5 y –1 )t t = 1.0 × 10 4 y (ii) For A = 10%...
View Full Document

{[ snackBarMessage ]}

Page1 / 15

chapter20_part1 - Chapter 20 Nuclear Chemistry 908...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online