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14 Chemical Kinetics Solutions to Exercises (b) For a reaction this slow, the graph of [A] vs t is very shallow. If the reaction takes 6 days to complete 4 half-lives, the half-life is 1.5 days or 18 hours. Data taken for a period of 2 hours, or 0.08 days, covers a very small portion of the curve and appears linear. [A] 0 1 2 3 4 5 6 t, days 14.104 (a) = 0.693/k = 0.693/7.0 10⁻⁴ -1 = 990 = 9.9 10² (b) k = 0.693 = 56.3 0.693 min 60 = 2.05 10⁻⁴ -1 14.105 Analyze. Given rate constants for the decay of two radioisotopes, determine half-lives, decay rates, and amount remaining after three half-lives. Plan. Determine reaction or- der. Based on reaction-order, select the appropriate relationships for (a) rate constant and half-life and (c) rate-constant, time and concentration. In this example, mass is a measure of concentration. Solve. Decay of radioiosotopes is a first-order process, since only one species is in- volved and the decay is not initiated by collision. (a) For a first-order process, = 0.693/k. t1/2 = 0.693/1.6 X 10⁻³ yr⁻¹ = 433.1 = 4.3 10² yr t1/2 = 0.693/0.011 = 63.00 = 63 days (b) For a given sample size, half of the sample decays in 433 years, whereas half of the sample decays in 63 days. 125[ decays at a much faster rate. (c) For a first order process, - In[A]₀ = -kt. = -kt + In[A]₀. [A]₀ = 1.0 mg; t = 3 t = 3 = 3(433.1 yr) = 1.299 10³ = 1.3 X 10³ yr In [Am]ₜ = -1.6 X 10⁻³ yr⁻¹ (1.299 10³ yr) - In (1.0) = -2.079 - 0 = -2.08 = 0.125 = 0.13 mg or, mass remaining = 1.0 mg/2³ = 0.125 = 0.13 mg For the same size starting sample and number of elapsed half-lives, the same mass, 0.13 mg will remain. (The difference is that the elapsed time of 3 half- lives (for is 3(63) = 189 days = 0.52 yr, 433 yr for (d) Again, for a first order process, In[A]₀ = -kt. In[A]ₜ = -kt + [A]₀ = 1.0 mg; t = 4 days. 424