Question

I have a test tomorrow and really need help seeing how to do numbers 1,2,3 and 4. If you can start with telling me what equation to use, then solving using that equation, it would be much appreciated.

facweb.bhc.edu Practice Kinetics Solve the following use appropriate units and sig figs. QUESTIONS Half lives and rate constants 1. Calculate the first-order rate constant for the radioactive decay of radium. The 1590 years. 2. Calculate the rate constant for the decay of ca The halflife is 5730 years. 3. Calculate the half-life of a radioactive isotope with rate constant k 0.0866 day 1 4. Calculate the half life of an isotope with rate constant k 0.0563 yi. Integrated rate laws 5. Calculate the percent carbon-14 remaining in an artifact 20000 years old. 6. Calculate the age of an artifact with 25.0% of its carbon-14 remaining. 7. Calculate the percent carbon-14 remaining after 10 half lives. 8. Calculate the time for 99.9% of radioactive iodine to decay. Half-life is tw, 8.00 days. Rate constants and activation energy 9. Calculate the activation energy for a reaction if the rate constant increases from 87.1 at 500 K to 15 io. Calculate the activation energy if the rate constant increases from k 12 at 300 K to k 1200 at 500 K.

Answer 1

T_1/2 = 1590 years k = 0.693/1590 = 4.36 times 10^-4 years rate constant = 4.36 times 10^-4 yars^-1 c-14 decay follows first -order kinetics t_1/2 = 5730 years k = 0.693/t_1/2 = 0.693/5730 = 1.21 times 10^-4 years^-1 rate constant k = 0.0866 day^-1 t_1/2 = 0.693/k = 0.693/0.0866 = 8 days rate constant0.0563 y^-1 t_1/2 = 0.693/k = 0.693/0.0563 = 12.3 years t = 20,000 years ln[R]_t/[R]_0 = -kt k = 1.2 times 10^-4 y^-1 where [R]_t rightarrow concentration at t time [R]_0 rightarrow concentration at t = 0 ln[R]_t/[R] = -20000 times 1.2 times 10^-4 [R]_t = e^-2.42 = 0.089 % of -14 remaining = c-14 conc. Remaining/initial conc. Times 100 = 0.089 times 100 = 8.9% t = p k = 1.21 times 10^-4 y^-1 [R]_t = 25 [R]_0 ln[R]_t/[R]_0 = -kt t = 1.145698 times 10^4 = 11456.98 years 10 half lines means t = 10 times t_1/2 = 10 times 5730 = 57300 years [R]_t/[R]_0 = e^-6.9333 = 0.000974 % of carbon-14 remaining after 10 half lines = [R]_t/[R]_0 times 100 = 0.000974 times 100 = 0.0974% t_1/2 = 8.0 days k = 0.693/t_1/2 = 0.693/8 = 0.0866 day^-1 [R]_0 = [R]_0 at 999% of radioactive decay only 0.01% will be left [R]_t = 0.0001[R]_0 t = -1/k ln[R]_t/[R]_0 = 106.35 days k = 87.1 T = 500k question is not fully available can be calculated using k_1 = 12,T_1 = 300k k_2 = 1200 T_2 = 500k ln(12/1200) =(1/500-1/300)E_9/8.314 E_9 = 28715.5 j/mol = 28.716 kj/mol