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ercise 13.13, the reaction of sucrose with water was In Practice Ex order with r

ID: 229828 • Letter: E

Question

ercise 13.13, the reaction of sucrose with water was In Practice Ex order with respect to sucrose. The rate constant under the conditions of the experiments was 6.17 × 10-4 s-1. Calculate the value of t1/2 for this reaction in minutes. How many minutes would it take for three-quarters of the sucrose to react? (Hint: What fraction of the sucrose remains?) From the answer to Practice Exercise 13.15, determine the half-life of an active ingredient that has a shelf life of 2.00 years. The radioactive isotope, phosphorus-32, has a half-life of 14.26 days. What percent of phosphorus-32 will remain after 60 days?

Explanation / Answer


Rate constant k = 6.17 * 10^-4 s^-1

we have the formula t1/2 = 0.693/k

t1/2 = 0.693/(6.17*10^-4) = 1123.18 sec

k = 2.303/t log(a/a-x)

t = time for the reaction

a = initial concentration

a-x = final concentration after time t

6.17 * 10^-4 = 2.303/t*log(1/(1-3/4))

t = 2247.235 Sec

t = 2247.235/60 = 37.454 min

t1/2 of P-32 = 14.26 days

t1/2 = 0.693/k

14.26 = 0.693/k

k = 4.856*10^-2

k = 2.303/tlog(a/a-x)

4.856 * 10^-2 = 2.303/60*log(a/a-x)

(a/a-x) = 18.412

let us consider a = 100 then

a-x = remaining concentration of P-32 = 100/18.412 = 5.4312