Only Part B Please! Half-life for First and Second Order Reactions The half-life
ID: 996784 • Letter: O
Question
Only Part B Please!
Half-life for First and Second Order Reactions The half-life of a reaction, t_1/2, is the time it takes for the reactant concentration [A] to decrease by half. For example, after one half-life the concentration falls from the initial concentration [A]_0 to [A]_0/2, after a second half-life to [A]_0/4, after a third half-life to [A]_0/8, and so on. on. For a first-order reaction, the half-life is constant. It depends only on the rate constant k and not on the reactant concentration. It is expressed as t_1/2 = 0.693/k For a second-order reaction, the half-life depends on the rate constant and the concentration of the reactant and so is expressed as t_1/2 = 1/k[A]_0 A certain first-order reaction (Ay products) has a rate constant of 4.50 Times 10^-3 s^-1 at 45 degree C. How many minutes does it take for the concentration of the reactant, [A], to drop to 6.25% of the original concentration? Express your answer numerically in minutes. A certain second-order reaction (B rightarrow products) has a rate constant of 1.90 Times 10^-3 M^-1 middot s^-1 at 27 degree C and an initial half-life of 210 s. What is the concentration of the reactant B after one half-life? Express the molar concentration numerically.Explanation / Answer
K = 2.303/t log[A0]/[A]
4.5*10-3 = 2.303/t log100/100-6.25
4.5*10-3 = 2.303/t log100/93.75
4.5*10-3 = 2.303*0.028/t
t = 0.064484/4.5*10-3 =14.33Sec
Part -B
t1/2 = 1/K[B]
210 = 1/1.9*10-3 *[B]
[B] = 210*1.9*10-3 = 0.399M
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