A certain first-order reaction (A-products) has a rate constant of 8.40x10-3 s1

Question

A certain first-order reaction (A-products) has a rate constant of 8.40x10-3 s1 at 45 ° C. How many minutes does it take for the concentration of the reactant, AI, to drop to 6.25% of the original concentration? Express your answer with the appropriate units. View Available Hint(s) alue Units Submit Part B A certain second-order reaction (B-products) has a rate constant of 1.55x10-3 Ms-1 at 27 o C and an initial half-life of 250 s. What is the concentration of the reactant B after one half-life? Express your answer with the appropriate units. View Available Hint(s) alue Units

Explanation / Answer

A)

we have:

[A]o = 100 M (This is initial concentration)

[A] = 6.25 % of 100 M = 6.25 M

k = 8.4*10^-3 s-1

use integrated rate law for 1st order reaction

ln[A] = ln[A]o - k*t

ln(6.25) = ln(100) - 8.4*10^-3*t

1.8326 = 4.6052 - 8.4*10^-3*t

8.4*10^-3*t = 2.7726

t = 330 s

t = 330/60

t = 5.5 minutes

Answer: 5.5 minutes

B)

use relation between rate constant and half life of 2nd order reaction

k = 1/([B]o*half life)

[B]o = 1/(k*half life)

= 1/(1.55*10^-3 * 250)

= 2.58 M

After one half life, concentration will be come half of this

So,

[B] = [B]o/2

= 2.58/2

= 1.29 M

Answer: 1.29 M

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