5. A first-order reaction has a rate constant of 3.00 x 103s. The time required

Question

5. A first-order reaction has a rate constant of 3.00 x 103s. The time required for the reaction to D) 201 s 6. At 25°C the rate constant for the first-order decomposition of a pesticide solution is 6.40x 10 be 75.0% complete is A) 95.8s B) 462s C) 231s E) 41.7 s min-1. If the starting concent 62.0 min at 25°C? g concentration of pesticide is 0.0314 M, what concentration will remain after A) 1.14 x 10 M B) 47.4 M c) 1.25 x10M D) 2.11 x 10M E) 2,68 x 10 AM A certain first-order reaction is 25% complete in 42 min. What isthe half-life of the reaction? A) 21 min B) 42 min C) 84 min D) 120 min E) 101 min 7 The rate law for the reaction 3A 2B is rate-k[A] with a rate constant of 0.0447 h-i, what is the half-life of the reaction? A) 0.0224 h B) 0.0645 h C) 15.5 h D) 22.4 h 8. E) 44.7 h 232U (15n,X)Es Inthe following reaction, identifyx. A) 7 9 B)3 C)4n D) 7 positrons E) 15p

Explanation / Answer

5. For a first order reaction,

rate constant k = 3 x 10^-3 s-1

using,

ln[A]t = ln[A]o - kt

t = time

So,

for [A]t = 25

[A]o = 100

we get,

time taken t = [ln(100) - ln(25)/3 x 10^-3) = 462 s

Answer : B) 462 s

6. [A]o = 0.0314 M

t = 62 min

k = 6.4 x 10^-3 min-1

So,

ln[A]t = ln(0.0314) - 6.4 x 10^-3 x 62

amount remaining after 62 min [A]t = 2.11 x 10^-2 M

Answer : D) 2.11 x 10^-2 M

7. [A]o = 100

[A]t = 75

t = 42 min

so,

k = [(ln[A]o - ln[A]t]/t

   = [ln(100) - ln(75)]/42 = 6.85 x 10^-3 min-1

Half-life t1/2,

t1/2 = ln2/k

       = ln2/6.85 x 10^-3 = 101 min

answer : E) 101 min

8. k = 0.0447 h-1

half life t1/2 = ln2/k

                   = ln2/0.0447 = 15.5 h

Answer : C) 15.5 h

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