Question

The rearrangement of methyl isonitrile (CH3NC) to acetonitrile (CH3NC) is a first-order reaction and has a rate constant of 5.11x10-5s-1 at 472k . If the initial concentration of CH3NC is 3.00×10-2M :
(A) What is the half-life (in hours) of this reaction?
t1/2=________________hr
(B) How many hours will it take for the concentration of methyl isonitrile to drop to 6.25% of its initial value?
t=____________________hr

Answer 1

Explanation:

Reaction is first order.

Rate Constant (k) = 5.11x10-5s-1

Temperature = 472k

Initial Concentration = 3.00×10-2M

(A) What is the half-life (in hours) of this reaction?

Formular for half life of a first order reaction is;

t1/2 = 0.693 / k

t1/2 = 0.693 / (5.11x10-5)

t1/2 = 0.1386 x 10 ^ 5 s

Upon converting to hours by dividing the value by 3600;

t1/2 = (0.1386 x 10 ^ 5) / 3600 = 3.85 hours

(B) How many hours will it take for the concentration of methyl isonitrile to drop to 6.25% of its initial value?

6.25% of initial concentration;

(6.25 / 100) * 3.00×10-2 = 0.1875 x 10 ^ -2M

ln[A] = ln[A]o − kt

[A] = Final Concentration

[A]o = Initial Concentration

upon making t subject of interest;

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

Inserting the values;

t = [  In(3.00×10-2) - In(0.1875 x 10 ^ -2) ] / 3.85

t = 2.7726 / 3.85

t = 0.72 hours