Step-by-step 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