Question

A certain first-order reaction (A→productsA→products) has a rate constant of 9.90×10−3 s−1s−1 at 45 ∘C∘C. How many minutes does it take for the concentration of the reactant, [A][A] , to drop to 6.25%% of the original concentration? Express your answer with the appropriate units

Answer 1

Answer:

4.66667 minutes

Explanation:

Rate constant, k = 9.90×10−3 s−1

Time = ?

Initial concentration, [A]o = 100

Final concentration, [A] = 6.25

The integral rate law for first order reactions is given as;

ln[A] = ln[A]o − kt

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

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

t = [ln(100) - ln(6.25)] / 9.90×10−3

t = 2.77 / 9.90×10−3

t = 0.28006 ×103

t = 280 seconds

t = 4.66667 minutes (Upon conversion  by dividing by 60)