Question

A certain first-order reaction (A→products) has a rate constant of 7.20×10−3 s−1 at 45 ∘C. How many minutes does it take for the concentration of the reactant, [A], to drop to 6.25% of the original concentration?
Express your answer with the appropriate units.

Answer 1

Answer:

6,41 min

Explanation:

For the reaction:

A → products

kinetics first-order reaction law is:

ln[A] = ln[A]₀ -kt

Where [A] is concentration of reactant, [A]₀ is initital concentration of reactant, k is rate constant and t is time.

If the concentration of A is 6,25% you can assume:

[A] = 6,25; [A]₀= 100. Replacing:

ln(6,25) = ln(100) -7,20×10⁻³s⁻¹t

-2,7726 = -7,20×10⁻³s⁻¹t

385s = t

In minutes:

385s×$\frac{1min}{60s}$ = 6,41 min

I hope it helps!