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elena-14-01-66 [18.8K]
3 years ago
8

a factor of 8, it means that A is now only 1/8 of its original concentration. A first order reaction, where A → products, has a

rate constant of 1.56 × 107 s −1 . At some time, a concentration of 1.06 × 10−6 M of species A is introduced into the reactor. How long does it take for the concentration of A to fall by a factor of 8?
Chemistry
1 answer:
Trava [24]3 years ago
4 0

Answer:

It takes 1.32x10⁻⁷s for the concentration of A to fall by a factor of 8

Explanation:

The equation that represents a first-order kinetics is:

Ln ([A] / [A]₀] = -kt

<em>Where [A] is actual concentration, [A]₀ is initial concentration, K is rate constant (For the given problem, 1.57x10⁷s⁻¹ and t is time.</em>

<em />

As you want the time when you have [A] in a factor of 8 = [A] / [A]₀ = 1/8

Replacing:

Ln ([A] / [A]₀] = -kt

Ln (1/8) = -1.57x10⁷s⁻¹*t

t = 1.32x10⁻⁷s

<h3>It takes 1.32x10⁻⁷s for the concentration of A to fall by a factor of 8</h3>

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