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RSB [31]
3 years ago
7

A first-order reaction has a half-life of 29.2 s . how long does it take for the concentration of the reactant in the reaction t

o fall to one-sixteenth of its initial value?
Chemistry
1 answer:
Dmitriy789 [7]3 years ago
5 0

Answer is: it takes 116,8 seconds to fall to one-sixteenth of its initial value

<span> The half-life for the chemical reaction is 29,2 s and is independent of initial concentration.
c</span>₀ - initial concentration the reactant.

c - concentration of the reactant remaining at time.

t = 29,2 s.<span>
First calculate the rate constant k:
k = 0,693 ÷ t = 0,693 ÷ 29,2 s</span> = 0,0237 1/s.<span>
ln(c/c</span>₀) = -k·t₁.<span>
ln(1/16 </span>÷ 1) = -0,0237 1/s · t₁.

t₁ = 116,8 s.

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