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andreyandreev [35.5K]
3 years ago
14

1/2-life of pyruvic acid in the presence of aminotransferase enzyme (which converts it to alanine) was found to be 221 s. how lo

ng will it take for the concentration of pyruvic acid to fall to 1/64 of its initial value in this first order reaction?
Chemistry
1 answer:
Ugo [173]3 years ago
6 0

Answer:

After 1326s, the concentration of pyruvic acid fall to 1/64 of its initial concentration.

Explanation:

The first order kinetics reaction is:

ln [A] = ln [A]₀ - kt

<em>Where [A] is concentration after t time, [A]₀ is intial concentration and k is reaction constant.</em>

To convert half-life to k you must use:

t(1/2) = ln 2 / K

221s = ln 2 / K

K = ln 2 / 221s

<h3>K = 3.1364x10⁻³s⁻¹</h3>

If [A] = 1/64, [A]₀ = 1:

ln [A] = ln [A]₀ - kt

ln (1/64) = ln 1 - 3.1364x10⁻³t

4.1588 = 3.1364x10⁻³s⁻¹t

1326s = t

<h3>After 1326s, the concentration of pyruvic acid fall to 1/64 of its initial concentration.</h3>

<em />

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