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What is the rate constant of a first-order reaction that takes 143 seconds for the reactant concentration to drop to half of its initial value?

1 Answer

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Answer:

k = 4.85 x 10⁻³ / s

Step-by-step explanation:

For a first order reaction we have two very important equations we need to memorize:

ln ( N/N₀ ) = -kt

and,

0.693/ k = t₁/₂

where k is the rate constant and t i is the halflife which can be derived from the first equation.

0.693/ k = t₁/₂ ⇒ k = 0.693 / t₁/₂

k = 0.693 / 143 s = 4.85 x 10⁻³ / s

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