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3 years ago

7

A first-order decomposition reaction has a rate constant of 0.00440 yr−1. How long does it take for [reactant] to reach 12.5% of

its original value? Be sure to report your answer to the correct number of significant figures.

1 answer:

mina [271]3 years ago

8 0

Answer:

473 year

Explanation:

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given:

To reach 12.5% of reactant means that 0.125 of $[A_0]$ is decomposed. So,

$\frac {[A_t]}{[A_0]}$ = 0.125

t = ?

$\frac {[A_t]}{[A_0]}=e^{-k\times t}$

$0.125=e^{-0.00440\times t}$

t = 473 year

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