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stellarik [79]
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.
Chemistry
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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Mass of solute = 0.0036 g

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