Question

A first-order decomposition reaction has a rate constant of 0.00140 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.

Answer 1

Reactants take 504.87 yr to reach 12.5% of their original value in first-order decomposition reaction.

Equation for the first-order decomposition reaction:-

$A_{t} =A_{0} e^{-kt}$....(1)

Here,  $A_{t}$ is the final concentration, t is the time,  $A_{0}$ is the initial concentration, and k is the rate constant.

Given:-

$A_{t} =0.125A_{0}$

k= $0.00140yr^{-1}$

Substitute the above value in equation (1) as follows:-

$0.125A_{0} =A_{0} e^{-kt} \\0.125A_{0} =A_{0} e^{-k\times0.00140 yr^{-1} }\\ln(0.125)/(-0.00140)=t\\t=504.87 year$

So, 504.87 yr does it take for the reactant to reach 12.5% of its original value.

Find more information about first- order decomposition reaction here:-

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