Question

A decomposition reaction has a half life of 578 years. (a) What is the rate constant for this reaction? (b) How many years does it take for the reactant concentration to reach 12.5% of its original value? Report your answer to 3 significant figures.

Answer 1

Answer:

$0.001199 year^{-1}$ is the rate constant for this reaction.

It will take $1.73\times 10^3 years$ to concentration to reach 12.5% of its original value.

Explanation:

A decomposition reaction follows first order kinetics:

Half life of the reaction = $t_{1/2}=578 years$

Rate constant of the reaction = k

For first order reaction, half life and rate constant are linked with an expression :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{578 years}=0.001199 year^{-1}$

$0.001199 year^{-1}$ is the rate constant for this reaction.

Initial concentration of reactant =$[A_o]$ =  x

Final concentration of reactant after time t =$[A]$ =  12.5% of x = 0.125x

The integrated law of first order reaction :

$[A]=[A_o]\times e^{-kt}$

$0.125x=x\times e^{-0.001199 year^{-1}\times t}$

t = 1,734.31 years =$1.73\times 10^3 years$

It will take $1.73\times 10^3 years$ to concentration to reach 12.5% of its original value.