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How many half-lives are required for the concentration of reactant to decrease to 1.56% of its original value?4247.56.56

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Answer:

6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.

Step-by-step 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:

Concentration is decreased to 1.56 % which means that 0.0156 of
[A_0] is decomposed. So,


\frac {[A_t]}{[A_0]} = 0.0156

Thus,


\frac {[A_t]}{[A_0]}=e^(-k* t)


0.0156=e^(-k* t)

kt = 4.1604

The expression for the half life is:-

Half life = 15.0 hours


t_(1/2)=\frac {ln\ 2}{k}

Where, k is rate constant

So,


k=\frac {ln\ 2}{t_(1/2)}


(4.1604)/(t)=\frac {ln\ 2}{t_(1/2)}


t = 6* t_(1/2)

6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.

User Callum Watkins
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