If iodine has a 13.3 hour half life, 3.12 mg of the original sample will still be present after 39.9 hours.
The reaction rate must decrease to half its initial value before the process has reached its half life. The split of that first order chemical reaction is a quantity related to the rate constant of the reaction.
Briefing:
Given, the iodine half-life is 13.3 hours. As a result, half-life t1/2 Equals ln 2/ K
k = 0.693 .... t1/2
≈ 0.693 /13.3 hours
= 5.21 *10-2
The rate law sets the flow timetime is set by the rate law, which varies for each emotion. When only one reactant's quantity influences the speed of chemical reaction, this is referred to as a first-order contact. Because of this, it is occasionally referred to as unimolecular process. The rate is as follows for this type of reaction: k= 1/t ln [A]0 / [A] is the rate constant for the first order reaction. 5.21 * 10-2 hours are in 39.9 hours.
By including all the numbers in the initial order reaction's rate constant equation, we obtain [A].
t = 3.12mg
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