Question
For a certain first-order reaction the rate constan is 2.9 s-1. what is the fraction of the reactant remaining after 4 half lives?
Answers
DrBob222
Two ways. The long way is
k = 0.693/t<sub>1/2</sub>. Solve for t<sub>1/2</sub>. Substitute
4*t<sub>1/2</sub> into the equation below for t. Solve for
ln(No/N) = kt
Solve for No/N and take the reciprocal.
The short way is 2<sup>4</sup> = 16 so there is 1/16 left.
k = 0.693/t<sub>1/2</sub>. Solve for t<sub>1/2</sub>. Substitute
4*t<sub>1/2</sub> into the equation below for t. Solve for
ln(No/N) = kt
Solve for No/N and take the reciprocal.
The short way is 2<sup>4</sup> = 16 so there is 1/16 left.
jahil
thanks so in any other question i can still use to as my base number raised to whatever half life is provided?
DrBob222
Yes, but you must be careful what it stands for. In this case it stands for the reciprocal of the fraction left; i.e.,
2<sup>4</sup> = 16 so there is 1/16 left.
2<sup>4</sup> = 16 so there is 1/16 left.