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15.27

Posted: Wed Mar 07, 2018 12:22 am
part (c) says: How much time must elapse for the concentration of A to decrease to 15% of its initial concentration? How do we use the 15% in our calculations? I looked at the answers and sort of got what they've done but I'm still a bit confused

Re: 15.27

Posted: Wed Mar 07, 2018 9:11 am
Personally (I don't have the solutions manual) I knew that to find $t_{1/2}$ you divide ln 2 by k, so generally you divide ln(reciprocal of the fraction) by k. 15% is equivalent to the fraction 15/100 which equals 3/20, so the reciprocal is 20/3. I found the solution by taking ln(20/3) and dividing by k.

Re: 15.27

Posted: Thu Mar 08, 2018 10:03 am
Since it's a first order reaction, we use t1/2 =$\frac{ln2}{k}$ and solve for k with the given half life. Once you have k, we can use the integrated rate law (ln[A] = -kt + ln[A]0). If the concentration of A decreases to 15% of the original, we can use [A]0 = 1 and [A] = 0.15 to solve for t.