## 7B.7

$\frac{d[R]}{dt}=-k[R]; \ln [R]=-kt + \ln [R]_{0}; t_{\frac{1}{2}}=\frac{0.693}{k}$

Posts: 97
Joined: Tue Feb 12, 2019 12:15 am

### 7B.7

The question says Substance A decomposes in a first order reaction and its half life is 355s. How much time must be elapsed for the concentration of A to decrease to a) one-eight of its initial concentration; b) one fourth of its initial concentration; c) 15% of its initial concentration ; d) one-ninth of its initial concentration?
The answer to b is [A]/[A]0=1/4= (1/2)^2, so the time elapsed 2 half lives. t= 2(355s)=710 s
The answer to d is t=ln([A]0/1/9[A]0)/k= ln9/1.95x10^-3*s^-1=1.1x10^3s
What is the difference between calculating the time elapsed in method b versus in method d? Why can't we use the same method of b on problem d?

Chris Tai 1B
Posts: 102
Joined: Sat Aug 24, 2019 12:16 am

### Re: 7B.7

Since part b of the question is just asking for the time that it takes the concentration of substance to decrease to 1/4 its original concentration, you can just multiply the half life of the substance by 2. It will decrease to half of the concentration first (one half life) and then decrease to half of that half (or, one fourth - this is the second half life).
However, in part d, we're looking for 1/9 of the concentration, so we can't use this simple heuristic to determine the amount of time. So, we can use the first order rate law equation which is [A] = [A]0e^(-kt). Since we're looking for when [A] = [A]0/9, we substitute this in for [A] to get (1/9)[A]0 = [A]0e^(kt) or 1/9 = e^(kt). Finally, we can solve for t since we're given the k rate constant value.