## Solution manual error?

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

Regina Chi 2K
Posts: 51
Joined: Fri Sep 26, 2014 2:02 pm

### Solution manual error?

I think there is a slight error on 14.29B... Instead of t=(ln([A]o/[A]t))/0.17 min-1 and t=(ln(0.015/0.005))/k for the last step, shouldn't it be t=(ln([A]t/[A]o))/k and t=(ln(0.005/.015)/0.17 min-1???

Niharika Reddy 1D
Posts: 127
Joined: Fri Sep 26, 2014 2:02 pm

### Re: Solution manual error?

The solutions manual is correct. The reason the concentrations are switched is because the equation was rearranged so the negative sign was "dropped". A property of logs is you can bring the exponent of the log down and out to the front, and vice versa. They put the negative sign back up as the exponent of the log, and since a number to the -1 power is the number/fraction's reciprocal, the concentrations were flipped when "dropping" the negative sign.

The second method you listed above would work as well as long as you leave the negative sign out front, which I think you forgot, since the first order integrated rate law (rearranged) is:
ln([A]t/[A]o)=-kt