An experiment was conducted in which the initial concentration of C2H5Cl was 8.80 x 10-5 M. At the temperature of the experiment, the reaction had a rate constant of 7.41 x 10-3 s-1. How long would it take for the concentration of C2H5Cl to be reduced to one eighth (1/8) of its original concentration?

Question

An experiment was conducted in which the initial concentration of C2H5Cl was 8.80 x 10-5 M.  At the temperature of the experiment, the reaction had a rate constant of 7.41 x 10-3 s-1.  How long would it take for the concentration of C2H5Cl to be reduced to one eighth (1/8) of its original concentration?

aaronq · Answer

What order is the reaction?

anonymous · Answer

@aaronq the reaction is first order sorry I forgot to include it

aaronq · Answer

This is just exponential decay, you can do it in various ways. The easiest is using the integrated rate-law equation:

 $ln[A]_t=-kt+ln[A]_o$

aaronq · Answer

so set $[A]_o=8.80 *10^{-5} M$ and $[A]_t=\dfrac{8.80 *10^{-5} M}{8}$

anonymous · Answer

So, the reaction is a first order one due to the rate constant ("s^-1"). The integrated equation would be $$[A]=[A ]_{0}e ^{-kt}$$
that once we inssert the values:$$[8.8x10^{-5}]=[1.1x10^{-5}]e ^{-(7.41x10^{-3})t}$$

anonymous · Answer

we get t=280.62 s

aaronq · Answer

it works both ways

aaronq · Answer

but the set up above is incorrect,it should be

$\large \dfrac{8.8*10^{-5}}{8}=8.8*10^{-5}*e^{-kt}$

anonymous · Answer

Yes, sorry, introduced it wrong, but the answer is the same it should be 
$$[1.1x10^{-5}]=[8.8x10^{-5}]e ^{-(7.41x10^{-3})t}$$

which gives us t=280.62 s