Question

The decomposition of sulfonyl chloride, SOCl2 is a first order reaction with a half life of 8.75 hours. Calculate the concentration of SOCl2 after 17.0 hours have passed in an experiment where the initial concentration of sulfonyl chloride was .0837M.

Answer 1

Answer:

      $[SOCl_2]=0.0218M$

Explanation:

The equations for a first order reaction are:

      $\dfrac{d[A]}{dt}=-k[A]$

      $[A]=[A_0]e^{-kt}$

      $t\frac{1}{2}=\dfrac{\ln 2}{k}$

1. Calculate the constant of reaction, k:

Use the equation

                                $t\frac{1}{2}=\dfrac{\ln 2}{k}$

     $8.75h=\dfrac{\ln 2}{k}$

     $k=\dfrac{\ln 2}{8.75h}$

     $k\approx 0.0792168h^{-1}$

2. Calculate the concentration after 17.0 hours

Use the equation

                               $[A]=[A_0]e^{-kt}$

      $[SOCl_2]=0.0837M\cdot e^{-0.0792168h^{-1}\times 17.0h}$

      $[SOCl_2]=0.0218M$