Question

In their notebook you see that it takes 9 hours for a sixth of a 0.5M solution of BC2 to react. Unfortunately, you have somewhere you need to be in 3.5 hours and will not be able to monitor the reaction properly. What concentration of BC2 should you originally use in order to have 0.42 M remaining in 3.5 hours

Answer 1

Complete Question

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Answer:

The concentration of $BC_2$ that should used originally is $C_Z_o = 0.4492M$

Explanation:

     From the question we are told that

         The necessary elementary step is  

                  $2BC_2 ----->4C + B_2$

          The  time taken for sixth of 0.5 M of reactant to react $t = 9 hr$

           The time available is $t_a = 3.5 hr$

             The desired concentration to  remain $C = 0.42M$

Let Z be the reactant ,   Y be the first product and X the second product

Generally the elementary rate  law is mathematically as

                    $-r_Z = kC_Z^2 = - \frac{d C_Z}{dt}$

Where k is the rate constant ,  $C_Z$ is the concentration of Z

From the elementary rate law we see that the reaction is second order (This because the concentration of the reactant is raised to power 2 )

 For second  order reaction

            $\frac{1}{C_Z} - \frac{1}{C_Z_o} = kt$

Where $C_Z_o$ is the initial concentration of Z which a value of   $C_Z_o = 0.5M$

       From the question we are told that it take  9 hours  for the concentration of  the reactant to become

                 $C_Z = C_Z_o - \frac{1}{6} C_Z_o$

                  $C_Z = 0.5 - \frac{0.5}{6}$

                       $= 0.4167 M$

So      

                     $\frac{1}{0.4167} - \frac{1}{0.50} = 9 k$

                          $0.400 = 9 k$

                =>    $k = 0.044\ L/ mol \cdot hr^{-1}$

  For   $C_Z = 0.42M$

                $\frac{1}{0.42} - \frac{1}{C_Z_o} = 3.5 * 0.044$

                $2.38 - 0.154 = \frac{1}{C_Z_o}$

                           $2.226 = \frac{1}{C_Z_o}$

                            $C_Z_o = \frac{1}{2.226}$

                             $C_Z_o = 0.4492M$