Answer :
Complete Question
The complete question is shown on the first uploaded image
Answer:
The concentration of [tex]BC_2[/tex] that should used originally is [tex]C_Z_o = 0.4492M[/tex]
Explanation:
From the question we are told that
The necessary elementary step is
[tex]2BC_2 ----->4C + B_2[/tex]
The time taken for sixth of 0.5 M of reactant to react [tex]t = 9 hr[/tex]
The time available is [tex]t_a = 3.5 hr[/tex]
The desired concentration to remain [tex]C = 0.42M[/tex]
Let Z be the reactant , Y be the first product and X the second product
Generally the elementary rate law is mathematically as
[tex]-r_Z = kC_Z^2 = - \frac{d C_Z}{dt}[/tex]
Where k is the rate constant , [tex]C_Z[/tex] 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
[tex]\frac{1}{C_Z} - \frac{1}{C_Z_o} = kt[/tex]
Where [tex]C_Z_o[/tex] is the initial concentration of Z which a value of [tex]C_Z_o = 0.5M[/tex]
From the question we are told that it take 9 hours for the concentration of the reactant to become
[tex]C_Z = C_Z_o - \frac{1}{6} C_Z_o[/tex]
[tex]C_Z = 0.5 - \frac{0.5}{6}[/tex]
[tex]= 0.4167 M[/tex]
So
[tex]\frac{1}{0.4167} - \frac{1}{0.50} = 9 k[/tex]
[tex]0.400 = 9 k[/tex]
=> [tex]k = 0.044\ L/ mol \cdot hr^{-1}[/tex]
For [tex]C_Z = 0.42M[/tex]
[tex]\frac{1}{0.42} - \frac{1}{C_Z_o} = 3.5 * 0.044[/tex]
[tex]2.38 - 0.154 = \frac{1}{C_Z_o}[/tex]
[tex]2.226 = \frac{1}{C_Z_o}[/tex]
[tex]C_Z_o = \frac{1}{2.226}[/tex]
[tex]C_Z_o = 0.4492M[/tex]
