Answer:
It will take 54 seconds for the concentration of [tex]ClCH_2CH_2Cl[/tex] to decrease to 10.0% of its initial value.
Explanation:
Initial concentration of [tex]ClCH_2CH_2Cl[/tex] = [tex][A_o]=1.41 M[/tex]
Final concentration of [tex]ClCH_2CH_2Cl[/tex] after t time = [tex][A]=10\%of [A_o]=0.1[A_o][/tex]
t = ?
Rate constant of the reaction = k [tex]=0.118 M^{-1}s^{-1}[/tex]
Integrated rate law for second order kinetics is given by:
[tex]\frac{1}{[A]}=kt+\frac{1}{[A_o]}[/tex]
[tex]\frac{1}{0.1\times 1.41 M}=0.118 M^{-1}s^{-1}\times t+\frac{1}{1.41 M}[/tex]
Solving for t :
t = 54 seconds
It will take 54 seconds for the concentration of [tex]ClCH_2CH_2Cl[/tex] to decrease to 10.0% of its initial value.
