Question

At a certain temperature this reaction follows second-order kinetics with a rate constant of 0.118 M^-1* s^-1:
ClCH2CH2Cl (g) ----------> CH2CHCI(g) + HCl (g)
a) Suppose a vessel contains ClCH2CH2Cl at a concentration of 1.41 M. Calculate how long it takes for the concentration of CICH2CH2Cl to decrease to 10.0% of its initial value. You may assume no other reaction is important. Round your answer to 2 significant digits.

Answer 1

Answer:

It will take 54 seconds for the concentration of $ClCH_2CH_2Cl$ to decrease to 10.0% of its initial value.

Explanation:

Initial concentration of $ClCH_2CH_2Cl$ = $[A_o]=1.41 M$

Final concentration of $ClCH_2CH_2Cl$ after t time = $[A]=10\%of [A_o]=0.1[A_o]$

t = ?

Rate constant of the reaction = k $=0.118 M^{-1}s^{-1}$

Integrated rate law for second order kinetics is given by:

$\frac{1}{[A]}=kt+\frac{1}{[A_o]}$

$\frac{1}{0.1\times 1.41 M}=0.118 M^{-1}s^{-1}\times t+\frac{1}{1.41 M}$

Solving for t :

t = 54 seconds

It will take 54 seconds for the concentration of $ClCH_2CH_2Cl$ to decrease to 10.0% of its initial value.