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Nastasia [14]
3 years ago
13

At a certain temperature this reaction follows second-order kinetics with a rate constant of 0.118 M^-1* s^-1:

Chemistry
1 answer:
Anestetic [448]3 years ago
7 0

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.

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