At a certain temperature this reaction follows first-order kinetics with a rate constant of 2.01 s^−1 CICH2CH2Cl (g) → CH2CHCI (g) + HCl(g)
Suppose a vessel contains CICH2CH2Cl at a concentration of 1.34 M. Calculate how long it takes for the concentration CICH2CH2Cl to decrease to 20.0% of its initial value. You may assume no other reaction is important. Round your answer to 2 significant digits.

0.80 seconds (2 significant figures)

Explanation:

The equation of the reaction is given as;

CICH2CH2Cl (g) --> CH2CHCI (g) + HCl(g)

Rate constant (k) = 2.01 s^-1

From the units of the rate constant, this is a first order reaction.

Initial Concentration = 1.34 M

t = ?

Final concentration = 20% of 1.34 = 0.268 M

The integrated rate law for a first order reaction is given as;

ln[A] = ln[A]o - kt

ln(0.268) = ln(1.34) - 2.01(t)

-2.01(t) = - 1.6094

t = 0.8007 ≈ 0.80 seconds (2 significant figures)

it is a solid

explanation: