Question

At a certain temperature this reaction follows second-order kinetics with a rate constant of 0.0569·M⁻¹s⁻¹:

ClCH₂CH₂Cl(g) → CH₂CHCl(g) + HCl(g)

Suppose a vessel contains ClCH₂CH₂Cl at a concentration of 0.730M. Calculate the concentration of ClCH₂CH₂Cl in the vessel 320. seconds later. You may assume no other reaction is important.

Round your answer to 2 significant digits.

_ M

Answer 1

Answer:

Given,

rate =(0.0569  s ^-1 ) [ClCH₂CH₂Cl ]

K   = 0.0569s^-1

[A0]   = 0.730M

t     = 320 sec

let us consider,

K   = 2.303/t log[A0]/[A]

Substitute values in the expression

0.0569 = 2.303/320 log0.730/[A]

log0.730/[A]   = 0.0569*0.320/2.303

log0.730/[A]   =0.00791

0.730/[A]         = 10^0.00791

0.730/[A]         = 1.01837

[A]                 = 0.730/1.01837

[A]              = 0.72 M