Question

in second order reaction 20% of the a substance is dissociated in 40 minutes. The time taken by the 80% of its dissociation is

Answer 1

Answer:

The time taken by the 80% of its dissociation is 640 min.

Explanation:

20 % of dissociation means 805 is left undissociated that [A]

$[A]=0.8 [A_o]$

The second order rate equation:

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

$k\times 40 min=\frac{1}{0.8[A_o]}-\frac{1}{[A_o]}$

$k=\frac{1}{[A_o]\times 160}$

Time taken to 80% dissociation that is 20 % of the amount will remain undissociated.

$[A]=0.2 [A_o]$

Let the time taken be t'

$\frac{1}{[A_o]\times 160}\times t'=\frac{1}{0.2[A_o]}-\frac{1}{[A_o]}$

t' = 640 minutes

The time taken by the 80% of its dissociation is 640 min.

Answer 2

REFER THE ATTACHMENT ...........

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