Question

For a reaction of first order it takes 20 minutes for the concentration to drop from 0.1m to 0.6m the time required for the concentration to drop from 0.6 m to 0.36m will be?

Answer 1

Answer: 20 min

Explanation:

a=1 M

a-x=0.6

k=2.303/t log a/a-x

k=0.0256

a=0.6

a-x=0.36

T=2.303/k Loga/a-x

T=20 min

Answer 2

Answer:  Time required for the concentration to drop from 0.6 m to 0.36m will be 20 minutes.

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant  

t = time taken for decomposition = 20 min

a = initial amount of the reactant  = 1

a - x = amount left after decay process = 0.6

Putting in the values we get:

$20=\frac{2.303}{k}\log\frac{1}{0.6}$

$k=\frac{2.303}{20}\log\frac{1}{0.6}$

$k=0.025min^{-1}$

The time required for the concentration to drop from 0.6 m to 0.36 m will be:

$t=\frac{2.303}{0.025}\log\frac{0.6}{0.36}$

$t=20.4min$

Thus time required for the concentration to drop from 0.6 m to 0.36 m will be 20 minutes.