Question

A first order reaction takes 10 min for 25% decomposition. calculate t1/2 for the reaction

Answer 1

Given:

Time = 10 min

Decomposition = 25 %

To find:

Half life period

Solution:

For a first order reaction,

k = 2.303 / time log ( R ( 0 )  / R )

Where,

R ( 0 ) - Initial concentration

R - Final concentration

As it takes 10 min,

k = 2.303 / 10 log ( 100 / 100 - 25 )

2.303 / 10 log (  4 / 3 )

2.303 / 10 ( log 4 - log 3 )

Substituting the log values,

We get,

2.303 / 10 ( 0.6021 - 0.4771 )

Hence,

k = 2.88 * 10^-2 min^-1

Half life period = 0.693 / k

Substituting,

0.693 / 2.88 * 10^-2

Half life period ≅ 24 min

Answer 2

Answer:

Step-by-step explanation:

Solution:

For a first order reaction,

k = 2.303 / time log ( R ( 0 )  / R )

Where,

R ( 0 ) - Initial concentration

R - Final concentration

As it takes 10 min,

k = 2.303 / 10 log ( 100 / 100 - 25 )

2.303 / 10 log (  4 / 3 )

2.303 / 10 ( log 4 - log 3 )

Substituting the log values,

We get,

2.303 / 10 ( 0.6021 - 0.4771 )

Hence,

k = 2.88 * 10^-2 min^-1

Half life period = 0.693 / k

Substituting,

0.693 / 2.88 * 10^-2

Half life period ≅ 24 min

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