Chemical Kinetics and Nuclear Chemistry

The expression for first-order reaction is
t =

2.303

k

‌log

[A0]

[A]

When the compound is decomposed to 1/8th of its initial value then the time taken is
t

1

8

= (

2.303

k

)‌log‌

1

( 1 8 )

= (

2.303

k

)‌log‌8 (1)
When the compound is decomposed to 1/10th of its initial value then the time taken is
t

1

10

= (

2.303

k

)‌log‌

1

( 1 10 )

= (

2.303

k

)‌log‌10 (2)
Dividing Eq. (1) by Eq. (2), we get

t 1 8

t 1 10

=

log‌8

log‌10

= log(23) = 3 × 0.3 = 0.9
So, the value of

[t 1 8 ]

[t 1 10 ]

×10 = 9