Question

show that in a first order reaction time required for completion of 99.9% is 10 times of t1/2 of the reaction . when reaction is completed 99.9% [R]n = [R]0 -0.999[R]0​

Answer 1

Answer:

1/2

=

K

2.303

log2 ......(1)

t

99.9%

=

K

2.303

log

0.1

100

=

K

2.303

log10

3

=

K

2.303

3log10 ......(2)

From equation (1) & (2)

t

99.9%

=10×t

1/2