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
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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
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