Question

For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

Show that for a first order reaction, time required for 99% completion is twice the time required for 90% completion of the reaction.

Answer 1

99% completion means that,

x = 99% of [R]₀

or, [R] = [R]₀ − 0.99 [R]₀ = 0.01[R]₀

For a first-order reaction,

t = `2.303/k log  [R]_0/[[R]]`

∴ t_99% = `2.303/k log  [R]_0/(0.01[R]_0)`

= `2.303/k log 10^2`

= `2 xx 2.303/k`

90% completion means that,

[R] = [R]₀ − 0.99[R]₀

= 0.1[R]₀

∴ t_90% = `2.303/k log  [R]_0/(0.1[R]_0)`

= `2.303/k log 10`

= `2.303/k`

∴ `t_(99%)/t_(90%) = ((2 xx 2.303/k)/((2.303/k)))`

= 2

or, t_99% = 2 × t_90%

Answer 2

If the initial concentration is [A]₀, for 99% completion,

[A] = `[A]_0 - [A]_0 xx 99/100`

= `[A]_0 xx 1/100`

The time required for 99% completion is given by:

t_99% = `2.303/k log_10  ([A]_0)/([A]_0 xx 1/100)`

= `2.303/k log_10  100/1`    ...(i)

For 90% completion,

[A] = `[A]_0 - [A]_0 xx 90/100`

= `[A]_0 xx 10/100`

Time required for 90% completion is given by:

t_90% = `2.303/k log_10  ([A]_0)/([A]_0 xx 10/100)`

= `2.303/k log_10  100/10`    ...(ii)

Dividing eq. (i) by eq. (ii), we get,

`t_(99%)/t_(90%) = (log_10  100/1)/(log_10  100/10)`

`t_(99%)/t_(90%)` = 2

∴ t_99% = 2 × t_90%

Answer 3

For a first-order reaction,

t = `2.303/k log_10  [A]_0/([A])`

For 99% completion,

[A]₀ = 100

[A] = 100 − 99 = 1

∴ t₉₉ = `2.303/k log_10  100/1`

∴ t₉₉ = = `(2 xx 2.303)/k`    ...(i)

For 90% completion,

[A]₀ = 100,

[A] = 100 − 90 = 10

∴ t₉₀ = `2.303/k log_10  100/10`

∴ t₉₀ = = `(1 xx 2.303)/k`    ...(ii)

From equations (i) and (ii),

t₉₉ = 2 × t₉₀.

Thus, the time required for 99% completion for a first-order reaction is twice the time required for 90% completion.