Question

In a first order reaction, 20% of the reactant is consumed in 30 minutes. Calculate the following:

1. The half life period of the reaction.
2. The time required for completing 93.75% of the reaction.

Answer 1

Given: In 30 minutes, 20% is consumed, so 80% remains.

∴ `([R])/[R]_0` = 0.80

By using first-order integrated rate law:

k = `2.303/t log ([R]_0/([R]))`

= `2.303/30 log (100/80)`

= `2.303/30 log (100/80)`

= `2.303/30 log (1.25)`

= `(2.303 xx 0.0969)/30`

= `0.2233/30`

= 0.007443 min⁻¹

For a first-order reaction:

t_1/2 = `0.693/k`

= `0.693/0.007443`

= 93.1 minutes

When 93.75% is completed, 6.25% remains:

`[R]_0/([R]) = 100/6.25` = 16

Use the formula:

t = `2.303/k log ([R]_0/([R]))`

= `2.303/0.007443 log (16)`

= `2.303/0.007443 xx log (2^4)`

= `2.303/0.007443 xx 4  log (2)`

= `(2.303 xx 4 xx 0.3010)/0.007443`

= `2.773/0.007443`

= 372.5 minutes