Question

A first order reaction is 50% completed after 30 minutes.This implies that the time required to complete 90% of the reaction is?

Answer 1

`"100 min"`

Explanation:

The integrated rate law for a first-order reaction looks like this

`color(blue)(ln( A/A_0) = - k * t)" "`, where#

`A` - the concentration at a given time `t`
`A_0` - the initial concentration
`k` - the rate constant, usually expressed in `"s"^(-1)` for first-order reactions

Now, I'll assume that you're not familiar with the equation that describes the half-life of a first-order reaction.

As you know, the half-life of a chemical reaction tells you how much time is needed for the concentration of a reactant to reach half of its initial value.

Simply put, the time needed for `50%` of a reaction to be completed will represent that reaction's half-life.

In your case, you know that your first-order reaction is `50%` completed after `30` minutes. This means that after `30` minutes, you will have

`A = 1/2 * A_0 ->` half of the initial concentration remains after one half-life

Plug this into the above equation and solve for `k`, the rate constant

`ln( (1/2 * color(red)(cancel(color(black)(A_0))))/(color(red)(cancel(color(black)(A_0)))) ) = - k * "30 min"`

This is equivalent to

`- k* "30 min" = ln(1/2)`

`-k * "30 min" = overbrace(ln(1))^(color(blue)(=0)) - ln(2)`

`k = ln(2)/"30 min" = 2.31 * 10^(-2)"min"^(-1)`

Now that you know the rate constant for this reaction, use the same equation again, only this time use

`A = 1/10 * A_0 ->` equivalent to `90%` completion

and solve for `t`, the time needed to get here

`ln( (1/10 * color(red)(cancel(color(black)(A_0))))/(color(red)(cancel(color(black)(A_0)))) ) = - 2.31 * 10^(-2)"min"^(-1) * t`

This is equivalent to

`t = (color(red)(cancel(color(black)(-))) ln(10))/(color(red)(cancel(color(black)(-))) 2.31 * 10^(-2)"min"^(-1)) = "99.68 min"`

Rounded to one sig fig, the answer will be

`t = color(green)("100 min")`