Question

For a first order reaction if the time for 50%, 75% and 87.5% changes are t1, t2 and t3 respectively then what would be the value of t1: t2: t3 ?

Answer 1

Recall, for a first order elementary reaction,

`ln[A]_"t" = -kt + ln[A]_0`

`=> ln(([A]_"t")/([A]_0))=-kt`

Without more data, I couldn't give you a real value, but I could give you one as a factor of `k` (the rate constant).

We can conceptually assume that `[A]_0 = 1M` to derive our data.

Hence,

`ln(([A]_"t")/([A]_0))=-kt`

`=> t_1 = -ln(0.5)/k approx 0.693/k`

`t_2 = -ln(0.75)/k approx 0.288/k`, and

`t_3 = -ln(0.875)/k approx 0.134/k`

Moreover,

`t_1:t_2:t_3 approx 5.17:2.15:1`

Answer 2

`t_1:t_2:t_3 = 1:2:3`

Explanation:

Consider a first-order reaction

`"A →Products"`

The time required for the initial concentration `["A"]_0` to drop to

• `1/2["A"]_0` is one half-life
• `1/4["A"]_0` is two half-lives
• `1/8["A"]_0` is three half-lives

Also,

`t_½ = ln2/k`. so each half-life takes the same amount of time.

The ratios are

`t_1:t_2:t_3 = "1 half-life:2 half-lives:3 half-lives = 1:2:3"`

Here is a typical plot for a first-order reaction.

The initial concentration is 1.7.

It takes

• 400 s for the concentration to drop by 50 % (to 0.85)
• 800 s for the concentration to drop by 75 % (0.42)
• 400 s for the concentration to drop by 87.5 % (to 0.21)

The times are in the ratio `"1:2:3"`.