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 ?

2 Answers
Mar 8, 2018

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#

Mar 9, 2018

Answer:

#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.

edu.hioa.no

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"#.