Question

If the half life period of a first order reaction in a is 2 minutes, how long will it take a to reach 25% of its initial concentration

Answer 1

Answer:

=> 0.83 min = 49.81 sec

Explanation:

=>Half life of 1st order reaction (t½)

= 0.693/K = 2 min

where k is the rate constant.

=> K = 0.693/2 = 0.3465

By Rate law

=> K = 2.303 × log [a/a-x] /t

=> (t¼)= 2.303 × log[a/(a-0.25a)]/0.3465

=> (t¼) = 2.303×log(4/3) /0.3465

=> (t¼) = 0.83 min = 49.81 sec

--------------------------

Hope this is your required answer

Proud to help you.

Answer 2

Answer: 6.65 min

Explanation:

Calculation of rate constant(k)

k = 0.693/t = 0.693/2 min = 0.3465 min^−1

Calculation of time taken to reach 25 percent of initial concentration

[A]o = 100%, [A] = 25%, k = 0.3465 min^−1.

t = (2.303/k)×log([A]o)/[A] = (2.303/0.3465)×0.6021 min = 4.0min

Calculation of time taken to reach 10 percent of initial concentration.

[A]o = 100%, [A] = 10%, k = 0.3465 min^−1

t = (2.303/k)×log([A]0/[A]

t = (2.303/0.3465)×log(100/10) = 2.303/0.3465min^−1 = 6.65 min