Question

Half life period of a first order reaction is 600s what percent of A remains after 30 minutes

Answer 1

Answer:

The half life of a first order reaction is given by the equation,

t(1/2) = 0.693/k

Where “k” is the rate constant.

From the equation, we can calculate the rate constant.

60 = 0.693/k

k = 0.693/60 = 0.01155 min^(-1)

Now, use the equation

k= (2.303/t)log(initial conc./final conc.)

After consuming 90% of the reactant, the final concentration will be 10% of the initial concentration.

Assume, initial concentration is x

Final concentration is 10% of x = (10/100)x = 0.1x

Therefore

Initial conc./final conc. = x/0.1x = 10

Substitute this in the above formula

0.01155 = (2.303/t)log(10)

0.01155 = 2.303/t

t = 199.4 minutes.

Therefore it takes 199.4 minutes to consume 90% of the reactant.