Question

In a first order reaction. 10% of the reactant is consumed in 25 minutes. Calculate:
(a) The half-life period of the reaction.
(b) The time required for completing 87.5% of the reaction.

Answer 1

Answer:

let initial reactantant is 100 %

reactant consumed = 10% of 100% = 10%

t= 25 min

t= 2.303/k log R1/R2

25 = 2.303/k log 100/10

k= 2.303/25

(i) half life for first irder reaction = 0.693/k . Put the value of k and solve it further.

(ii)  t = 2.303/ k log 100/ 12.5

Answer 2

Answer:

t=2.303/k*log(100/100-10)

k=2.303/25*log(100/90)

k=4.215X10^-3

t1/2=0.693/k

t1/2=0.693/4.215X10^-3

=164.40min

second part:

x=87.5

100-x=100-87.5=12.5

t=2.303/k*log(100/100-87.5)

=2.303/4.215X10^-3*log(12.5)

=8.2hrs

Explanation:

i calculated this.

hopefully correct.

still i want someone to confirm my answer