Question

The rate of a first order reaction is 0.04 mol/L/s at 10 min and 0.03 mol/L/s at 20 min after initiation. Find the half life of the reaction.



answer only if you know​

Answer 1

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24.088 min

step-by-step explanation:

we know that,

for the 1st order reaction,

rate constant of concentration at different time intervals is given by

K = 2.303/(t2-t1) log{ a-x1/a-x2 }

now,

for 1st order reaction,

rate (dx/dt) = K[A]

=> (dx/dt)1 = K(a-x1)

=> 0.04 = K(a-x1)

=> a-x1 = 0.04/K

after 10 min i.e., (t1)

Similarly,

for 20 minutes,

a-x2 = 0.03/K

after 20 min i.e., (t2)

.°. a-x1/a-x2 = 0.04/0.03 = 4/3

.°. K = 2.303/(20-10) log {4/3}

= 2.303 × 0.1249/10

=> K = 0.02877

Now,

half - life,

t50 = 0.693/K

= 0.693 min/ 0.02877

= 24.088 min

Answer 2

24.088 min

step-by-step explanation:

we know that,

for the 1st order reaction,

rate constant of concentration at different time intervals is given by

K = 2.303/(t2-t1) log{ a-x1/a-x2 }

now,

for 1st order reaction,

rate (dx/dt) = K[A]

=> (dx/dt)1 = K(a-x1)

=> 0.04 = K(a-x1)

=> a-x1 = 0.04/K

after 10 min i.e., (t1)

Similarly,

for 20 minutes,

a-x2 = 0.03/K

after 20 min i.e., (t2)

.°. a-x1/a-x2 = 0.04/0.03 = 4/3

.°. K = 2.303/(20-10) log {4/3}

= 2.303 × 0.1249/10

=> K = 0.02877

Now,

half - life,

t50 = 0.693/K

= 0.693 min/ 0.02877

= 24.088 min