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 1

$\huge\bf\mathscr\pink{Your\: Answer}$

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 }

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 ${min}^{-1}$

Now,

half - life,

t50 = 0.693/K

= 0.693 min/ 0.02877

= 24.088 min

Answer 2

The rate of a 1st order of reaction $(A _{1})$ = 0.04 mol/L/s

At time $(t_{1})$ = 10 min

The rate of a 2nd order of reaction $(A _{2})$ = 0.03 mol/L/s

At time $(t_{2})$ = 20 min

Now..

K = $\dfrac{2.303}{t_{2}\:-\:t_{1}}$ .log $\dfrac{(A_{1})}{(A_{2})}$

So..

t = $t_{2}$ - $t_{1}$

t = 20 - 10

t = 10 min

k = $\dfrac{2.303}{10}$ .log $\dfrac{(0.04)}{(0.03)}$

k = $\dfrac{2.303}{10}$ × 0.124

k = $\dfrac{0.028}{min}$

k = 0.028 min-¹

For the 1st order reaction.. half life $t_{ \frac{1}{2} }$ is given as;

$t_{ \frac{1}{2} }$ = $\dfrac{ 0.693}{k}$

$t_{ \frac{1}{2} }$ = $\dfrac{0.693}{0.02876}$ (From above)

$t_{ \frac{1}{2} }$ = 24.096 min