Question

In a first order reaction 60% of the reactant decomposes in 45 minutes. Calculate the half

life of the reaction.



please please give me the answer it's urgent please help me​

Answer 1

Answer:

In a first order reaction 60% of the reactant decomposes in 45 minutes. Calculate

the half-life for the reaction

Answer 2

Given: In a first order reaction 60% of the reactant decomposes in 45 minutes

To find: Half life of the reaction

Explanation: Let the amount at the starting of the reaction be r.

60% of the reactant decomposes. Therefore, percentage left after reaction= 40%

Amount left= 40r/100

= 2r/5

= 0.4 r

Time= 45 minutes

= 45*60 seconds

Let rate constant be k.

Formula for first order reaction:

$t = \frac{2.3}{k} log( \frac{amount \: at \: start}{amount \: left} )$

$t = \frac{2.3}{k} log( \frac{r}{0.4r} )$

$k = \frac{2.3}{45 \times 60} log(2.5)$

$k = \frac{2.3 \times 0.39}{45 \times 60}$

k = 0.00033 s^-1

Half life of reaction= 0.693/k

= 0.693/0.00033

= 2100 seconds

= 2100/60 minutes

= 35 minutes

Therefore, the half life of the reaction is 35 minutes.