Question

In a first order reaction time required to complete 40%is 10min calculate time required to complete 80%of the reaction.​

Answer 1

Hola

Your answer:

For a first order reaction, $t=\frac{2.303}{k} log[\frac{a}{a-x} ]$

a = 100%

x = 40%

a - x = 40%

Now,  t = $\frac{2.303}{k} log\frac{100}{40}$

t = $\frac{2.303}{k} \:\:log[2.5]}$

log 2.5 = 0.4 (approx)

t = 10 mins

10 x 60 = 600 seconds

Now, 600 =  $\frac{2.303}{k} \times0.4$

k = $\frac{2.303}{600} \times0.4$ = 0.00154 /sec (approx)

Now, time taken when 80% is complete,

t = $\frac{2.303}{0.00154} \:\:log[\frac{100}{100-80}] =\:\frac{2.303}{0.00154}\:\:log[5]$

log 5 = 0.699 (approx)

$\frac{2.303}{0.00154} \times0.699 = 1045\:\:seconds$ = 17.42 minutes

Please mark me brainliest