Question

A reaction 75% complete in 90 min, find the time in min taken to complete 60% of reaction.
[Given log 2 = 0.3, log2.5 =0.4]​

Answer 1

Answer:

72 mins

Explanation: Let x mins be the time required for 60 prcnt of the reaction.

75 prcnt can also be written as 3/4

and 60 prcnt can be written as 3/5

ok so 3/4 in 90 mins

3/5 in x mins

cross multiply both of them and you get

x * 3/4 = 3/5 * 90

after solving this you would get x= 72 mins

Hope this helps you :)

Answer 2

ANSWER.

The time ( in min) taken to complete 60% of

reaction = 60 minutes.

EXPLANATION.

$\sf \to \: a \: reaction \: 75\% \: complete \: in \: 90 \: min \\ \\ \sf \to \: let \: a_{0} \: = initial \: rate \: \\ \\ \sf \to \: a \: = final \: rate \\ \\ \sf \to \: t \: = time \: taken$

$\sf \to \: 75\% \: reaction \: complete \: in \: 90 \: min \\ \\ \sf \to \: let \: initial \: rate \: of \: reaction \: = a_{0} \: = 100 \\ \\ \sf \to \: let \: final \: rate \: of \: reaction \: = a = 25 \\ \\ \sf \to \: t \: = 90 \: min \\ \\ \sf \to \: from \: first \: order \: reaction \\ \\ \sf \to \: k \: = \frac{2.303}{t} log( \frac{a_{0}}{a} )$

$\sf \to \: k \: = \dfrac{2.303}{90} log( \frac{100}{25} ) \\ \\ \\ \sf \to \: k \: = \frac{2.303}{90} log(4) \\ \\ \sf \to \: k \: = \frac{2.303}{90} 2 log(2) \\ \\ \sf \to \: k \: = \frac{2.303 \times 2 \times 0.3}{90} \\ \\ \sf \to \: k \: = 0.01535$

$\sf \to \: find \: time \: taken \: to \: complete \: 60\% \: reaction \\ \\ \sf \to \: a_{0} \: = 100 \\ \\ \sf \to \: a \: = 40 \\ \\ \sf \to \: to \: find \: t$

$\sf \to \: t \: = \dfrac{2.303}{k} log( \dfrac{ a_{0} }{a} ) \\ \\ \sf \to \: t \: = \frac{2.303}{0.01535} \times log( \frac{100}{40} ) \\ \\ \sf \to \: t \: = \frac{2.303}{0.01535} \times log( \frac{5}{2} ) \\ \\ \sf \to \: t \: = \frac{2.303}{0.01535} \times log(2.5) \\ \\ \sf \to \: t \: = \frac{2.303 \times 0.4}{0.01535} \\ \\ \sf \to \: t \: = 60.1 \: = 60 \: minutes$