Question

3)If 40% of first order reaction was completed in 50 minutes, then 50% of the same reaction would be completed in approximately. [Given : log 3 = 0.47, log 5 = 0.7]

79 minute

65 minute

61 minute

74 minute​

Answer 1

Answer:

see the attachment

Explanation:

hope it will help you

Answer 2

Answer:

50% of the reaction would be completed in 65sec.

Option 2 is correct.

Explanation:

As we all know, the rate constant of a first-order reaction is denoted by k.

The current concentration of the first-order reaction is denoted by [A].

The initial concentration of the first-order reaction is denoted by [$A_{0}$].

The time taken by the reaction is denoted by t.

Now,

40% of the first-order reaction was completed in 50 minutes means, [A] is equal to 60% of [$A_{0}$],

Then,

By the equation,

$[A]=[A]_{0} e^{-kt}$

60% of [$A_{0}$]$=[A]_{0} e^{-kt}$

$\frac{60}{100} \times[A]_{0}=[A]_{0} e^{-k\times50}$

$k=\frac{ln\frac{5}{3}}{50}s^{-1}$

Now,

50% of the same first-order reaction would be completed in $t_{0}$ sec, [A] is equal to 50% of [$A_{0}$]

Then,

By the equation,

$[A]=[A]_{0} e^{-kt}$

50% of [$A_{0}$]$=[A]_{0} e^{-kt}$

$\frac{50}{100} \times[A]_{0}=[A]_{0} e^{-\frac{ln\frac{5}{3}}{50}\times t_{0} }$                      ∴ $k=\frac{ln\frac{5}{3}}{50}s^{-1}$

$t_{0}=\frac{log2 \times50}{log\frac{5}{3} }\\t_{0}=\frac{0.3 \times50}{0.23}$

$t_{0}$ ≈ 65minutes

So, 50% of the first-order reaction would be completed in 65min.