Question

If the half life period of the first order reaction is 10 min. The fraction of the reaction completed at 20th min. Is

Answer 1

Answer:

Refer to the attachment..

Answer 2

Answer:

The fraction of the reaction completed at $20th$ min. is $99.66\%$.

Explanation:

Given,

The half-life period of the first-order reaction, $t_{1/2}$ = $10 min$.

The fraction of the reaction completed at the $20th min$ , x =?

Let the initial amount, a = $100\%$

As we know,

• The half-life is the time taken by the initial concentration to reach half of its original value.
• The formula of a half-life has shown below:
• $t_{1/2} = \frac{0.693}{k}$

Here, k = The rate constant

Therefore,

• $k = \frac{0.693}{t_{1/2} }$

After putting the value of $t_{1/2}$ in the equation, we get:

• $k = \frac{0.693}{10 }$ = $0.0693 min^{-1}$

As we know, the rate law expression for the first-order reaction is given by:

• $t = \frac{2.303}{k} log\frac{a}{a-x}$

Now, after putting all the given values in the equation for the period $20 min$, we get:

• $20 = \frac{2.303}{0.0693} log\frac{100}{100-x}$
• $20 = 33.23 log\frac{100}{100-x}$
• $log\frac{100}{100-x} = 0.6$
• $log100 = 60 - 0.6x$
• $2 = 60 - 0.6x$                ( $\because$ $log 100 = 2$ )

• $0.6x = 58$
• $x = 99.66\%$

Hence, the fraction of the reaction completed at $20th min$, x = $99.66\%$.