Question

The rate constant for a first order reaction is $20 \min ^{-1}$ The time required for the initial concentration of the reactant to reduce to its $\frac{1}{32}$ level is ___$\times 10^{-2} \min$ (Nearest integer) 
(Given: $\ln 10=2.303$ $\log 2=0.3010$ )

Hint:

For a first-order reaction, the time required to reach a certain concentration can be calculated using the formula \( t = \frac{\ln \left( \frac{C_0}{C} \right)}{k} \).

Answer 1

Correct Answer: 17

$C=\frac{C_o}{2^n}=\frac{C_o}{32}$
$n=5$
$t=5t_{1/2}$
$=\frac{5\times 0.693}{20}=\frac{0.693}{4}$
$=0.17325\min =17.325\times 10^{-2}\min$
So, the correct answer is 17.

Answer 2

The integrated rate equation for a first order reaction is:

\[ \ln \left( \frac{C_0}{C} \right) = kt \]

Where \( C_0 \) is the initial concentration, \( C \) is the concentration at time \( t \), \( k \) is the rate constant, and \( t \) is the time.

For the reaction to reduce to \( \frac{1}{32} \) of its original concentration, we have:

\[ \frac{C}{C_0} = \frac{1}{32} \]

Taking the natural logarithm:

\[ \ln \left( \frac{C_0}{C} \right) = \ln 32 = 5 \ln 2 = 5 \times 0.693 = 3.465 \]

Thus, the time \( t \) is:

\[ t = \frac{3.465}{k} = \frac{3.465}{20} = 0.17325 \, \text{min} \]

Therefore, the time required for the concentration to reduce to \( \frac{1}{32} \) is \( 17.325 \times 10^{-2} \) min.