Question

The rate of a first order reaction is \( 0.04 \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1} \) at 10 minutes and \( 0.03 \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1} \) at 20 minutes after initiation. Half-life of the reaction is \( \_\_\_\_ \) minutes. (Given \( \log 2 = 0.3010, \, \log 3 = 0.4771 \))

Answer 1

Correct Answer: 24

Given that the rate of a first-order reaction is decreasing over time, we can use the integrated rate law for first-order kinetics:

Rate = \( k[A]_0 e^{-kt} \)

At time \( t = 10 \) min:

0.04 = \( k[A]_0 e^{-k \times 600} \)

At time \( t = 20 \) min:

0.03 = \( k[A]_0 e^{-k \times 1200} \)

Dividing equation (2) by equation (1):

\( \frac{0.03}{0.04} = e^{-k \times (1200 - 600)} \)

\( \frac{4}{3} = e^{-600k} \)

Taking natural log:

\( \ln \left( \frac{4}{3} \right) = 600k \)

Now solve for \( t_{\frac{1}{2}} \) using:

\( t_{\frac{1}{2}} = \frac{\ln 2}{k} \)

\( t_{\frac{1}{2}} = \frac{600 \ln 2}{\ln \left( \frac{4}{3} \right)} \) sec.

Now using the given values:

\( t_{\frac{1}{2}} = 600 \times \frac{\log 2}{\log 4 - \log 3} = 10 \times \frac{0.3010}{0.6020 - 0.4771} \) min

\( t_{\frac{1}{2}} = 24.08 \) min

\( t_{\frac{1}{2}} = 24 \)