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

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

$t_{\frac{1}{2}} = 24$