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

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

Answer 2

The problem asks for the half-life of a first-order reaction, given the reaction rates at two different times.

Concept Used:

For a first-order reaction, the rate of reaction is directly proportional to the concentration of the reactant at that instant. Let the reaction be \( A \rightarrow \text{Products} \).

The rate law is given by:

\[ \text{Rate} = k[A] \]

where \(k\) is the rate constant and \([A]\) is the concentration of the reactant.

The integrated rate law for a first-order reaction is:

\[ k = \frac{2.303}{t_2 - t_1} \log_{10}\left(\frac{[A_1]}{[A_2]}\right) \]

where \([A_1]\) and \([A_2]\) are the concentrations of the reactant at times \(t_1\) and \(t_2\), respectively.

Since the rate is directly proportional to the concentration (\(\text{Rate} \propto [A]\)), we can substitute the rates in place of the concentrations in the integrated rate law:

\[ k = \frac{2.303}{t_2 - t_1} \log_{10}\left(\frac{\text{Rate}_1}{\text{Rate}_2}\right) \]

The half-life (\(t_{1/2}\)) of a first-order reaction is related to the rate constant \(k\) by the formula:

\[ t_{1/2} = \frac{0.693}{k} = \frac{2.303 \log_{10}(2)}{k} \]

Step-by-Step Solution:

Step 1: List the given information.

• Time 1, \(t_1 = 10\) minutes.
• Rate at \(t_1\), \(\text{Rate}_1 = 0.04 \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1}\).
• Time 2, \(t_2 = 20\) minutes.
• Rate at \(t_2\), \(\text{Rate}_2 = 0.03 \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1}\).

Step 2: Calculate the rate constant (\(k\)) of the reaction.

We use the integrated rate law expressed in terms of rates. The time interval is \(t_2 - t_1 = 20 - 10 = 10\) minutes.

\[ k = \frac{2.303}{10} \log_{10}\left(\frac{0.04}{0.03}\right) \] \[ k = \frac{2.303}{10} \log_{10}\left(\frac{4}{3}\right) \]

Using the logarithm property \(\log(\frac{a}{b}) = \log(a) - \log(b)\):

\[ k = \frac{2.303}{10} (\log_{10}(4) - \log_{10}(3)) \]

Using the property \(\log(a^n) = n\log(a)\):

\[ k = \frac{2.303}{10} (2\log_{10}(2) - \log_{10}(3)) \]

Substitute the given values for \(\log(2)\) and \(\log(3)\):

\[ k = \frac{2.303}{10} (2 \times 0.3010 - 0.4771) \] \[ k = \frac{2.303}{10} (0.6020 - 0.4771) \] \[ k = \frac{2.303}{10} (0.1249) \] \[ k \approx 0.02877 \, \text{min}^{-1} \]

The unit of the rate constant is min^-1 because the time interval used was in minutes.

Step 3: Calculate the half-life (\(t_{1/2}\)) of the reaction.

Using the formula for the half-life of a first-order reaction:

\[ t_{1/2} = \frac{0.693}{k} \]

Alternatively, to avoid intermediate calculations and potential rounding errors, we can use the expression involving logarithms:

\[ t_{1/2} = \frac{2.303 \log_{10}(2)}{k} = \frac{2.303 \log_{10}(2)}{\frac{2.303}{10} (\log_{10}(4/3))} \]

The \(2.303\) terms cancel out:

\[ t_{1/2} = \frac{10 \log_{10}(2)}{\log_{10}(4/3)} \]

Final Computation & Result:

Substitute the log values into the simplified expression:

\[ t_{1/2} = \frac{10 \times 0.3010}{2\log_{10}(2) - \log_{10}(3)} \] \[ t_{1/2} = \frac{3.010}{2 \times 0.3010 - 0.4771} \] \[ t_{1/2} = \frac{3.010}{0.6020 - 0.4771} \] \[ t_{1/2} = \frac{3.010}{0.1249} \] \[ t_{1/2} \approx 24.099 \, \text{minutes} \]

Rounding to the nearest significant figures, the half-life is approximately 24.1 minutes.

The half-life of the reaction is 24.1 minutes.