Question

The reaction represented by \( A \rightarrow B \) follows first-order kinetics. At a given temperature, 20% of the reaction is completed in 223 s. The time taken to complete 50% of the reaction at the same temperature is _________ s (rounded off to the nearest integer).

Hint:

For first-order reactions, the time to complete a given percentage of the reaction can be calculated using the integrated rate law. The time for 50% completion (half-life) is independent of the initial concentration, which makes it a useful parameter for kinetics studies.

Answer 1

For a first-order reaction, the time required to complete a given percentage of the reaction is given by the equation:

\[ \ln \left( \frac{C_0}{C} \right) = k t \] where:
- \( C_0 \) is the initial concentration,
- \( C \) is the concentration at time \( t \),
- \( k \) is the rate constant,
- \( t \) is the time.

We are given that 20% of the reaction is completed in 223 seconds. To find the rate constant \( k \), we use the following equation for first-order kinetics:

\[ \ln \left( \frac{1}{1 - f} \right) = k t \] For \( f = 0.20 \) (since 20% of the reaction is completed), we have:

\[ \ln \left( \frac{1}{1 - 0.20} \right) = k \times 223 \] \[ \ln \left( \frac{1}{0.80} \right) = k \times 223 \] \[ \ln (1.25) = k \times 223 \] \[ 0.2231 = k \times 223 \] Solving for \( k \):

\[ k = \frac{0.2231}{223} \approx 9.99 \times 10^{-4} \, \text{s}^{-1} \] Now, to find the time to complete 50% of the reaction, we use the equation:

\[ \ln \left( \frac{1}{1 - 0.50} \right) = k \times t_{50\%} \] \[ \ln (2) = k \times t_{50\%} \] \[ 0.6931 = 9.99 \times 10^{-4} \times t_{50\%} \] Solving for \( t_{50\%} \):

\[ t_{50\%} = \frac{0.6931}{9.99 \times 10^{-4}} \approx 693 \, \text{s} \] Thus, the time to complete 50% of the reaction is between 685 and 705 seconds.

Answer: 685 to 705 s.