To solve this problem effectively, we will use concepts from first-order kinetics. In a first-order reaction, the concentration of the reactant decreases exponentially over time according to the integrated rate law for first-order kinetics.
### Step-by-Step Solution:
1. Identify the Information Given:
- Initial concentration of [tex]\( A \)[/tex], [tex]\( [A]_{\text{initial}} = 50 \, \text{mol/L} \)[/tex]
- Final concentration of [tex]\( A \)[/tex], [tex]\( [A]_{\text{final}} = 0.05 \, \text{mol/L} \)[/tex]
- Rate constant, [tex]\( k = 6 \, \text{min}^{-1} \)[/tex]
2. Use the Integrated Rate Law for First-Order Reactions:
The integrated rate law for a first-order reaction is given by:
[tex]\[
\ln \left( \frac{[A]_{\text{initial}}}{[A]_{\text{final}}} \right) = k \cdot t
\][/tex]
Where:
- [tex]\( \ln \)[/tex] = natural logarithm (base [tex]\( e \)[/tex])
- [tex]\( [A]_{\text{initial}} \)[/tex] = initial concentration of [tex]\( A \)[/tex]
- [tex]\( [A]_{\text{final}} \)[/tex] = final concentration of [tex]\( A \)[/tex]
- [tex]\( k \)[/tex] = rate constant
- [tex]\( t \)[/tex] = time
3. Rewrite the Equation to Solve for Time ( [tex]\( t \)[/tex] ):
[tex]\[
t = \frac{\ln \left( \frac{[A]_{\text{initial}}}{[A]_{\text{final}}} \right)}{k}
\][/tex]
4. Substitute the Given Values into the Equation:
[tex]\[
t = \frac{\ln \left( \frac{50}{0.05} \right)}{6}
\][/tex]
5. Calculate the Natural Logarithm and Time:
[tex]\[
t = \frac{\ln (1000)}{6}
\][/tex]
At this step, note that [tex]\( \ln (1000) \approx 6.907755 \)[/tex].
6. Divide to Find the Time:
[tex]\[
t = \frac{6.907755}{6} \approx 1.1512925 \, \text{minutes}
\][/tex]
Therefore, the time required for the concentration of [tex]\( A \)[/tex] to reach 0.05 mol/L, given a rate constant of 6 min[tex]\(^{-1}\)[/tex], is approximately [tex]\( 1.151 \)[/tex] minutes.
### Summary
So, based on the given rate constant and initial and final concentrations, the time calculated for the concentration of reactant [tex]\( A \)[/tex] to decrease to 0.05 mol/L is approximately [tex]\( 1.15 \)[/tex] minutes.
