Step 1: Understanding the Concept:
Chemical kinetics is the study of reaction rates and the molecular pathways by which reactions occur.
A first-order reaction is defined as a process where the reaction rate is directly proportional to the concentration of a single reactant.
The mathematical hallmark of such reactions is that the time required for a certain percentage of completion is independent of the starting concentration.
In this problem, we are looking for the time (\(t\)) required for the concentration to drop to 10% (one-tenth) of its initial magnitude.
Understanding the logarithmic nature of the integrated rate law is essential to solving such problems efficiently in competitive exams.
Step 2: Key Formula or Approach:
For a first-order chemical reaction, the integrated rate equation expressing the relationship between time and concentration is:
\[ k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]} \]
Rearranging this to solve for time (\(t\)):
\[ t = \frac{2.303}{k} \log_{10} \frac{[A]_0}{[A]} \]
Where:
\( [A]_0 \) = Initial concentration at time \( t=0 \).
\( [A] \) = Final concentration remaining at time \( t \).
\( k \) = Rate constant (given as \( 2.303 \times 10^{-3} \text{ s}^{-1} \)).
Step 3: Detailed Explanation:
First, we define the relationship between the initial and final concentrations as specified in the prompt.
The problem states that the concentration reduces to \(\frac{1}{10}\)th of the original.
Mathematically, this means:
\[ [A] = \frac{1}{10} [A]_0 \]
This can be rewritten to isolate the ratio required for the logarithmic term:
\[ \frac{[A]_0}{[A]} = 10 \]
Now, we substitute the provided rate constant \( k = 2.303 \times 10^{-3} \text{ s}^{-1} \) and our concentration ratio into the rearranged equation:
\[ t = \frac{2.303}{2.303 \times 10^{-3} \text{ s}^{-1}} \log_{10}(10) \]
We simplify the fraction first:
The numerical value 2.303 appears in both the numerator and denominator, effectively cancelling each other out.
\[ t = \frac{1}{10^{-3} \text{ s}^{-1}} \times \log_{10}(10) \]
From basic logarithmic identities, we know that \(\log_{10}(10) = 1\).
The expression now becomes:
\[ t = \frac{1}{10^{-3}} \times 1 \]
\[ t = 10^3 \text{ s} \]
Performing the exponentiation:
\[ t = 1000 \text{ s} \]
This result signifies that given the specified rate constant, the reactant will take exactly 1000 seconds to reach 10% of its initial amount.
It is important to note that the units of the rate constant (s\(^{-1}\)) correctly yield the units for time in seconds (s).
Step 4: Final Answer:
The exact time required for the reduction is determined to be 1000 s.
Comparing this to the given options, we find it matches (C).