Question

A first-order reaction has a specific rate constant (\( k \)) equal to \( 2.303 \times 10^{-3} \text{ s}^{-1} \). Calculate the exact time required for the initial concentration of the reactant to be reduced to exactly \( \frac{1}{10} \)th of its original value.

• A. \( 100 \text{ s} \)
• B. \( 2303 \text{ s} \)
• C. \( 1000 \text{ s} \)
• D. \( 693 \text{ s} \)

Hint:

For first-order kinetics calculations, remember these common logarithmic milestones to speed up your calculation: \( \log_{10}(2) \approx 0.3010 \), \( \log_{10}(3) \approx 0.4771 \), and \( \log_{10}(10) = 1 \).

Answer 1

The Correct Option is C

Concept: For a first-order chemical reaction, the integrated rate equation expressing the relationship between the time elapsed (\( t \)), the rate constant (\( k \)), and concentration parameters is given by: \[ t = \frac{2.303}{k} \log_{10}\left( \frac{[A]_0}{[A]} \right) \] Where \( [A]_0 \) represents the initial reactant concentration and \( [A] \) represents the final remaining concentration at time \( t \).

Step 1: Identify the concentration values and parameters from the problem structure.
The problem states that the final concentration drops to one-tenth of its initial value: \[ [A] = \frac{1}{10}[A]_0 \quad \Rightarrow \quad \frac{[A]_0}{[A]} = 10 \] The rate constant parameter value is provided as: \[ k = 2.303 \times 10^{-3} \text{ s}^{-1} \]

Step 2: Substitute values into the first-order kinetic formula.
Plug the isolated terms directly into our integrated rate equation: \[ t = \frac{2.303}{2.303 \times 10^{-3}} \log_{10}(10) \]

Step 3: Simplify the expression to find the final time.
Cancel out the common scalar value \( 2.303 \) from the fraction and use the logarithmic identity \( \log_{10}(10) = 1 \): \[ t = \frac{1}{10^{-3}} \cdot (1) = 10^3 = 1000 \text{ s} \]