Question

The rate constant for the first order reaction is \(1.15 \times 10^{-3} \, \text{s}^{-1}\). How long will 5 g of the reactant take to reduce to 3 g?

• A. 424 sec
• B. 414 sec
• C. 434 sec
• D. 444 sec

Hint:

For a first-order reaction, use the equation \( \ln\left(\frac{[A]_0}{[A]}\right) = kt \) to calculate the time required for a specific concentration.

Answer 1

The Correct Option is D

Step 1: Understanding the first-order rate equation.
For a first-order reaction, the relation is given by: \[ \ln\left(\frac{[A]_0}{[A]}\right) = kt \] Where: - \([A]_0\) is the initial concentration of the reactant - \([A]\) is the concentration at time \(t\) - \(k\) is the rate constant - \(t\) is the time
Step 2: Substituting the values.
We know that \( [A]_0 = 5 \, \text{g} \), \( [A] = 3 \, \text{g} \), and \( k = 1.15 \times 10^{-3} \, \text{s}^{-1} \). Substituting these values into the equation, we get: \[ \ln\left(\frac{5}{3}\right) = (1.15 \times 10^{-3})t \] Solving for \(t\), we get \( t = 444 \, \text{sec} \).
Step 3: Conclusion.
The correct answer is (D) 444 sec.