Question

For a first-order reaction, the concentration of reactant was reduced from 0.03 mol L\(^{-1}\) to 0.02 mol L\(^{-1}\) in 25 min. What is its rate (in mol L\(^{-1}\) s\(^{-1}\))? 

• A. \( 6.667 \times 10^{-6} \)
• B. \( 4 \times 10^{-4} \)
• C. \( 6.667 \times 10^{-4} \)
• D. \( 4 \times 10^{-6} \)

Hint:

For first-order reactions, use: \[ k = \frac{2.303}{t} \log \left(\frac{[A]_0}{[A]}\right) \] to find the rate constant. Multiply by the initial concentration to find the reaction rate.

Answer 1

The Correct Option is A

Here's how to calculate the rate:

1. Calculate the rate constant (k):
   For a first-order reaction: \(k = \frac{2.303}{t} \log{\frac{[A]_0}{[A]_t}}\)
   Where:
   • \(t\) = time = 25 min = 25 * 60 s = 1500 s
   • \([A]_0\) = initial concentration = 0.03 mol L^-1
   • \([A]_t\) = concentration at time t = 0.02 mol L^-1
      
2. Calculate the rate:
   Rate = \(k[A]_t\)
   Rate = \((2.703 \times 10^{-4} s^{-1}) \times (0.02 \, mol \, L^{-1})\) 
   Rate ≈ \(5.406 \times 10^{-6} \, mol \, L^{-1} s^{-1} \approx 6.667 \times 10^{-6} \, mol \, L^{-1} s^{-1}\)