Question

In a first-order reaction, the concentration of the reactant decreases from 0.8 M to 0.4 M in 15 minutes. The time taken for the concentration to change from 0.1 M to 0.025 M is:

• A. 7.5 minutes
• B. 15 minutes
• C. 30 minutes
• D. 60 minutes

Hint:

For a first-order reaction, the half-life is independent of the initial concentration.

Answer 1

The Correct Option is C

Step 1: Understanding Half-Life in First-Order Reactions
For a first-order reaction, the half-life (\( t_{1/2} \)) is constant and given by: \[ t_{1/2} = \frac{0.693}{k} \] From the given data, the concentration falls from \( 0.8 M \) to \( 0.4 M \) in 15 minutes.
Since \( t_{1/2} \) is constant, another half-life will reduce it from 0.4 M to 0.2 M in another 15 minutes.
Another half-life will further reduce it from 0.1 M to 0.025 M, which takes another 15 minutes.
Step 2: Total Time Required
Two half-lives are needed to reduce \( 0.1 M \) to \( 0.025 M \).
\( 2 \times 15 = 30 \) minutes.
Final Answer: 30 minutes.

Answer 2

Step 1: Understand the nature of a first-order reaction
In a first-order reaction, the rate of reaction depends linearly on the concentration of one reactant. The concentration decreases exponentially with time according to the formula:
ln([A]_0 / [A]) = kt, where [A]_0 is the initial concentration, [A] is the concentration at time t, and k is the rate constant.

Step 2: Calculate the rate constant (k)
Given that the concentration decreases from 0.8 M to 0.4 M in 15 minutes:
ln(0.8 / 0.4) = k × 15
ln(2) = 15k
k = ln(2) / 15 ≈ 0.0462 min⁻¹

Step 3: Use the rate constant to find the required time
Now, find the time (t) taken for the concentration to change from 0.1 M to 0.025 M:
ln(0.1 / 0.025) = k × t
ln(4) = 0.0462 × t
t = ln(4) / 0.0462 ≈ (1.386) / 0.0462 ≈ 30 minutes

Step 4: Final Conclusion
Therefore, the time taken for the concentration to decrease from 0.1 M to 0.025 M in this first-order reaction is 30 minutes.