Question

75% of a first order reaction was completed in 32 min. 50% of the reaction would have been completed in:

• A. 24 min
• B. 16 min
• C. 18 min
• D. 22 min

Hint:

For first-order reactions, you can use these common shortcuts: $t_{75\%} = 2 \times t_{50\%}$ $t_{87.5\%} = 3 \times t_{50\%}$ $t_{99.9\%} \approx 10 \times t_{50\%}$

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For first-order reactions, the half-life (\(t_{1/2}\)) is a constant. It does not depend on the initial concentration.
This means the time required to reduce the reactant concentration from \(100%\) to \(50%\) is the same as the time required to reduce it from \(50%\) to \(25%\).
If a reaction is \(75%\) complete, it means only \(25%\) of the reactant remains.
Going from \(100%\) to \(25%\) involves exactly two half-life cycles (\(100 \to 50 \to 25\)).
This characteristic "halving" behavior allows us to solve kinetics problems using simple ratios instead of complex logarithms.
Step 2: Key Formula or Approach:
1. General Formula: \(t = n \times t_{1/2}\), where \(n\) is the number of half-lives.
2. Specifically for first order: \(t_{75%} = 2 \times t_{50%}\).
Step 3: Detailed Explanation:
The reaction is first order.
Amount of reactant remaining after \(75%\) completion = \(100% - 75% = 25%\).
Fraction of reactant remaining = \(25/100 = 1/4 = (1/2)^2\).
The exponent represents the number of half-lives that have elapsed. So, \(n = 2\).
We are given that \(t_{75%} = 32\) minutes.
Using the relation \(t_{75%} = 2 \times t_{50%}\):
\[ 32 = 2 \times t_{50%} \implies t_{50%} = 16 \text{ minutes} \]
Thus, \(50%\) completion (which is the definition of one half-life) takes \(16\) minutes.
Step 4: Final Answer:
\(50%\) of the reaction is completed in \(16\) minutes.