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

Concept: For a first-order reaction, the time required for a specific percentage of the reaction to complete is related to the half-life ($t_{1/2}$). A key property of first-order kinetics is that the half-life is constant, regardless of the starting concentration.
• Half-life ($t_{50\%}$ or $t_{1/2}$): The time taken for the reactant concentration to reduce to 50% of its initial value.
• $t_{75\%$ Relation:} Completing 75% of a reaction means two half-lives have passed (reducing 100% $\to$ 50% $\to$ 25% remaining).
• General Formula: $t_{75\%} = 2 \times t_{50\%}$

Step 1: Understanding the relationship between $t_{75\%}$ and $t_{50\%}$.
In a first-order reaction:
• After the 1st half-life ($t_{1/2}$), 50% of the reactant is left.
• After the 2nd half-life ($t_{1/2}$), 50% of the remaining 50% is consumed (i.e., another 25%).
• Total consumed = 50% + 25% = 75%. Therefore, the time taken for 75% completion ($t_{75\%}$) is exactly equal to two half-lives ($2 \times t_{1/2}$).

Step 2: Substituting the given values.
Given: $t_{75\%} = 32 \text{ min}$ Using the relation: \[ 32 \text{ min} = 2 \times t_{50\%} \]

Step 3: Calculating the time for 50% completion.
\[ t_{50\%} = \frac{32 \text{ min}}{2} \] \[ t_{50\%} = 16 \text{ min} \]