If 75% of a first-order reaction gets completed in 32 minutes, the time taken for 50% completion of this reaction is:

The correct option is 16 minutes. Let's understand why this is the correct answer step-by-step.

For a first-order chemical reaction, the rate constant is given by the formula:

k = 2.303 t log 10 [A] 0 [A]

where:
- k is the rate constant,
- t is the time taken,
- [A]0 is the initial concentration of the reactant,
- [A] is the concentration of the reactant remaining at time t.

Step 1: Write the equation for 75% completion of the reaction.
When 75% of the reaction is completed, the remaining amount of reactant is:
[A] = [A] 0 - 0.75 [A] 0 = 0.25 [A] 0 = [A] 0 4
The time taken (t75%) is given as 32 minutes. Substituting these values into the first-order rate equation, we get:
k = 2.303 32 log 10 [A] 0 [A] 0 / 4
Simplifying the log term:
k = 2.303 32 log 10 4
Since log10(4)=log10(22)=2log10(2):
k = 2.303 32 × 2 log 10 2
Which simplifies to:
k = 2 × 2.303 log 10 ( 2 ) 32 = 2.303 log 10 ( 2 ) 16 --- (Equation 1)

Step 2: Write the equation for 50% completion (half-life, t50% or t1/2).
For 50% completion, the remaining concentration of reactant is:
[A] = 0.5 [A] 0 = [A] 0 2
Using the rate formula:
k = 2.303 t 50 % log 10 [A] 0 [A] 0 / 2
Which simplifies to:
k = 2.303 log 10 ( 2 ) t 50 % --- (Equation 2)

Step 3: Compare Equation 1 and Equation 2.
Since the rate constant k remains constant for a given reaction at a constant temperature, we can set the two equations equal to each other:
2.303 log 10 ( 2 ) 16 = 2.303 log 10 ( 2 ) t 50 %
Canceling the common term 2.303log10(2) from both sides gives:
t 50 % = 16 minutes

Alternative Shortcut:
For any first-order reaction, the time required for 75% completion is always exactly twice the half-life (t50%) because 75% completion represents the passage of two successive half-lives (from 100% to 50%, and then from 50% to 25%).
t 75 % = 2 × t 50 %
32 = 2 × t 50 %
t 50 % = 16 minutes