Class NEET 2026 ALL Q #1978
COMPETENCY BASED
APPLY
4 Marks 2026 NTA-RE-NEET-2026 MCQ SINGLE
$2A\rightarrow B$ is a zero-order reaction, where $k=1.0~mol~L^{-1}min^{-1}$. If the initial concentration of A is 2 M, then the time taken to complete 75% of the reaction will be
(A) 2.0 min
(B) 1.5 min
(C) 0.75 min
(D) 1.0 min
Correct Answer: C

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Detailed Solution

Step 1: Identify the Rate Law for Zero-Order Reaction

For a zero-order reaction of the type $nA \rightarrow Products$, the integrated rate equation is given by: $$[A]_t = [A]_0 - n \cdot k \cdot t$$ Where $n$ is the stoichiometric coefficient of the reactant $A$. Here, $n=2$.

Step 2: Determine the change in concentration

The initial concentration $[A]_0 = 2.0\ M$. The reaction is 75% complete, meaning 75% of the initial concentration has reacted. $$Amount\ reacted = 0.75 \times 2.0\ M = 1.5\ M$$ The remaining concentration $[A]_t$ is: $$[A]_t = 2.0 - 1.5 = 0.5\ M$$

Step 3: Calculate the time taken

Substitute the values into the rate equation: $$0.5 = 2.0 - (2 \times 1.0 \times t)$$ $$2 \times 1.0 \times t = 2.0 - 0.5$$ $$2t = 1.5$$ $$t = \frac{1.5}{2} = 0.75\ min$$

Final Answer: 0.75 min

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student must apply the integrated rate law formula for a zero-order reaction while accounting for the stoichiometric coefficient.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the execution of a specific mathematical procedure (integrated rate law calculation) to solve for time.
Syllabus Audit: In the context of NEET, this is classified as COMPETENCY. This tests the student's ability to handle stoichiometry in kinetics, a common trap in competitive exams.