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Chemistry Question 72 – JEE-MAIN 2025

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For the reaction A products. The concentration of A at 10 minutes is _______ ×103 mol L1 (nearest integer). The reaction was started with 2.5 mol L1 of A. The graph shows t1/2 (min) vs [A]0 (mol L1) is a straight line with slope = 76.92.

Analyze the relationship between half-life and initial concentration from the given graph to determine the order of the reaction.

Video Walkthrough
Step 1: Determine Reaction Order and Rate Constant✦ Active

The graph shows that t1/2 is directly proportional to [A]0, i.e., t1/2=slope×[A]0. This relationship is characteristic of a zero-order reaction.

For a zero-order reaction, the half-life is given by t1/2=[A]02k. Comparing the two expressions, we get:

[A]02k=slope×[A]012k=slope

Given slope = 76.92 min (mol L1)1. Therefore, the rate constant k is:

k=12×slope=12×76.92=1153.840.00650026 mol L1 min1
Step 2: Calculate Concentration at 10 minutes○ Expand

The integrated rate law for a zero-order reaction is [A]t=[A]0kt. Given [A]0=2.5 mol L1 and t=10 minutes, we can calculate [A]10:

[A]10=2.5(0.00650026 mol L1 min1×10 min)
[A]10=2.50.0650026=2.4349974 mol L1
Step 3: Convert to Required Units and Round○ Expand

The question asks for the concentration in ×103 mol L1 (nearest integer). Convert the calculated concentration:

2.4349974 mol L1=2.4349974×1000×103 mol L1
=2434.9974×103 mol L1

Rounding to the nearest integer gives 2435.

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