JEE Main 2019ChemistryChemical KineticsHardMCQ

JEE Main 2019 — Chemical Kinetics Question with Solution

JEE Main 2019 (09 Jan Shift 1)

Question

The following results were obtained during kinetic studies of the reaction.

2A+B→ product
 
Experiment A in mol L-1 B  in mol L-1 Initial rate of reaction in mol L-1 min -1
I 0.10 0.20 6.93×10-3
II 0.10 0.25 6.93×10-3
III 0.20 0.30 1.386×10-2

The time (in minutes) required to consume half of A is

Choose an option

Show full solutionCorrect option: B
Correct answer
B10

Step-by-step explanation

Let's assume x and y are the order of the reaction with respect to A and B, respectively.

From equation 1 and 2,

0.10.1x0.20.25y=6.93×10-36.93×10-3⇒y=0

From equation 1 and 3 and put the value y=0,

0.10.2x0.20.30=6.93×10-31.386×10-2    ∴y=0

12x=12⇒x=1

The rate law will be R=K[A]1[B]0

Hence, the reaction is of the first order.

K=RA=6.93×10-30.1=6.93×10-2

Here t12 for the first-order reaction is given by, t12=0.693K=0.693(6.93×10-2)=10

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About this question

This is a previous-year question from JEE Main 2019, covering the Chemical Kinetics chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.