JEE Main 2019ChemistryChemical KineticsMediumMCQ

JEE Main 2019 — Chemical Kinetics Question with Solution

JEE Main 2019 (12 Apr Shift 2)

Question

NO2 required for a reaction is produced by the decomposition of N2O5 in CCl4 as per the equation,
2N2O5g→4NO2g+O2g.
The initial concentration of N2O5 is 3.00 mol L-1 and it is 2.75 mol L-1 after 30 minutes. The rate of formation of NO2 is:

Choose an option

Show full solutionCorrect option: A
Correct answer
A1.667×10-2  mol L-1  min-1

Step-by-step explanation

Rate of reaction in terms of rate of disappearance of N2O5(s)
-dN2O5dt=12dNO2dt=2dO2dt=rate of Reaction
-(-3+1.75)30=rate of reaction
--0.2530=8.3×10-3mol L-1min-1
=253000=8.3×10-3=12× rate of formation of O2
8.3×10-3=12dNO2dt
1.66×10-2m/min =d[NO2]dt

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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.