Question bank/JEE Main/Chemistry

JEE MainChemistryModerateChemical Kinetics

A graph between log [IMAGE] and log a (abscissa) a
being the initial concentration of A in the reaction. For reaction A→
Product, the rate law is

Show answer and solution
t1/2=ka(1−n)t_{1/2} = {ka}^{(1 - n)}, n being order k = Rate constant
log⁡t1/2=logk+(1−n)log⁡a\log t_{1/2} = logk + (1 - n)\log a
1−n=−11 - n = - 1
n=2n = 2

Source: Grass-G / CH-15 Chemistry Paper 11 Nov..docx

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