Question

URGENT!URGENT!URGENT! I WILL MARK BRAINLIEST FOR PROPER EXPLANATION AND ANSWER.

Answer 1

#General formula of half life period (t1/2) for any order reaction is given by
t 1/2 = [1 /(n-1) K ] ✖ { 1/ [At] ^(n-1) - 1/ [Ao]^(n-1)}
where n= order of reaction, k= rate constant,, [At ] is final concentration of reactant A ,,ND ,,[Ao] is initial concentration of reactant A

#let's put values of n one by one

#first n=0
t1/2 = [ 1/ (0-1)k ] ✖ { 1 / [At ]^ (0-1) -- 1/ [Ao]^(0-1) }
hence t1/2=[ 1/- K ] ✖ { 1/ [At ]^(-1) -- 1/ [Ao]^(-1) }
hence t1/2 = [1/-k] ✖ { [At ] - [ Ao ] }
this shows that half life time period is directly proportional to the[ Ao] initial consentration

#if put n=1 it will
t1/2= 0.693/ K that means it is not depend on [Ao]

#now put n=2
t1/2 = [1 / k] { 1 / [At] - 1 / [Ao] }
here in second order of reaction half life period is inversely proportional to [ Ao]

# see in question in that half life period changing inverse of initial concentration raised to power 1
hence answer (C) 2

#if we put n=3
then t1/2 = [1 / 2k] { 1 / [At ]^( 2) - 1 / [Ao]^(2) }
here is also half life period is changing inverse to [Ao] but raised to power not 1 but 2

#HENCE CORRECT OPTION IS( C ) 2 nd order