If the rate of a chemical reaction is expressed in the following manner : Rate = K[A]2[B], then the order of this reaction would be–
(A) 2
(B) 3
(C) 1
(D) 0
`"log"_(10) =(E_(2))/(2.303xxR)[(T_(2)-T_(1))/(T_(1)T_(2))]` Given `k_(1)=9.5xx10^(-5)s^(-1),k_(2)=1.9xx10^(-4)s^(-1),` R=8.314J `"mol"^(-1)K^(-1),` `T_(1)-407 K and T_(2)=420K` Substituting the values in Arrhenius equation `"log"_(10)(1.9xx10^(-4))/(9.5xx10^(-5))=(E_(a))/(2.303xx8.314)[(420-407)/(420xx407)]` Applying now log `k_(1)` = log `A-(E_(a))/(2.3030RT_(1))` log `9.5xx10^(-5)="log"A-(75782.3)/(2.303xx8.314xx407)` or log `(A)/(9.5xx10 ^(-5))=(757.82.3)/(2.3030xx8.314xx407)=9.7246` `A=5.04xx10^(5)s^(-1)`
2 Answers 1 viewsUnit of `[("litre")/("mol")]^(n-1)xxsec^(-1)" "....(i)` Given unit of k = `"litre"^(2)"mol"^(-1)"sec"^(-1)" "....(ii)` n-1=2 n=3
2 Answers 1 viewsCorrect Answer - A Unit of rate and rrte constant are same for zero order reaction.
2 Answers 1 viewsCorrect Answer - B `(k_(T_(2)))/(k_(T_(1)))=(mu)^(DeltaT//10)` `(10^(-3))/(k_(T_(1)))=(mu)^(10//10)=2` `k_(T_(1))=(10^(-3))/(2)=0.5xx10^(-3)=5xx10^(-4)"min"^(-1)`
2 Answers 1 viewsCorrect Answer - A For zero order reaction, x=kt `:. (a)/(2)=kxxt_(1//2),i.e.,t_(1//2)=(a)/(2k)" ".....(i)` For frist order reaction, `t_(1//2)=(log_(e)2)/(k)" ".....(ii)` From eqs. (i) and (ii), `(a)/(2k)=(log_(e)2)/(k)` `a=log_(e)4M`
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