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Match the following:

  Column A  

 

 Column B
(I)     On increasing A_0, t_{\frac{1}{2}} increases (P)     First order reaction
(II)     On doubling A_0, t_{\frac{1}{2}} doubles (Q)     Order of reaction is less than one
(III)     On changing A_0, t_{\frac{1}{2}} remains independent (R)    

Second order reaction
(IV)     On increasing A_0, t_{\frac{1}{2}} decreases (S)     Zero order reaction
(V)     On doubling A_0, t_{\frac{1}{2}} become half  (T)     Order of reaction is more than one

 

Option: 1

(Q), (II)-(S), (III)-(P), (IV)-(T), (V)-(R)
 


Option: 2

 (I)-(Q), (II)-(S), (III)-(P), (IV)-(T), (V)-(R)


Option: 3

 (I)-(S), (II)-(S), (III)-(T), (IV)-(R), (V)-(P)

 


Option: 4

 (I)-(Q), (II)-(R), (III)-(P), (IV)-(T), (V)-(S)


Answers (1)

best_answer

General formula for half life of n^{\text {th }} order reaction:

t_{\frac{1}{2}}=\frac{2^{n-1}-1}{\left(A_0\right)^{n-1}(n-1) K}
Thus we can say that t_{\frac{1}{2}} is directly proportional to \left(A_0\right)^{1-n}
We can conclude from above relation that:
For zero order reaction, t_{\frac{1}{2}} is directly proportional to \left(A_0\right)^1
For first order reaction, t_{\frac{1}{2}} is independent of\left(A_0\right)
For second order reaction, t_{\frac{1}{2}}  is directly proportional to \left(A_0\right)^{-1}, thus we can say that t_{\frac{1}{2}} is inversely proportional to \left(A_0\right)
Thus,t_{\frac{1}{2}} increases on increasing initial concentration of reactant \left(A_0\right), when order of reaction is less than one. And, t_{\frac{1}{2}}  decreases on increasing initial concentration of reactant \left(A_0\right), when the order of reaction is more than one.

Posted by

vishal kumar

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