KVPY Sample Paper KVPY Stream-SX Model Paper-1

For a given reaction A\[\to \]Product, rate is \[1\times {{10}^{-\,4}}M{{s}^{-\,1}}\] when [A] = 0.01 M and rate is\[1.41\times {{10}^{-\,4}}M{{s}^{-\,1}}\] when [A] = 0.02 M. Hence, rate law is -

A) \[-\frac{d\,[A]}{dt}=k\,{{[A]}^{2}}\]

B) \[-\frac{d\,[A]}{dt}=k\,[A]\]

C) \[-\frac{d\,[A]}{dt}=\frac{k}{4}[A]\]

D) \[-\frac{d\,[A]}{dt}=k\,{{[A]}^{1/2}}\]

Correct Answer: D

Solution :

[D]

\[A\xrightarrow{{}}\,\,\operatorname{Product}\]
We know, Rate \[=K{{[conc.]}^{n}}\]
\[1\times {{10}^{-4}}=K{{[.01]}^{n}}\]	??.. (i)
\[1.41\times {{10}^{-4}}=K{{[.02]}^{n}}\]	??.. (ii)
(i)/(ii) \[\frac{1}{1.41}={{\left( \frac{1}{2} \right)}^{n}}\]
\[n=\frac{1}{2}\]
Then \[\frac{-d(A)}{dt}=K{{[A]}^{1/2}}\]