KVPY Sample Paper KVPY Stream-SX Model Paper-17

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 :

\[A\xrightarrow{{}}\,\,\text{Product}\]
We know, \[\text{Rate}=K\,\,{{[\text{conc}\text{. }\!\!]\!\!\text{ }}^{n}}\]
\[1\times {{10}^{-\,4}}=K\,\,{{[.01]}^{n}}\]	..... (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 \[\frac{-\,d\,(A)}{dt}=K\,\,{{[A]}^{1/2}}\]