KVPY Sample Paper KVPY Stream-SX Model Paper-9

Graph between log K and \[\frac{1}{T}\] [Where K is rate constant \[({{S}^{-1}})\] and T is temperature (K)] is a straight line with OX = 5, \[\theta ={{\tan }^{-1}}\left[ -\frac{1}{2.303} \right].\] Hence \[Ea\]and log A respectively will be:

A) \[2.303\times 2\,cal,5\]

B) \[\frac{2}{3.303}cal,{{e}^{5}}\]

C) 2 cal, 5

D) none of these

Correct Answer: C

Solution :

\[K=A{{e}^{\frac{Ea}{RT}}}\]

\[Log\,\,K=\log \,A-\frac{Ea}{2.303R}\times \frac{1}{T}\]

Slope=\[\tan \theta =\left[ -\frac{1}{2.303} \right]=-\frac{{{E}_{a}}}{2.303R}\]

So, \[Ea=R=2cal/mol\]