question_answer

The activation energy of a reaction can be determined from the slope of which of the following graphs?         [NEET 2015 ]

A) In \[K\,vsT\]

B) \[\frac{\text{ln}K}{T}vsT\]

C) In \[Kvs\frac{l}{T}\]

D) \[\frac{T}{\ln \,k}vs\frac{l}{T}\]

Correct Answer: C

Solution :

[c] By Arrhenius equation

\[K=A{{e}^{-{{E}_{a}}/RT}}\]

where, \[{{E}_{a}}\] = energy of activation

Applying log on both the side,

\[\ln \,k=\ln A-\frac{{{E}_{a}}}{RT}\,\,\,\,\,\,\,\]              (i)

or\[\log \,k=-\frac{{{E}_{a}}}{2.303RT}+\log A\,\]          (ii)

This equation is of the form of \[y=mx+c\] i.e. the equation of a straight line. Thus, if a plote of  \[\log k\,vs\frac{1}{T}\] is a straight line, the validity of the equation is confirmed.

Slope of the line \[=-\frac{{{E}_{a}}}{2.303R}\]

Thus, measuring the slope of the line, the value of \[{{E}_{a}}\] can be calculated.