question_answer

When a graph is plotted between log K and \[\frac{1}{T}\] for a first order reaction we get straight line. The slope of the line is equal to:

A)  \[-\frac{{{E}_{a}}}{2.303}\]                      

B)  \[-\frac{{{E}_{a}}}{2.303\,R}\]

C)  \[-\frac{2.303}{{{E}_{a}}\,R}\]                  

D)  \[-\frac{{{E}_{a}}\,}{R}\]

Correct Answer: B

Solution :

\[\because \]   \[\log \,K=\log A-\frac{{{E}_{n}}}{2.303\,RT}\] When a graph is plotted between log K and %, for first order reaction a straight line is obtained and the slope of the line is equal to \[-\frac{{{E}_{n}}}{2.303\,R}\]