question_answer

When a graph is plotted between log K and\[\left( \frac{1}{T} \right),\] the slope of line obtained represents: (K = rate constant, T = temperature)

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

B)  \[-\frac{{{E}_{a}}}{R}\] 

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

D)         \[\log A\]

Correct Answer: C

Solution :

The Arrhenius equation can be written as       \[\log \,K=\log A-\frac{E}{2.303RT}\] On comparing this equation with general equation of a straight line, \[y=mx+c\] We get, \[y=\log \,K,\]  \[x=\frac{1}{T},\] \[m=-\frac{E}{2.303R},\]               \[c=\log A\] i.e., if we plot a graph between log K (at Y-axis) and \[\frac{1}{T}\](at X-axis), then the slope of the line obtained will be equal to \[-\frac{e}{2.303\,R}.\]