Question

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

• A. ln K vs T
• B. $\frac{lnK}{T}vsT$
• C. ln K vs $\frac{1}{T}$
• D. $\frac{T}{lnK}vs\frac{l}{T}$

Answer 1

The correct option is C ln K vs $\frac{1}{T}$

By Arrhenius equation
$k=A{e}^{-\frac{E_a}{RT}}$
Where, Ea = energy of activation
Applying log on both the side.
ln k =lnA $-\frac{E_a}{RT}....\left(i\right)$
Or $logk=-\frac{E_a}{2.303RT}+logA...\left(ii\right)$
Compare the above equation w.r.t. straight line equation of y =mx +c. Thus, if a plote on ln $kvs\frac{1}{T}$ is a straight line, the validity of the equation is confirmed.
Slope of the line $=-\frac{E_a}{R}$
Thus, measuring the slope of the line, the value of $E{­}_a$ can be calculated.