Question

The Arrhenius plots of two reactions, I and II are shown graphically
The graph suggests that

• A. $E_I > E_{II}$ and $A_I > A_{II}$
• B. $E_{II} > E_{I}$ and $A_{II} > A_{I}$
• C. $E_{II} > E_{II}$ and $A_{I} > A_{II}$
• D. $E_{II} > E_{I}$ and $A_{I} > A_{II}$

Answer 1

Answer: (A)

The temperature dependance of rate of a chemical reaction is expressed by Arrhenius equation,
$k = Ae^{-E_a/KT}$
$ ln \,k = ln\,A - \frac{E_a}{RT}$
If we compare the above equation with the equation of a straight line, $y = mx + c$
$\therefore $ For the plot between $\log \,k$ versus $1/T$.
slope $ = - E_a/R$
and intercept $ = ln \,A$
$\therefore E_1 > E_{II} $ and $A_I > A_{II}$