Question

A straight line with negative slope was found in graph of $lo{g}_{e}Kvs.\frac{1}{T}$ with an angle of ${45}^o$ with x-axis. The energy of activation (in $cal$) is:

• A. 2
• B. 1
• C. 4
• D. None of these

Answer 1

Answer: (A)

2

As we know, the Arrhenius equation is given as follows:

$k=A{e}^{-E_a/RT}$

$lnk=lnA-E_a/RT$

The slope of ln k v/s 1/T is as follows:

$m=tan{45}^o=1=E_a/R$

$E_a=R=2cal$