Question

What will be the value of activation energy ( $ E_a $ in $ kJ $ ) and rate constant ( $ k\, in\, min^{-1} $ ) for the given equation at $ 27^{\circ}C $ ? The equation : $ lnk = - 2576/T+ 12.1 $

• A. $ E_{a} = 21.416\, kJ $ and $ k =0.335 \times 10^{2 }\, min^{-1} $
• B. $ E_{a} = 11.46 \,kJ $ and $ k =0.335\times10^{2 } \,min^{-1} $
• C. $ E_{a} = 20.23\, kJ $ and $ k =0.43 \times 10^{2 }\, min^{-1} $
• D. $ E_{a} = 21.416\, kJ $ and $ k =1.44 \times 10^{2 }\, min^{-1} $

Answer 1

Answer: (A)

ln $k=-\frac{2576}{T}+12.1 \ldots\left(i\right)$
Arrhenius equation is :
$k=Ae^{-E_{a} / RT}$
Taking logarithm on both sides
ln $k=ln \, A-\frac{E_{a}}{RT} \ldots\left(ii\right)$
Comparing equation $\left(i\right)$ and $\left(ii\right)$ , we get
ln $A=12.1, \frac{E_{a}}{R}=2576$
$E_{a}=2576\times8.314\times10^{-3}$
$=21.416\, kJ$
ln $k=\frac{-2576}{300}+12.1$
$=-8.59+12.1=3.51$
$k=33.448 \, min^{-1}$
$=0.335 \times10^{2} min^{-1}$