Question

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

• A. $E_a=21.416kJ$ and $k=0.335\times 10^{2}mi{n}^{-1}$
• B. $E_a=11.46kJ$ and $k=0.335\times 10^{2}mi{n}^{-1}$
• C. $E_a=20.23kJ$ and $k=0.43\times 10^{2}mi{n}^{-1}$
• D. $E_a=21.416kJ$ and $k=1.44\times 10^{2}mi{n}^{-1}$

Answer 1

Answer: (A)

ln $k=-\frac{2576}{T}+12.1\dots \left(i\right)$
Arrhenius equation is :
$k=A{e}^{-E_a/RT}$
Taking logarithm on both sides
ln $k=lnA-\frac{E_a}{RT}\dots \left(ii\right)$
Comparing equation $\left(i\right)$ and $\left(ii\right)$ , we get
ln $A=12.1,\frac{E_a}{R}=2576$
$E_a=2576\times 8.314\times 10^{-3}$
$=21.416kJ$
ln $k=\frac{-2576}{300}+12.1$
$=-8.59+12.1=3.51$
$k=33.448mi{n}^{-1}$
$=0.335\times 10^{2}mi{n}^{-1}$