Question

The rate constant for a certain reaction is measured at two different temperatures:

Temperature K
376.0°C 4.8 x 10^8
280°C 2.3 x 10^8

Assuming the rate constant obeys the Arrhenius equation, calculate the activation energy for this reaction. Round your answer to significant digits.

Answer 1

Answer: 22.78KJ/ mol

Explanation:

The computation of activation energy $E_0$ follows by formula:

$\ln (\dfrac{K_2}{K_1}) = (\dfrac{E_a }{ R}) (\dfrac{1}{T_1} -\dfrac{ 1}{T_2})$

Given:$T_1 = 376.0^{\circ}C =376+273K= 649 K$

$K_1 = 4.8 \times10^8$

$T_2 = 280.0^{\circ}C = 273+280K=553 K ,\ \ \ K_2 = 2.3 \times 10^8$

$\ln(\dfrac{2.3 \times10^8 }{ 4.8 \times 10^8}) = (\dfrac{E_a }{ 8.314}) (\dfrac{1}{649} - \dfrac{1}{553})\\\\\Rightarrow\ \ln(0.479) = (\dfrac{E_a }{ 8.314}) (-\dfrac{96}{358897})\\\\\Rightarrow\ -0.73605468= (\dfrac{E_a }{ 8.314}) (-\dfrac{96}{358897})\\\\\Rightarrow\E_a=-0.73605468\times 8.314\times\dfrac{-358897}{96}=22878.033\\\\\Rightarrow\E_a=22878.033\ J/mol=22.78 KJ /mol$

Hence, the activation energy for this reaction =22.78KJ/ mol