Answer:
[tex]Ea=44418.56\ J/mol[/tex]
Explanation:
Using the expression,
[tex]\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )[/tex]
Where,
[tex]k_1\ is\ the\ rate\ constant\ at\ T_1[/tex]
[tex]k_2\ is\ the\ rate\ constant\ at\ T_2[/tex]
[tex]E_a[/tex] is the activation energy
R is Gas constant having value = 8.314 J / K mol
[tex]k_2=4\times k_1[/tex]
[tex]T_1=25\ ^0C[/tex]
[tex]T_2=50\ ^0C[/tex]
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
[tex]T_1[/tex] = (25 + 273.15) K = 298.15 K
[tex]T_2[/tex] = (50 + 273.15) K = 323.15 K
So,
[tex]\ln \dfrac{k_{1}}{4\times k_1} =-\dfrac{E_{a}}{8.314} \left (\dfrac{1}{298.15}-\dfrac{1}{323.15} \right )[/tex]
[tex]Ea=-\frac{\ln \frac{1}{4}\times \:8.314}{\left(\frac{1}{298.15}-\frac{1}{323.15}\right)}=44418.56\ J/mol[/tex]
