Respuesta :
Answer:
The activation energy for this reaction = 23 kJ/mol.
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 = 8.314×10⁻³ kJ / K mol
[tex]k_2=2.3\times 10^8[/tex]
[tex]T_2=280\ ^0C[/tex]
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (280 + 273.15) K = 553.15 K
[tex]T_2=553.15\ K[/tex]
[tex]k_1=4.8\times 10^8[/tex]
[tex]T_1=376\ ^0C[/tex]
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (376 + 273.15) K = 649.15 K
[tex]T_1=649.15\ K[/tex]
So,
[tex]\left(\ln \left(\:\frac{4.8\times \:\:\:10^8}{2.3\times \:\:\:10^8}\right)\right)\:=-\frac{E_a}{8.314\times \:10^{-3}\ kJ/mol.K}\times \:\left(\frac{1}{649.15\ K}-\frac{1}{553.15\ K}\right)[/tex]
[tex]E_a=-\frac{10^{-3}\times \:8.314\ln \left(\frac{10^8\times \:4.8}{10^8\times \:2.3}\right)}{-\frac{96}{359077.3225}}\ kJ/mol[/tex]
[tex]E_a=-\frac{\frac{8.314\ln \left(\frac{4.8}{2.3}\right)}{1000}}{-\frac{96}{359077.3225}}\ kJ/mol[/tex]
[tex]E_a=22.87\ kJ/mol[/tex]
The activation energy for this reaction = 23 kJ/mol.
