Respuesta :
Answer:
The activation energy for the decomposition = 33813.28 J/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]
Wherem
[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
Thus, given that, [tex]E_a[/tex] = ?
[tex]k_2=4.0\times 10^3s^{-1}[/tex]
[tex]k_1=1.5\times 10^3s^{-1}[/tex]
[tex]T_1=5\ ^0C[/tex]
[tex]T_2=25\ ^0C[/tex]
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (5 + 273.15) K = 278.15 K
T = (25 + 273.15) K = 298.15 K
[tex]T_1=278.15\ K[/tex]
[tex]T_2=298.15\ K[/tex]
So,
[tex]\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)[/tex]
[tex]E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}[/tex]
[tex]E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}[/tex]
[tex]E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}[/tex]
[tex]E_a=33813.28\ J/mol[/tex]
The activation energy for the decomposition = 33813.28 J/mol
