Respuesta :
Answer:
[tex]155.65\ s^{-1}[/tex] is the rate constant at a temperature of 30.0 °C.
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] = 52 kJ/mol = 52000 J/mol (As 1 kJ = 1000 J)
[tex]k_2=?[/tex]
[tex]k_1=77.0s^{-1}[/tex]
[tex]T_1=20\ ^0C[/tex]
[tex]T_2=30\ ^0C[/tex]
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (20 + 273.15) K = 293.15 K
T = (30 + 273.15) K = 303.15 K
[tex]T_2=303.15\ K[/tex]
So,
[tex]\ln \frac{77.0}{k_2}\:=-\frac{52000}{8.314}\times \left(\frac{1}{293.15}-\frac{1}{303.15}\right)[/tex]
[tex]\frac{77}{k_2}=e^{-\frac{52000}{8.314}\left(\frac{1}{293.15}-\frac{1}{303.15}\right)}[/tex]
[tex]k_2=77e^{\frac{520000}{738852.06466}}[/tex]
[tex]k_2=155.65\ s^{-1}[/tex]
[tex]155.65\ s^{-1}[/tex] is the rate constant at a temperature of 30.0 °C.
