Respuesta :
Answer:
T = 42.08 °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] = 45.6 kJ/mol = 45600 J/mol (As 1 kJ = 1000 J)
[tex]k_2=2\times k_1[/tex]
[tex]k_1=0.0160s^{-1}[/tex]
[tex]T_1=30\ ^0C[/tex]
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (30 + 273.15) K = 303.15 K
[tex]T_1=303.15\ K[/tex]
So,
[tex]\ln \dfrac{k_{1}}{2\times k_{1}} =-\dfrac{45600}{8.314} \left (\dfrac{1}{303.15}-\dfrac{1}{T_2} \right )[/tex]
[tex]\ln \dfrac{1}{2} =-\dfrac{45600}{8.314} \left (\dfrac{1}{303.15}-\dfrac{1}{T_2} \right )[/tex]
[tex]8.314\ln \left(2\right)=-45600\left(\frac{1}{303.15}-\frac{1}{T_2}\right)[/tex]
[tex]8.314\ln \left(2\right)=-150.42058+\frac{45600}{T_2}[/tex]
[tex]144.65775 =\frac{45600}{T_2}[/tex]
[tex]T_2=\frac{45600}{144.65775}[/tex]
[tex]T_2=315.23\ K[/tex]
Conversion to °C as:
T(K) = T( °C) + 273.15
So,
315.23 = T( °C) + 273.15
T = 42.08 °C
Answer:
Explanation:
2t2 is 1 1t1 = 11 11t2
1212t12 = 0 0t2=2 der fo it 2