Respuesta :
Activation Energy Ea = 249 x 10^3 J/mol
Frequency Factor = 1.6 Ă— 1014 sâ’1
Rate of reation is twice as fast with temperature change so k1/k2 = 1/2.
Temperature T1 = 730.6, T2 =?
Gas constant R = 8.314 J/ mole-K
We have Arrhenius equation ln (k1 / k2) = (Ea / R) x [(1/T1) - (1/T2)]
ln(1/2) = ((249 x 10^3) / 8.314) x [(1/T2) - (1/730.6)]
ln(1/2) = 29.95 x 10^3 x [(1/T2) - 0.0013687]
-0.69315 / (29.95 x 10^3) = (1/T2) - 0.0013687
(1/T2) - 0.0013687 = - 2.3 x 10^-5
1 / T2 = 0.0013455
T2 = 743.18 k
Answer:
The temperature at which the rate of the reaction would be twice as fast as when the reaction runs at 730.6 K is 743.12 K.
Explanation:
According to the Arrhenius equation,
[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]
[tex]\log \frac{K_2}{K_1}=\frac{E_a}{2.303R}\times [\frac{T_2-T_1}{T_1T_2}][/tex]
[tex]K_2[/tex] = rate of reaction at [tex]T_2[/tex]
[tex]K_1[/tex] = rate of reaction at [tex]T_1[/tex]
[tex]E_a[/tex] = activation energy
R = gas constant
We have:
[tex]K_1=1.6\times 10^{14} s^{-1}[/tex]
[tex]K_2=2\times K_1[/tex]
[tex]T_1=730.6 K, T_2=?[/tex]
[tex]E_a=249 kJ/mol =249000 J/mol[/tex]
[tex]\log \frac{2K_1}{K_1}=\frac{249000 kJ/mol}{2.303\times 8.314 J/mol K}\times [\frac{T_2-730.6 K}{730.6 K\times T_2}][/tex]
[tex]0.3010=\frac{249000 J/mol}{2.303\times 8.314 J/mol K}\times [\frac{T_2-730.6 K}{730.6 K\times T_2}][/tex]
[tex]\frac{0.3010\times 2.303\times 8.314 J/mol K\times 730.6 K\times T_2}{249000 J/mol}=T_2-730.6 K[/tex]
[tex]T_2=743.12 K[/tex]
The temperature at which the rate of the reaction would be twice as fast as when the reaction runs at 730.6 K is 743.12 K.
