A certain reaction has an activation energy of 74.38 kJ/mol. At what Kelvin temperature will the reaction proceed 7.00 times faster than it did at 313 K?
See your earlier post.
To determine the Kelvin temperature at which the reaction will proceed 7.00 times faster than it did at 313 K, we can use the Arrhenius equation:
k2 = k1 * exp((-Ea/R) * (1/T2 - 1/T1))
Where:
k1 and k2 are the rate constants at temperatures T1 and T2 respectively,
Ea is the activation energy,
R is the gas constant (8.314 J/(mol*K)),
T1 is the initial temperature (313 K), and
T2 is the unknown temperature.
First, we need to convert the activation energy from kJ to J:
Ea = 74.38 kJ/mol * (1000 J/1 kJ) = 74,380 J/mol
Next, we'll rewrite the equation to solve for T2:
ln(k2/k1) = (-Ea/R) * (1/T2 - 1/T1)
To find T2, we'll rearrange the equation:
1/T2 = (ln(k2/k1) / (-Ea/R)) + 1/T1
Now, let's substitute the values into the equation:
1/T2 = (ln(7.00) / (-74,380 J/mol / (8.314 J/(mol*K)))) + 1/313 K
1/T2 = (-2.944 / -8.947) + 1/313
1/T2 = 0.3288 + 0.0032
1/T2 = 0.3320
Now, we'll solve for T2:
T2 = 1 / 0.3320
T2 ≈ 3.012 K
Therefore, the reaction will proceed 7.00 times faster at approximately 3.012 Kelvin.