A certain reaction has an activation energy of 25.14 kJ/mol. At what Kelvin temperature will the reaction proceed 8.00 times faster than it did at 305 K?
To determine the Kelvin temperature at which the reaction will proceed 8.00 times faster, we can use the Arrhenius equation. The Arrhenius equation relates the rate constant (k) of a reaction to the temperature (T) and the activation energy (Ea). The equation is as follows:
k = A * e^(-Ea/RT)
Where:
- k is the rate constant,
- A is the pre-exponential factor or frequency factor,
- Ea is the activation energy,
- R is the gas constant (8.314 J/(mol·K)),
- T is the temperature in Kelvin.
Since the problem doesn't provide the pre-exponential factor (A), we can assume it remains constant. Therefore, we can compare the rates (k1 and k2) at two different temperatures using the following ratio:
k2 / k1 = e^(-(Ea/R) * (1/T2 - 1/T1))
We are given that the reaction proceeds 8.00 times faster at T2 when compared to T1. Therefore, we can set up the equation as follows:
8 = e^(-(Ea/R) * (1/T2 - 1/T1))
Plugging in the given values:
Ea = 25.14 kJ/mol
R = 8.314 J/(mol·K)
T1 = 305 K
Let's solve for T2, the temperature at which the reaction proceeds 8.00 times faster.
8 = e^(-(25.14 * 10^3 J/mol) / (8.314 J/(mol·K) * (1/T2 - 1/305 K)))
Simplifying the equation:
ln(8) = -(25.14 * 10^3 J/mol) / (8.314 J/(mol·K)) * (1/T2 - 1/305 K)
ln(8) = -3.03 * 10^3 / T2 + 3.03 * 10^3 / 305
Now, let's rearrange the equation to solve for T2:
-3.03 * 10^3 / T2 = ln(8) - 3.03 * 10^3 / 305
T2 = -3.03 * 10^3 / (ln(8) - 3.03 * 10^3 / 305)
Calculating the right-hand side of the equation:
T2 ≈ -3.03 * 10^3 / (2.079 - 3.03 * 10^3 / 305)
T2 ≈ -3.03 * 10^3 / (2.079 - 9.984)
T2 ≈ -3.03 * 10^3 / (-7.905)
T2 ≈ 384 K
Therefore, the reaction will proceed 8.00 times faster at approximately 384 Kelvin.