The activation energy of a certain reaction is 34.9kJ/mol . At 23 ∘C , the rate constant is 0.0190s−1. At what temperature in degrees Celsius would this reaction go twice as fast?
Use the Arrhenius equation. The problem gives you k1, make k2 twice that.
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ln(2)=34900/8.314 - (1/296 - 1/T1)
To determine the temperature at which the reaction would go twice as fast, you can use the Arrhenius equation. The Arrhenius equation relates the rate constant (k) to the activation energy (Ea) and the temperature (T):
k = Ae^(-Ea/RT)
Where:
k = rate constant
A = pre-exponential factor (frequency factor)
Ea = activation energy
R = gas constant (8.314 J/(mol·K))
T = temperature in Kelvin
To find the temperature at which the reaction goes twice as fast, we'll use the fact that the rate constant doubles.
Let's label the given information:
Ea = 34.9 kJ/mol
k1 = 0.0190 s^-1 (at 23 °C)
k2 = 2 * k1 (twice as fast)
First, convert the given temperature (23 °C) to Kelvin:
T1 = 23 + 273.15 = 296.15 K
Next, rearrange the Arrhenius equation to solve for T:
k1 = Ae^(-Ea/RT1)
k2 = Ae^(-Ea/RT2)
Since k2 = 2 * k1, we get:
2 * k1 = Ae^(-Ea/RT2)
Now, divide the second equation by the first equation:
(2 * k1) / k1 = Ae^(-Ea/RT2) / Ae^(-Ea/RT1)
Simplifying further:
2 = e^((Ea/R) * (1/T1 - 1/T2))
To solve for T2, take the natural logarithm of both sides:
ln(2) = (Ea/R) * (1/T1 - 1/T2)
Rearrange the equation to solve for T2:
1/T2 = (ln(2) * R) / Ea * (1/T1)
T2 = 1 / [(ln(2) * R) / Ea * (1/T1)]
Plug in the given values:
Ea = 34.9 kJ/mol = 34,900 J/mol
T1 = 296.15 K
R = 8.314 J/(mol·K)
Calculate T2:
T2 = 1 / [(ln(2) * 8.314 J/(mol·K)) / (34,900 J/mol) * (1/296.15 K)]
T2 ≈ 362 K
Lastly, convert T2 back to degrees Celsius:
T2 ≈ 362 - 273.15 = 88.85 °C
Therefore, the reaction would go twice as fast at approximately 88.85 °C.