If the rate of the reaction is 3.5 x 10-3 at 330 K, and the rate of reaction at 285 K is 5.2 x 10-5 what is the activation energy Ea in KJ/mole?
Substitute into the Arrhenius equation.
(3.5x10^-3/5.3x10^-5)/(1/330)-1/285)=.0035
Write the Arrhenius equation for me then show what you have substituted for each symbol.
To determine the activation energy (Ea) in KJ/mole, you can use the Arrhenius equation:
k = A * e^(-Ea/RT)
Where:
- k is the rate constant
- A is the pre-exponential factor (frequency factor)
- Ea is the activation energy
- R is the ideal gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin
Given the rates of reaction (k1 and k2) at two different temperatures (T1 and T2), we can use the Arrhenius equation to find the activation energy Ea.
1) Convert the rates of reaction to rate constants:
k1 = 3.5 x 10^(-3) (at 330 K)
k2 = 5.2 x 10^(-5) (at 285 K)
2) Convert the temperature to Kelvin:
T1 = 330 K
T2 = 285 K
3) Rearrange the Arrhenius equation to solve for Ea:
Ea = -(ln(k1/k2) * R) / (1/T1 - 1/T2)
4) Substitute the given values into the equation:
Ea = -(ln(3.5 x 10^(-3) / 5.2 x 10^(-5)) * 8.314) / (1/330 - 1/285)
5) Calculate the value using a calculator:
Ea = -(ln(67.31) * 8.314) / (0.00303 - 0.00351)
Ea ≈ 51.77 KJ/mol
Therefore, the activation energy (Ea) is approximately 51.77 KJ/mol.