A second order reaction has a rate constant of 8.7x10^-4/(M*s) at 30 degrees C. At 40 degrees C the rate constant is 1.8x10^-3/(M*s). What are the activation energy and frequency factor for this reaction? Predict the value of the rate constant at 4 degrees C
To determine the activation energy and frequency factor for the reaction, we can use the Arrhenius equation:
k = Ae^(-Ea/RT)
Where:
k = rate constant
A = frequency factor
Ea = activation energy
R = gas constant (8.314 J/(mol*K))
T = temperature in Kelvin
To solve for the activation energy (Ea) and frequency factor (A), we can use the given rate constants at two different temperatures: 30 degrees C and 40 degrees C.
Step 1: Convert temperatures to Kelvin.
T1 = 30 + 273 = 303 K
T2 = 40 + 273 = 313 K
Step 2: Set up equation using the Arrhenius equation and the rate constants.
k1 = 8.7x10^-4/(M*s)
k2 = 1.8x10^-3/(M*s)
Using T1 and k1:
k1 = A * e^(-Ea / (R * T1))
Using T2 and k2:
k2 = A * e^(-Ea / (R * T2))
Step 3: Rearrange the equations to solve for the activation energy (Ea) and frequency factor (A).
Divide the second equation by the first equation:
k2 / k1 = (A * e^(-Ea / (R * T2))) / (A * e^(-Ea / (R * T1)))
Cancel out the A:
k2 / k1 = e^(-Ea / (R * T2)) / e^(-Ea / (R * T1))
Simplify the equation:
k2 / k1 = e^(-Ea / (R * T2 + Ea / (R * T1))
Take the natural logarithm (ln) of both sides:
ln(k2 / k1) = - Ea / (R * T2) + Ea / (R * T1)
Combine terms:
ln(k2 / k1) = Ea / (R * T1) - Ea / (R * T2)
Step 4: Solve for the activation energy:
Ea = R * (T2 - T1) * ln(k2 / k1) / (T2 * T1)
Ea = 8.314 J/(mol*K) * (313 K - 303 K) * ln(1.8x10^-3/(M*s) / (8.7x10^-4/(M*s))) / (313 K * 303 K)
Ea ≈ 43.4 kJ/mol
Step 5: Solve for the frequency factor (A):
Using the first equation (k1 = A * e^(-Ea / (R * T1))) and rearranging:
A = k1 / e^(-Ea / (R * T1))
A = (8.7x10^-4/(M*s)) / e^(-43,400 J/mol / (8.314 J/(mol*K) * 303 K))
A ≈ 6.5x10^7 / M * s
Step 6: Predict the value of the rate constant at 4 degrees C (277 K).
Using the Arrhenius equation:
k = A * e^(-Ea / (R * T))
k = (6.5x10^7 / M * s) * e^(-43,400 J/mol / (8.314 J/(mol*K) * 277 K))
k ≈ 2.3x10^-5 / (M * s)
Therefore, the predicted value of the rate constant at 4 degrees C (277 K) is approximately 2.3x10^-5 / (M * s).