The aquation of tris(1, 10-phenanthroline)iron(II) in acid solution takes place according to the equation:
Fe(phen)3^2+ + 3H3O + 3H2O==> (Fe(H2O)6^2+ + 3 phenH^+
If the activation energy is 126 kJ/mol and frequency factor is 8.62 x 1017 s-1, at what temperature is the rate constant equal to 3.63 x 10-3 s-1 for the first-order reaction?
I believe ln k = (-Ea/RT) + ln A
Ea is the activation energy in jouls/mol
R is 8....
A is the frequency factor. Solve for T.
i keep getting 32.3005 for T, if you have worked it out did you use 8.314472 for R, also is -126kj/mol -12600j/mol?
Yes, 8.314 is R but 126 kJ/mol = 126,000 J/mol.
ok the answer i got now was 50 degrees celcius
To solve this problem, we can use the Arrhenius equation, which gives the relationship between the rate constant (k) of a reaction, the activation energy (Ea), the frequency factor (A), and the temperature (T):
k = A * e^(-Ea/RT)
Where:
- k is the rate constant
- A is the frequency factor
- Ea is the activation energy
- R is the gas constant (8.314 J/(mol*K))
- T is the temperature in Kelvin
We are given:
- k = 3.63 x 10^-3 s^-1 (rate constant)
- Ea = 126 kJ/mol (activation energy)
- A = 8.62 x 10^17 s^-1 (frequency factor)
We need to find the temperature (T).
Now, we can rearrange the equation to solve for T:
T = (-Ea / (R * ln(k / A)))
Let's substitute the given values and calculate the temperature T:
T = (-126,000 J/mol) / (8.314 J/(mol*K) * ln(3.63 x 10^-3 s^-1 / (8.62 x 10^17 s^-1)))
T = (-126,000 J/mol) / (8.314 J/(mol*K) * ln(3.63 x 10^-3 / 8.62 x 10^17))
T = (-126,000 J/mol) / (8.314 J/(mol*K) * ln(4.21 x 10^-21))
T = (-126,000 J/mol) / (8.314 J/(mol*K) * -49.279)
T = (-126,000 J/mol) / (-409.962 J/K)
T ≈ 307.510 K
Therefore, at a temperature of approximately 307.510 K, the rate constant is equal to 3.63 x 10^-3 s^-1 for the first-order reaction.