what is the energy of activation, ea, of a reaction if the rate constant, k, is 2.0 s-1 at 298 k and 24.2 s-1 at 372 k? report your answer in kj, but only enter the value, not the units.
What is the energy of activation, ea, of a reaction if the rate constant, k, is 2.0 s-1 at 298 K and 24.2 s-1 at 372 K?
Answer:
To determine the energy of activation (Ea) of a reaction, we can use the Arrhenius equation:
k = A \times e^{-\frac{Ea}{RT}}
Where:
- k is the rate constant
- A is the pre-exponential 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 rate constants k at two different temperatures, we can set up two equations:
-
2.0 = A \times e^{-\frac{Ea}{8.314 \times 298}}
-
24.2 = A \times e^{-\frac{Ea}{8.314 \times 372}}
By dividing the first equation by the second, we eliminate A:
\frac{2.0}{24.2} = \frac{e^{-\frac{Ea}{8.314 \times 298}}}{e^{-\frac{Ea}{8.314 \times 372}}}
\frac{2.0}{24.2} = e^{\frac{Ea}{8.314} \times \frac{1}{372} - \frac{1}{298}}
\frac{2.0}{24.2} = e^{\frac{Ea}{8.314} \times \frac{298-372}{298 \times 372}}
0.0826 = e^{-\frac{74.158 \times Ea}{242008}}
-2.4931 = -\frac{74.158 \times Ea}{242008}
Ea = \frac{(-2.4931) \times 242008}{74.158}
Ea \approx 814.58 kJ
Therefore, the energy of activation (Ea) of the reaction is approximately 814.58 kJ.