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

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1

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.

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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:

  1. 2.0 = A \times e^{-\frac{Ea}{8.314 \times 298}}

  2. 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.