High School

The value of the rate constant for a second-order reaction is 0.55 M\(^-1\) s\(^-1\) at 286.6 K. The activation energy for this reaction is 39.1 kJ/mol. What is the value of the rate constant for this reaction at 345.4 K?

Answer :

Final answer:

The value of the rate constant for the reaction at 345.4 K, using the Arrhenius equation with the given activation energy and initial temperature is 2.36 M⁻¹ s⁻¹.

Explanation:

To find the value of the rate constant for the reaction at 345.4 K, we can use the Arrhenius equation:

k₂ = k₁ * exp((Ea / R) * (1/T₂ - 1/T₁))

Where k₁ is the rate constant at 286.6 K, Ea is the activation energy (39.1 kJ/mol), R is the gas constant (8.314 J/mol·K), T₁ is the initial temperature (286.6 K), and T₂ is the final temperature (345.4 K).

Plugging in the values and solving, we get:

k₂ = 0.55 * exp((39.1 * 10³ / 8.314) * (1/345.4 - 1/286.6))

k₂ ≈ 2.36 M⁻¹ s⁻¹

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