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Given the same reactant concentrations, the reaction CO (g) + Cl₂(g) → COCl₂(g). At 250 C is 1.50 x 10³ times as fast as the same reaction at 150 C. Calculate the activation energy for this reaction. Assume the frequency factor is constant.

User INeal
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Final answer:

To calculate the activation energy for a reaction, use the Arrhenius equation. Given the rate constant at two different temperatures, you can set up an equation to calculate the activation energy. Plugging in the values, the activation energy for this reaction is approximately -5.38 J/mol.

Step-by-step explanation:

To calculate the activation energy for a reaction, we can use the Arrhenius equation:

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

Given that the reaction at 250°C is 1.50 x 10^3 times as fast as the same reaction at 150°C, we can use this information to calculate the activation energy.

We can set up the following equation:

(1.50 x 10^3) * k1 = k2

Where:

  • k1 is the rate constant at 150°C
  • k2 is the rate constant at 250°C

By rearranging the equation and plugging in the values:

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

Ea = -8.314 J/mol*K * ln((1.50 x 10^3 * (150 + 273))/(1 * 250 + 273))

Ea = -8.314 J/mol*K * ln(1437/752)

Ea = -8.314 J/mol*K * ln(1.91)

Ea ≈ -8.314 J/mol*K * 0.647

Ea ≈ -5.38 J/mol

Therefore, the activation energy for this reaction is approximately -5.38 J/mol.

User Nilay Singh
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