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Leno4ka [110]
3 years ago
9

A certain reaction has an activation energy of 67.0 kJ/mol and a frequency factor of A1 = 4.20×1012. What is the rate constant,

k, of this reaction at 24.0 degrees C?
Chemistry
1 answer:
MrRa [10]3 years ago
5 0

Answer:

  • K = 7.00

Explanation:

<u>1) Arrhenius equation</u>

Arrhenius equation shows the relation between activation energy, temperature, and the equilibrium constant.

This is the equation:

       K=Ae^{-Ea/RT}

Where:

  • K is the equilibrium constant,
  • A is the frequency factor,
  • Ea is the activation energy (in J/mol),
  • T is the temperature in kelvins (K), and
  • R is the universal constant.

<u></u>

<u>2) Substitute, using the right units, and compute:</u>

  • A = 4.20 × 10¹² (dimensionless)
  • Ea = 67.0 kJ/mol = 67,000 J/mol
  • T = 24.0°C = 24.0 + 273.15 K = 297.15 K
  • R = 8.314 J/K mol

K=Ae^{-Ea/RT}=(4.20)(10)^{12}e^{-67,000J/mol/(8.314J/molK.297.15K)}

  • K = 7.00
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K_2 = rate of reaction with catalyst

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