Question

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?

Answer 1

Answer:

• K = 7.00

Explanation:

1) Arrhenius equation

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.

2) Substitute, using the right units, and compute:

• 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