The first-order isomerization reaction: cyclopropane → propene has a rate constant of 1.10×10-4 s-1 at 470°C and 5.70×10-4 s-1 at 500°C. What is the activation energy, Ea, for the reaction?

Activation energy, Ea, for the reaction 261.7 kj/mole

Explanation:

Given

Rate constant at 470 ⁰C (K₁) = 1.10 x 10⁻⁴ s⁻¹ and rate constant at 500⁰C (K₂) = 5.7 x 10⁻⁴ s⁻¹

Temperature (T₁) = 470 + 273 = 743 K and

temperature (T₂) = 500 + 273 = 773 K

Activation energy (Eₐ) = ?

Universal gas constant (R) = 8.314 J. K⁻¹. mole⁻¹

We know   log =

              ⇒ log =

              ⇒ 0.714 =

              ⇒ Eₐ = j/mole = 261.7 Kj/mole

answer; i believe that the correct answer is (a) potassium(k)

idk what the question is about