Question

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?

Answer 1

Answer:

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$\frac{K_{2} }{K_{1} }$ = $\frac{E_{a} }{2.303XR}$$\frac{(T_{2}-T_{1} ) }{T_{2}T_{1} }$

              ⇒ log $\frac{5.7 X 10^{-4} }{1.1 X 10^{-4} }$ = $\frac{E_{a} }{2.303X8.314}$ $\frac{(773 - 743)}{773X743}$

              ⇒ 0.714 = $\frac{E_{a}X30 }{10996950.3}$

              ⇒ Eₐ = $\frac{7851822.5}{30}$ j/mole = 261.7 Kj/mole