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[tex]\frac{K_{2} }{K_{1} }[/tex] = [tex]\frac{E_{a} }{2.303XR}[/tex][tex]\frac{(T_{2}-T_{1} ) }{T_{2}T_{1} }[/tex]
⇒ log [tex]\frac{5.7 X 10^{-4} }{1.1 X 10^{-4} }[/tex] = [tex]\frac{E_{a} }{2.303X8.314}[/tex] [tex]\frac{(773 - 743)}{773X743}[/tex]
⇒ 0.714 = [tex]\frac{E_{a}X30 }{10996950.3}[/tex]
⇒ Eₐ = [tex]\frac{7851822.5}{30}[/tex] j/mole = 261.7 Kj/mole
The activation energy of the reaction is 1.37 * 10^5 J
The activation energy is the energy possessed by the particles of a body in order to be able to participate in chemical reaction.
Using the formula;
ln(k2/k1) = -Ea/R(1/T2 - 1/T1)
Substituting values, we have;
ln(5.70×10-4 /1.10×10-4) = -Ea/8.314(1/500 - 1/470)
1.65 = -Ea/8.314(0.0020 - 0.0021)
1.65 = 0.0001Ea/8.314
Ea = 1.65 * 8.314/0.0001
Ea = 1.37 * 10^5 J
The activation energy of the reaction is 1.37 * 10^5 J
Learn more about activation energy: https://brainly.com/question/9743981
