Consider the reaction 2HI(g)→H2(g)+I2(g). At 585 K, the rate constant is 9.64×10−5Lmol s. At 690. K, the rate constant is 2.83×10−3Lmol s.

Question

Consider the reaction 2HI(g)→H2(g)+I2(g). At 585 K, the rate constant is 9.64×10−5Lmol s. At 690. K, the rate constant is 2.83×10−3Lmol s. Use the Arrhenius equation to calculate the activation energy for the reaction. 

Answer 1

ln (k1/k2) = Ea/R (1/T2 - 1/T1)

ln (9.64x10-5/2.83x10-3) = Ea/8.314 J/molK (1/690 - 1/585)

-3.38 = Ea/8.314 (0.00145 - 0.00171)

Ea = 108,082 J/mol = 108 kJ/mole