A reaction has a rate constant of 0.000122s-1 at27oC and 0.228s-1 at 77oC. a)Determine the activation energy of the reaction. b)What is the value of the rate constant at17oC? Please exlain in details how you get the answer so I canunderstand.Thank you! A reaction has a rate constant of 0.000122s-1 at27oC and 0.228s-1 at 77oC. a)Determine the activation energy of the reaction. b)What is the value of the rate constant at17oC? Please exlain in details how you get the answer so I canunderstand.Thank you!
Explanation / Answer
a ) Formula : ln ( k1 / k2 ) = Ea / R ( 1 / T2 - 1 / T1 ) Where k1 is the rate constant at T1 k2is the rate constant at T2 Eais the activation energy Data : k1 = 0.000122s-1 T1 = 300 K k2 =0.228s-1 T2 = 350 K Upon substituting in the above formula , ln ( 0.000122 / 0.228 ) = Ea / 8.314 J / mol .K ( 1 /350 K - 1/ 300 K) -7.53 = Ea - 5. 72 * 10 ^ - 5 mol / J Ea = 7.53 / 5.72 * 10 ^ - 5 mol / J = 1.314 * 10 ^ 5 J / mol b ) ln ( k1/ 0.000122s-1 ) = 1.314 * 10 ^ 5 J / mol / 8.314 J / mol.K ( 1 / 300 K - 1/ 290 K) = 1.81 k1/ 0.000122s-1 = 65.55 k1 = 0.0079 s-1