A reaction has a rate constant of 0.000122s -1 at27 o C and 0.228s -1 at 77 o C.

Question

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                                   

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