Question

Day 9-10 Homework

± The Arrhenius Equation

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Review | Constants | Periodic Table The Arrhenius equation shows the relationship between the rate constant kkk and the temperature TTT in kelvins and is typically written as k=Ae−Ea/RTk=Ae−Ea/RT where RRR is the gas constant (8.314 J/mol⋅K8.314 J/mol⋅K), AAA is a constant called the frequency factor, and EaEaE_ a is the activation energy for the reaction. However, a more practical form of this equation is lnk2k1=EaR(1T1−1T2)ln⁡k2k1=EaR(1T1−1T2) which is mathmatically equivalent to lnk1k2=EaR(1T2−1T1)ln⁡k1k2=EaR(1T2−1T1) where k1k1k_1 and k2k2k_2 are the rate constants for a single reaction at two different absolute temperatures (T1T1T_1 and T2T2T_2).

Part A The activation energy of a certain reaction is 34.4 kJ/molkJ/mol . At 26  ∘C ∘C , the rate constant is 0.0150s−10.0150s−1 . At what temperature in degrees Celsius would this reaction go twice as fast? Express your answer with the appropriate units. View Available Hint(s) T2T_2 = nothingnothing Submit Part B Given that the initial rate constant is 0.0150s−10.0150s−1 at an initial temperature of 26  ∘C ∘C , what would the rate constant be at a temperature of 150.  ∘C ∘C for the same reaction described in Part A? Express your answer with the appropriate units. View Available Hint(s)

Answer 1