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16 Nov 2019

# The Arrhenius Equation The Arrhenius equation shows the relationship between the rate constant k and the temperature T in kelvins and is typically written as k = Ae-EA/RT where R is the gas constant (8.314 J/mol . K). A is a constant called the frequency factor, and Ea is the activation energy for the reaction. However, a more practical form of this equation is ln k2/k1 = Ea/R(1/T1 - 1/T2) which is mathmatically equivalent to ln k1/k2 = Ea/R(1/T2 - 1/T1) where k1 and k2 are the rate constants for a single reaction at two different absolute temperatures (T1 and T2). The activation energy of a certain reaction is 36.9kJ/mol. At 20 degree C, the rate constant is 0.0140s-1 . At what temperature in degrees Celsius would this reaction go twice as fast? Given that the initial rate constant is 0.0140s-1 at an initial temperature of 20 degree C, what would the rate constant be at a temperature of 180 degree C for the same reaction described in Part A?

The Arrhenius Equation The Arrhenius equation shows the relationship between the rate constant k and the temperature T in kelvins and is typically written as k = Ae-EA/RT where R is the gas constant (8.314 J/mol . K). A is a constant called the frequency factor, and Ea is the activation energy for the reaction. However, a more practical form of this equation is ln k2/k1 = Ea/R(1/T1 - 1/T2) which is mathmatically equivalent to ln k1/k2 = Ea/R(1/T2 - 1/T1) where k1 and k2 are the rate constants for a single reaction at two different absolute temperatures (T1 and T2). The activation energy of a certain reaction is 36.9kJ/mol. At 20 degree C, the rate constant is 0.0140s-1 . At what temperature in degrees Celsius would this reaction go twice as fast? Given that the initial rate constant is 0.0140s-1 at an initial temperature of 20 degree C, what would the rate constant be at a temperature of 180 degree C for the same reaction described in Part A?

Elin HesselLv2

23 Jan 2019