Question

The Arrhenius equation shows the relationship between the rate constant k and the temperature Tin kelvins and is typically written as

k=Ae−Ea/RT

where R is the gas constant (8.314 J/mol⋅K), Ais a constant called the frequency factor, and Ea is the activation energy for the reaction.

However, a more practical form of this equation is

lnk2k1=EaR(1T1−1T2)

which is mathmatically equivalent to

lnk1k2=EaR(1T2−1T1)

where k1 and k2 are the rate constants for a single reaction at two different absolute temperatures (T1and T2).

Part A

The activation energy of a certain reaction is 45.6 kJ/mol . At 30  ∘C , the rate constant is 0.0160s−1. At what temperature in degrees Celsius would this reaction go twice as fast?

Express your answer with the appropriate units.

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T2 =

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Part B

Given that the initial rate constant is 0.0160s−1 at an initial temperature of 30  ∘C , what would the rate constant be at a temperature of 170.  ∘C for the same reaction described in Part A?

Express your answer with the appropriate units.

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k2 =

Submit

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Answer 1