Question

The Arrhenius equation shows the relationship between the rate constant k and the temperature T in...

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

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 (T1 and T2).

Part A

The activation energy of a certain reaction is 34.1 kJ/mol . At 24  ∘C , the rate constant is 0.0140s−1 . At what temperature in degrees Celsius would this reaction go twice as fast?

Express your answer with the appropriate units.

Part B

Given that the initial rate constant is 0.0140s−1 at an initial temperature of 24  ∘C , what would the rate constant be at a temperature of 190.  ∘C for the same reaction described in Part A?

Express your answer with the appropriate units.

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Answer #1

The answer is given as follows

Ams Stur-D Amrhenins giren eguation FAr PAETA 3416 bJm 34 1 kJlm) = Ea = 20 297 K 21273 k= T2- o o1 Seel O,O28see 2 o01 Seel-217 T2 O O031 T2 K_ o. oo31 3p2.64K 39 ca PART) cmiven 24t23 29k TI 2 T210C -6 +243 k ma > 34100 Jmq 34.1 3 in rches ejuaban

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