Activation energy of a certain reaction

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SUMMARY

The activation energy for the reaction is established at 45.8 kJ/mol, with a rate constant of 0.0130 s-1 at 20 degrees Celsius. To determine the temperature at which the reaction rate doubles, the Arrhenius Equation is utilized: k = Ae-Ea/RT. The relationship between the rate constants at two temperatures is expressed as ln(k2/k1) = (Ea/R)(1/T1 - 1/T2), where k2/k1 equals 2. This allows for the calculation of the unknown temperature T2.

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The activation energy of a certain reaction is 45.8 kJ/mol. At 20 degrees celsius, the rate constant is 0.0130 s^(-1). At what temperature would this reaction go twice as fast?

Arrhenius Equation:k = Ae^(-E_a/RT)

ln(k_2 / k_1) = (E_a /R)(1/T_1 - 1/T_2)

Okay, so I'm not quite sure how to approach this problem. I think you're supposed to solve for k_2 and then sub it into find the activation energy and then find the temperature. But I don't know how to solve for k_2 if I'm not given a second temperature. Any help is appreciated!
 
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You are given that k2/k1=2.
 

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