Activation energy of a certain reaction

[sam.]

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!

 

[Gokul43201]

You are given that k2/k1=2.

 

[ravenprp]

To find the temperature at which the reaction would go twice as fast, we can use the Arrhenius Equation and set k_2 = 2k_1 (since we want the reaction to go twice as fast). This gives us:

ln(2k_1 / k_1) = (E_a/R)(1/T_1 - 1/T_2)

Simplifying, we get:

ln(2) = (E_a/R)(1/T_1 - 1/T_2)

We know the activation energy (E_a = 45.8 kJ/mol) and the rate constant at 20 degrees celsius (k_1 = 0.0130 s^(-1)). So we can plug these values in and solve for T_2:

ln(2) = (45.8 kJ/mol / 8.314 J/molK)(1/293.15 K - 1/T_2)

Solving for T_2, we get:

T_2 = 1/(1/293.15 K - ln(2) * 8.314 J/molK / 45.8 kJ/mol)

T_2 = 352.4 K

Therefore, at a temperature of 352.4 K, the reaction would go twice as fast as it does at 20 degrees celsius.