[student8607]

A certain first-order reaction has a rate constant of 1.0x10^-3s^-1 at 25 degrees C.

If the reaction rate doubles when the temperature is increased to 35 degrees C, what is the activation energy in kJ?

I am familiar with the equation to solve for Ea but how am I to determine my second rate constant just from knowing the reaction rate doubles?

[Yggdrasil]

You know that k₃₅ / k₂₅ = 2.  Can you write this ratio of rate constants in terms of activation energy?

[student8607]

If the rate doubles then the rate constant does as weel?

Ln(1.0x10-3 / 2.0x10-3) = -Ea/8.31 x (1/308 - 1/298)

?

[Yggdrasil]

I would assume that is what the problem means, although doubling the rate constant is not the only way to double the rate of reaction.

[student8607]

Ok, will go with that.
Thanks!