Calculate Ea: Activation Energy Solution

[magma_saber]

Homework Statement

Calculate E_a from given info:
ln k = 0 t(sec^-1) = 0.0033
ln k = -0.2 t(sec^-1) = 0.00335
ln k = -0.4 t(sec^-1) = 0.0034

Homework Equations

ln k = -(E_a/RT) + ln A

The Attempt at a Solution

I could find Ea if i had the temperature but I'm given the time. I think you can cancel out the ln A with the temperature can't you? But if you do that, Ea would be in J/K*mol, but Ea should be kJ/mol. This is all from a graph btw and the y intercept = -4515.7x + 14.749 and R² = 0.9964. Am i suppose to treat this as y=mx + b?

 

[GCT]

Do you have any temperature data? You are going to need it.

 

[magma_saber]

no temps are given. if they were, i would have been able to solve it.

 

[wywyluva]

same here