If he activation energy of a reaction is 4.86Kj, then waht is the percent increase in the rate constant of a raction when the temperature is increased from 45 to 75C

Answer 1

The reaction rate is assumed to be of the form
Rate = Constant * e^(-Ea/RT)
where T must be in Kelvin

Rate(75C)/Rate(45C) = e^-[Ea/348R - Ea/318R]
where R = molar gas constant
= 8.317 J/(mol*K)
and Ea = 4860 J/mol is the activation anergy

Rate(75C)/Rate(45C) = e^[(Ea/k)(1/318 - 1/348)]
= e^(584*2.71*10^-4)= e^.158 = 1.17

The percent rate increase is 17%

Check my math. I tend to be careless

Answer 2

I understand how to get the Constant but I don't get how you know that it is a 17% increase?

Answer 3

The constant does not matter. You can't even compute it, but you don't have to. They only ask for the PERCENT INCREASE in reaction rate.

The 17% increase comes from the 1.17 factor increase in the reaction rate

Answer 4

Thank you