The Arrhenius Equation?

Question

a) The activation energy of a certain reaction is 37.2 kJ/mol. At 20 degrees C, the rate constant is 0.0130 s^-1. At what temperature would this reaction go twice as fast? (I got part a - answer: T_2 =33.9 degrees C. Just need help on part b)

b) What would the rate constant be at a temperature of 100 degrees C?

Answer 1

Rate const. = A e^(-Ea/RT)

or

LnRate const. = lnA -Ea/RT

(a); you know rate const. at 293.15 K, Ea, so solve for A

Now using this A, put rate const. = 0.0260 s-1 (because the question tells you to), and solve for T

(b), you know A (from (a)), Ea (from question. Put T = 373.15 and solve for rate const.

Note: an equation like the Arrhenius eq. is a two-way street. You can use it to solve for ANY of the terms if you know the value of al lthe others.