Question

Need help on last step?

A certain reaction has an activation energy of 33.57 kJ/mol. At what Kelvin temperature will the reaction proceed 7.50 times faster than it did at 341 K?

I know I need the Arrhenius formula, but I get stuck with the ln part, I know I'm supposed to make it into the exponent as well as the other side of the equation, but how do I put that in my calculator?

Answer 1

\(ln(\dfrac{k_1}{k_2})=ln(7.5)=the \;rest\)

Answer 2

if you assume k2 =1, then k1= 7.5

Answer 3

yes so ln(7.50), do I just plug that in? When I do that, I get a super small number instead of one in the hundreds, let me redo my math and see if I messed up somewhere

Answer 4

yeah, you would just plug it in your calculator

Answer 5

make sure you assigned the temperatures accordingly, though.

Answer 6

would I be using (1/341-1/T2) or (1/T1-1/341) ?? I've been using the second one

Answer 7

we'll if k2= 1 and k1=7.5, then T2=341

if k1=1 and k2=7.5, then T1=341

Answer 8

ok is it 410.9 so 411

Answer 9

hm do you have to express it in the "correct" amount of sig figs?

Answer 10

if so then yes

Answer 11

well the question is as you see it, so I'm guessing they don't care much either way

Answer 12

What is this? Chemical engineering or chemistry?

Answer 13

we'll the least amount of sig figs is 3 anyways, so that works out to 411 kJ

Answer 14

thanks :)

Answer 15

no problem !