Question

The activation energy of a certain reaction is 40.5Kj/mol.the rate constant is Given 0.0130s^-1 the initial temperature is 20 C , what would the rate constant be at a temperature of 100 C ?

Answer 1

E=-RT ln (k/A)
40.5 = -0.08206(295.15) ln (0.0130/A)
e^(-40.5/(0.08206(293.15))) = 0.0130/A

0.07000234854... approx = A

use A to solve for k at 100C (373.15 K):
40.5=-RT ln (k/A)
40.5=-0.08206(373.15) ln (k/30/0.07000234854)
40.5/(-0.08206(373.15))=ln (k/30/0.07000234854)
e^(40.5/(-0.08206(373.15)))=k/0.07000234854
k=0.2627397387

pls. double check my calculations :))

Answer 2

ok thank let me do so

Answer 3

i think there's an error

Answer 4

line 2: 40.5 = -0.08206(293.15) ln (0.0130/A)

Answer 5

im getting a very big answer

Answer 6

last line: 0.01865088557

Answer 7

the rate constant should be at least close to 0.0130s^-1. not geater than 1.