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Part B A certain reaction with an activation energy of 205 kJ/mol was run at 505 K and again at 525 K. What is the ratio of f

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Answer #1

Given:

T1 = 505 K

T2 = 525 K

Ea = 205 KJ/mol

= 205000 J/mol

use:

ln(f2/f1) = (Ea/R)*(1/T1 - 1/T2)

ln(f2/f1) = (205000.0/8.314)*(1/505 - 1/525.0)

ln(f2/f1) = 24657*(7.544*10^-5)

ln(f2/f1) = 1.86

f2/f1 = e^(1.86)

f2/f1 = 6.424

Answer: 6

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