the activation energy of a certain reaction is 39.5kj/mol. at 29 degrees C the r
ID: 883354 • Letter: T
Question
the activation energy of a certain reaction is 39.5kj/mol. at 29 degrees C the rate constant is .0160s-1. at what temperature in degrees Celsius would this reaction go twice as fast? the activation energy of a certain reaction is 39.5kj/mol. at 29 degrees C the rate constant is .0160s-1. at what temperature in degrees Celsius would this reaction go twice as fast? the activation energy of a certain reaction is 39.5kj/mol. at 29 degrees C the rate constant is .0160s-1. at what temperature in degrees Celsius would this reaction go twice as fast?Explanation / Answer
T1= 273+29 = 302 K
T2= ?
Ea = 39.5 KJ/mol = 39,500 J/mol
use:
ln(k2/k1)=(Ea/R)(1/T1-1/T2)
ln(2)=(39,500 /8.314)* (1/302 - 1/T2)
ln(2)=(39,500 /8.314)* (1/302 - 1/T2)
1.459*10^-4 = (1/302 - 1/T2)
T2 = 316 K
In degree celsius, Temperature is 316-273 = 43 oC
Answer: 43 oC
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