The activation energy of a certain reaction is 49.6kJ/mol .?
The activation energy of a certain reaction is 49.6kJ/mol . At 23 ∘C , the rate constant is 0.0190s−1. At what temperature in degrees Celsius would this reaction go twice as fast?
Help me I've been stuck for awhile :(
- BIILv 77 years agoFavorite Answer
Right from the start I know that rates typically double for every 10 °C rise in T so the answer should be ~33 °C
Rate = Aexp-Ea1/RT
ln(k2/k1) = (Ea/R)[1/T1 -1/T2]
Ea in J mol^-1; R = 8.3145 J mol^-1;T in K; T1 =296 T2 to find;
rate2/rate1 = 2.00
ln(2.00) = 0.693 = [46600/8.3145](1/296 - 1/T2)
(I tend to lose myself with these calc so I tread carefully!)
0.693 ×8.3145/46600 = 1.237×10^-4 = 3.378×10^-3 -1/T2
1/T2 = (3.378×10^-3) - (1.237×10^-4) = 3.254×10^-3
T2 = 1/(3.254×10^-3) = 307 K or 34°C (Oh - you’re good drp :))