Part A: The activation energy of a certain reaction is 46.6kJ/mol. At 28 C, the
ID: 737858 • Letter: P
Question
Part A: The activation energy of a certain reaction is 46.6kJ/mol. At 28 C, the rate constant is .0180s^-1. At what temperature in degrees Celsius would this reaction go twice as fast?T2=_________
Part B: Given that the initial rate constant is .0180s^-1 at an initial temperature of 28 C, what would the rate constant be at a temperature of 130 C for the same reaction described in Part A?
k2=_________
Any help and shown work would be appreciated!
Explanation / Answer
A)K = Ae^(-Ea/RT) ln(K/A) = -Ea/RT At 28 oC, 0.018 = A*e^((46600/(8.314*(28+273)))*(-1)) =>A = 2199830.622 2*0.018 = 2199830.622*e^((46600/(8.314*T))*(-1)) =>e^((46600/(8.314*T))*(-1)) = 1.63649E-08 =>(46600/(8.314*T))*(-1) = -17.92812727 =>(46600/(8.314*T)) = 17.92812727 =>8.314*T = 2599.267582 =>T = 312.6374286K = 39.64 oC B)K = 2199830.622*EXP((46600/(8.314*(130+273)))*(-1)) = 2.0051