I have the answer to the first question I need to know how to get to the answer
ID: 810635 • Letter: I
Question
I have the answer to the first question I need to know how to get to the answer to the second question please!
± The Arrhenius Equation Part A The Arrhenius equation shows the relationship between the rate constant k and the temperature T in kelvins and is typically written as The activation energy of a certain reaction is 41.3kJ/mol. At 25 °C, the rate constant is 0.0160s-1.At what Celsius would this reaction go twice as fast? Express your answer with the appropriate units k = Ae-Ea/RT where R is the gas constant (8.314 J/mol K), A is a constant called the frequency factor,and Ea is the activation energy for the | T2 379 °C reaction Submit Hints My Answers Give Up Review Part However, a more practical form of this equation is Answer Requested which is mathmatically equivalent to Part B Given that the initial rate constant is 0.0160s1 at an initial temperature of 25 °C, what would the rate constar 160 ° C for the same reaction described in Part A? k2 where k1 and k2 are the rate constants for a single reaction at two different absolute temperatures (T1 and T2). Express your answer with the appropriate units Value UnitsExplanation / Answer
ln[K2/k1] = Ea/R[1/T1-1/T2]
ln[0.016*2/0.016] = 41.3*10^3/8.314[1/298-1/T2]
T2 = 310.92 K
ln[K2/k1] = Ea/R[1/T1-1/T2]
ln[K2/0.016] = 41.3*10^3/8.314[1/298-1/433]
ln K2 = 9.33
K2 = 2.14*10^9