Hello! I need help with part B (I got A). Part A The activation energy of a cert

Question

Hello! I need help with part B (I got A). Part A The activation energy of a certain reaction is 35.0 C . At 27 C, the rate constant is .015 s^-1. At what temperature in degrees Celsius would this reaction go twice as fast? Express your answer with the appropriate units. T2 = 42.6 Part B Given that the initial rate constant is at an initial temperature of 27 C, what would the rate constant be at a temperature of 100 C for the same reaction described in Part A? Express your answer with the appropriate units.

Explanation / Answer

log(k2/k1) = Ea/2.303R (1/T1 - 1/T2) log(k2/0.015) = 35 x 10^3/2.303*8.314 (1/300 - 1/373) k2/0.015 = 15.423 k2 = 0.23 s^-1 (taking unit of activation energy in kJ/mol)

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.