The Arrhenius Equation The Arrhenius equation shows the relationship between the

Question

The Arrhenius Equation The Arrhenius equation shows the relationship between the rate constant k and the temperature T in kelvins and is typically written as 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 reaction. However, a more practical form of this equation is ln k2/k1 = Ea/R(1/T1 - 1/T2) which is mathmatically equivalent to ln k1/k2 = Ea/R(1/T2 - 1/T1) where k1 and k2 are the rate constants for a single reaction at two different absolute temperatures (T1 and T2). The activation energy of a certain reaction is 36.9kJ/mol. At 20 degree C, the rate constant is 0.0140s-1 . At what temperature in degrees Celsius would this reaction go twice as fast? Given that the initial rate constant is 0.0140s-1 at an initial temperature of 20 degree C, what would the rate constant be at a temperature of 180 degree C for the same reaction described in Part A?

Explanation / Answer

Part A

Ea = 36.9 * 103 J/mol

T1 = 20o C = 293 K

k2/k1 = 2

ln (k2/k1) = Ea/R (1/T1 - 1/T2)

or, ln 2 = 36900/8.314 (1/293 - 1/T2)

or, T2 = 307 K = 34o C

Part B

T1 = 293 K

T2 = 180o C = 453 K

k1 = 0.0140 s-1

ln (k2/k1) = Ea/R (1/T1 - 1/T2)

or, ln (k2/0.014) = 36900/8.314 (1/293 - 1/453)

or, k2 = 2.947 s-1

or, k2 = 2.95 s-1 (approx.)

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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