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 which is mathmatically equivalent to 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 38.2 kJ/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? Express your answer with the appropriate units. Part B 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? Express your answer with the appropriate units.

Explanation / Answer

1)

the equation to use is

ln (k2/K1) = (Ea/R) ( 1/T1 - 1/T2)

given

k2 = 2 k1

also

E = 38.2 x 1000 J/mol

T1 = 20 C = 293 Kelvin

so

we get

ln(2k1/k1) = ( 38.2 x 1000 / 8.314 ) ( 1/293 - 1/T2)

ln2= ( 38.2 x 1000 / 8.314) ( 1/293 - 1/T2)

1/293 - 1/T2 = 1.5086 x 10-4

1/T2 = 3.26 x 10-3

T2 = 306.55

so

T2 = 306 .55 kelvin

T2 = 306.55 - 273 C

T2 = 33.55 C

so

the temperature should be T2 = 33.55 C

2)

Given

k1 = 0.014

T1 = 293 K

T2 = 180 + 273 = 453 K

Ea = 38.2 x 1000 J/mol

R = 8.314

so

we get

ln (k2/K1) = (Ea/R) ln ( 1/T1 - 1/T2)

ln(k2/k1) = ( 38.2 x 1000 / 8.314 ) ( 1/293 - 1/453)

ln (k2/K1) = 5.587

k2/k1 = 254.346

so

k2/0.014 = 254.346

k2 = 3.56

so

the rate constant should be 3.56 s-1

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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