The Arrhenius equation shows the relationship between the rate constant k and th

Question

The Arrhenius equation shows the relationship between the rate constant k and the temperature T in Kelvin and is typically written as k= Ae^-E_a/RT where R is the constant (8.314 J/mol K), A is a constant called frequency factor, and E_a is the activation energy for the reaction. However, a more practical form of this equation is ln K_2/K_1 =E_2/R (1/T_1 - 1/T_2) which is mathmatically equivalent to ln k_1/K_2 = E_a/R (1/T_2 - 1/T_1) where K_1 and K_2 are the rate constant for a single reaction at two different absolute temperatures (T_1 and T_2). The activation energy of a certain reaction is 35.6 kJ/mol. At 24 degree C, the rate constant is 0.0150s^-1. At what temperature in degrees Celsius would this reaction go twice as fast? Given that the rate constant is 0.0150s^-1 at an initial temperature of 24 degree C what would the rate constant be at a temperature of 150 degree C for the same reaction described in Part A?

Explanation / Answer

part A

ln(K2/K1) = Ea/R(1/T1-1/T2)

ln(2K1/K1) = (35.6*10^3/8.314)((1/297.15)-(1/T2))

T2 = 312.165 K

= 39.015 c

PART B

ln(K2/0.015) = (35.6*10^3/8.314)((1/297.15)-(1/423.15))

K2 = 1.095 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.