The rate constant of a chemical reaction increased from 0.100 s−1 to 2.80 s−1 upon raising the temperature from 25.0 ∘C to 39.0 ∘C.

Question

Part A Calculate the value of (1T2−1T1) where T1 is the initial temperature and T2 is the final temperature. Part B Calculate the value of ln(k1/k2) where k1 and k2 correspond to the rate constants at the initial and the final temperatures as defined in part A. Express your answer numerically. Part C What is the activation energy of the reaction? Express your answer numerically in kilojoules per mole.

Answer 1

The Arrhenius equation relates the rate constant to the activation energy and absolute temperature.

 K = A e^-Ea/RT where K is the rate constant, Ea the activation energy, A the frequency factor, R and T are the gas constant and absolute temperature respectively.

We are given the rate constants for each temperature, so

Equation 1: K1 = Ae^-Ea/RT1 and equation 2: K2 = A e^-Ea/RT2.

Part A: (39.0 + 273) – (25.0 + 273) = 14 °K

Part B: ln K1/K2 = ln (0.100s^-1/ 2.80 s^-1) = -3.33

Part C: We can’t get the activation energy directly as we were not given A, but if we divide equation 1 by equation 2, we obtain

K1/K2 = e^-Ea / 8.314 x 298°/ e^-Ea/8.314 x 312° A cancels out.

Ln(K1/K2)  =  -Ea/(8.314 x 298°) + Ea/8.314 x 312° = (Ea/8.314) x (1/312 – 1/298) factoring out Ea/8.134

-3.33 = ( Ea / 8.314) x (-0.000151) = Ea x(- 0.0000182)

Solve for Ea to get 183348 J / mole or 183 kJ/mole to 3 sig figs.

Please check math.