Question

The second-order rate constants for the reaction of oxygen atoms witharomatic hydrocarbons have been measured (R. Atkinson and J.N. Pitts, J. Phys. Chem.79, 295 (1975)). In the reaction with benzene the rate constantsare 1.44 ×107dm3mol−1s−1at 300.3 K, 3.03 ×107dm3mol−1s−1at 341.2 K,and 6.9 ×107dm3mol−1s−1at 392.2 K. Find the pre-exponential factor andactivation energy of the reaction.

Answer 1

A = 1,13x10¹⁰

Ea = 16,7 kJ/mol

Step-by-step explanation:

Using Arrhenius law:

ln k = -Ea/R × 1/T + ln(A)

You can graph ln rate constant in x vs 1/T in y to obtain slope: -Ea/R and intercept is ln(A).

Using the values you will obtain:

y = -2006,9 x +23,147

As R = 8,314472x10⁻³ kJ/molK:

-Ea/8,314472x10⁻³ kJ/molK = -2006,9 K⁻¹

Ea = 16,7 kJ/mol

Pre-exponential factor is:

ln A = 23,147

A = e^23,147

A = 1,13x10¹⁰

I hope it helps!